Class 8 Mathematics - BeTOPPERS IIT Foundation Series - 2022 Edition

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IIT FOUNDATION Class VIII

MATHEMATICS

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Published by

:

USN Eductech Private Limited Hyderabad, India.

PREFACE Our sincere endeavour in preparing this Book is to enable students effectively grasp & understand the Concepts of Mathematics and help them build a strong foundation in this Subject. From among hundreds of questions being made available in this Book, the Student would be able to extensively practice in each concept exclusively, throughout that Chapter. At the end of each Chapter, two or three Worksheets are provided with questions which shall cover the entire Chapter, helping each Student consolidate his / her learning. This Book help students prepare for their respective Examinations including but not limited to i.e. CBSE, ICSE, various State Boards and Competitive Examinations like IIT, NEET, NTSE, Science Olympiads etc. It is compiled by our inhouse team of experts who have a collective experience of more than 40 years in their respective subject matter / academic backgrounds. This books help students understand concepts and their retention through constant practice. It enables them solve question which are ‘fundamental / foundational’ as well questions which needs ‘higher order thinking’. Students gain the ability to concentrate, to be self-reliant, and hopefully become confident in the subject matter as they traverse through this Book. The important features of this books are: 1.

Lucidly presented Concepts: For ease of understanding, the ‘Concepts’ are briefly presented in simple, easy and comprehensible language.

2.

Learning Outcomes: Each chapter starts with ‘Learning Outcomes’ grid conveying what the student is going to learn / gain from this chapter.

3.

Bold-faced Key Terms: The key words, concepts, definitions, formulae, statements, etc., are presented in ‘bold face’, indicating their importance.

4.

Tables and Charts: Numerous strategically placed tables & charts, list out etc. summarizes the important information, making it readily accessible for effective study.

5.

Box Items: Are ‘highlighted special topics’ that helps students explore / investigate the subject matter thoroughly.

6.

Photographs, Illustrations: A wide array of visually appealing and informative photographs are used to help the students understand various phenomena and inculcate interest, enhance learning in the subject matter.

7.

Flow Diagrams: To help students understand the steps in problem-solving, flow diagrams have been included as needed for various important concepts. These diagrams allow the students visualize the workflow to solve such problems.

8.

Summary Charts: At the end of few important concepts or the chapter, a summary / blueprint is presented which includes a complete overview of that concept / chapter. It shall help students review the learning in a snapshot.

9.

Formative Worksheets: After every concept / few concepts, a ‘Formative Worksheet’ / ‘Classroom Worksheet’ with appropriate questions are provided from such concept/s. The solutions for these problems shall ideally be discussed by the Teacher in the classroom.

10. Conceptive Worksheets: These questions are in addition the above questions and are from that respective concept/s. They are advised to be solved beyond classroom as a ‘Homework’. This rigor, shall help students consolidate their learning as they are exposed to new type of questions related to those concept/s.

11. Summative Worksheets: At the end of each chapter, this worksheet is presented and shall contain questions based on all the concepts of that chapter. Unlike Formative Worksheet and Conceptive Worksheet questions, the questions in this worksheet encourage the students to apply their learnings acquired from that entire chapter and solve the problems analytically. 12. HOTS Worksheets: Most of the times, Summative Worksheet is followed by an HOTS (Higher Order Thinking Skills) worksheet containing advanced type of questions. The concepts can be from the same chapter or as many chapters from the Book. By solving these problems, the students are prepared to face challenging questions that appear in actual competitive entrance examinations. However, strengthening the foundation of students in academics is the main objective of this worksheet. 13. IIT JEE Worksheets: Finally, every chapters end with a IIT JEE worksheet. This worksheet contains the questions which have appeared in various competitive examinations like IIT JEE, AIEEE, EAMCET, KCET, TCET, BHU, CBSE, ICSE, State Boards, CET etc. related to this chapter. This gives real-time experience to students and helps them face various competitive examinations. 14. Different Types of Questions: These type of questions do appear in various competitive examinations. They include:

• Objective Type with Single Answer Correct

• Non-Objective Type

• Objective Type with > one Answer Correct

• True or False Type

• Statement Type - I (Two Statements)

• Statement Type - II (Two Statements)

• MatchingType - I (Two Columns)

• MatchingType - II (Three Columns)

• Assertion and Reasoning Type

• Statement and Explanation Type

• Roadmap Type

• FigurativeType

• Comprehension Type

• And many more...

We would like to thank all members of different departments at BeTOPPERS who played a key role in bringing out this student-friendly Book. We sincerely hope that this Book will prove useful to the students who wish to build a strong Foundation in Mathematics and aim to achieve success in various boards / competitive examinations. Further, we believe that as there is always scope for improvement, we value constructive criticism of the subject matter, as well as suggestions for improving this Book. All suggestions hopefully, shall be duly incorporated in the next edition. Wish you all the best!!!

Team BeTOPPERS

CONTENTS 1.

Number System

..........

01 - 16

2.

Square roots and Cube Roots

..........

17 - 30

3.

Ratio, Proportions and Variations

.........

31 - 40

4.

Comparing Quantities

..........

41 - 52

5.

Exponents and Radicals

..........

53 - 60

6.

Polynomials

..........

61 - 78

7.

Linear Equations and Inequations

..........

79 - 90

8.

Sets and Relations

..........

91 - 114

9.

Number Theory

..........

115 - 136

10.

Mensuration

..........

137 - 158

11.

Plane Geometry

..........

159 - 192

12.

Solutions

..........

193 - 444

.

Chapter -1

Number System

Learning Outcomes

By the end of this chapter, you will understand • • • • • •

Natural numbers Whole numbers Integers Rational numbers Irrational numbers Real Numbers

• • • • •

Even Numbers Odd Numbers Prime Numbers Composite Numbers Fractions

1. Introduction A number is a mathematical object used in counting and measuring. A notational symbol which represents a number is called a numeral. In addition to their use in counting and measuring, numerals are often used for labels (telephone numbers), for ordering (serial numbers), and for codes (ISBNs). In mathematics, the definition of number has been extended over the years to include such numbers as zero, negative numbers, rational numbers, irrational numbers and complex numbers. Certain procedures which take one or more numbers as input and produce a number as output are called numerical operations. Unitary operations take a single input number and produce a single output number. For example, the successor operation adds one to an integer, thus the successor of 4 is 5. More common are binary operations which take two input numbers and produce a single output number. Examples of binary operations include addition, subtraction, multiplication, division and exponentiation. The study of numerical operations is called arithmetic.

2. Review of Basic Numbers Natural Numbers The countable numbers like 1, 2, 3... ........ are called Natural Numbers. The set of natural numbers is denoted by N. N = { 1, 2, 3, 4…….}

Integers The set of numbers containing negatives of natural numbers along with whole numbers is called Integers. The set of integers is denoted by Z. Z = { .......–3, –2, –1, 0, 1, 2, 3 .......}

3. Rational and Irrational Numbers Rational Numbers The numbers of the form p/q where p and q are integers and q ≠ 0 , are called Rational Numbers. Rational Numbers are denoted by Q. Example:

3 5 23 5 8 , , , , ............. 4 9 19 7 13

Note: i) Zero is also a rational number, since

0 0. Rational Number ii) Every integer is also a rational number, since any integer can be expressed as

integer . 1

5 and 5 is a rational number.. 1 Equivalent Rational Numbers For example, 5 

If

p is a rational number and m is any non-zero q

Whole Numbers

integer, then

The numbers containing natural numbers and zero are called Whole Numbers. The set of whole numbers is denoted by W. W = { 0, 1, 2, 3, 4, …………}

In this case,

p mp  . q mq p mp and are called Equivalent q mq

rational numbers.

2 Example: A rational number 3/7 on multiplying both numerator and denominator with 5 , we get

3 5 15 3 15   . Here, and are called Equivalent 7 5 35 7 35 rational numbers.

Irrational Numbers Any number that cannot be expressed in the form of

p (which is neither terminating nor repeating q

decimal) is said to be an irrational number. Example:

2, 3, 5,etc,.

8th Class Mathematics Rational Numbers Between Two Rational Numbers Between two rational numbers there are infinitely many rational numbers. Method-1: Let a and b be two rational numbers

a b  a < q1 < b 2 q1 is the rational number between a and b. q1 

a  q1  a  q2  b 2 q2 is the rational number between a and q1. q2 

q1  b  a  q 2  q1  q3  b 2 q3 is the rational number between q1 and b. In this manner we can find infinite rational numbers between two given distinct rational numbers. q3 

Why is 2 an irrational number? The value of 2  1.41421356..... . The close observation of the value shows that it is a nonrepeating and non-terminating decimal. Further, it

p cannot be expressed in the form of . Hence, it is q an irrational number. Is  rational or irrational number? We know that  = 22/7

22 , which is equal The approximate value of  is 7 to 3.142857143...  is irrational number..

a c and be two rational numbers b d Step-1: Make denominators equal in both rational numbers. Step-2: If we have to find n rational numbers between Method-2:

ad cd and , then multiply numerators and denominabd bd tors by such a number so that difference between the numerators is at least n.

Properties of Rational and Irrational Numbers Properties Addition

Subtraction

Rational The sum of two rational numbers is a rational number.

Irrational The sum of two irrational numbers may or may not result in an irrational number.

Example: 4 + 2 = 6  rational

Example:

The differ ence of two rational numbers is a rational number. Example: 4 – 2 = 2  rational

Multiplication

The product of two rational numbers is a rational number. Example: 4 × 2 = 8  rational

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



2   2  0

rational 5 2  3 2  8 2  irrational The differenc e of two irrational numbers may or may not result in an irrational number. Example: 2  2  0  rational 5 2  3 2  2 2  irrational The product of two irrational numbers may or may not result in an irrational number. Example: 2  2 2  4  rational 2  5  10  irrational

Rational & Irrational The sum of a rational and an irrational number is an irrational number.

Example: 6  3  irrational The difference of a rational and an irrational number is an irrational number. 6  3  irrational

The product of a rational and an irrational number is an irrational number. Exception: Zero Example: 6  3  6 3  irrational

Number System Properties Division

3 Rational The quotient of two rational numbers is a rational number.

Irrational The quotient of two irrational numbers may or may not result in an irrational number.

Example: 4/2 = 2  rational

Example: 2 3



3 3

 1  rational

2  irrational 3

Rational & Irrational The quotient of a rational and an irrational number is an irrational number. Exception: Zero 6 Example: 2 3  3 irrational

Difference between Rational and Irrational Numbers Rational Numbers 1. Rational numbers can be expressed in the form of p/q

7  3.5 2

Example :

2. Rational numbers, when expressed as decimals will be terminating or non-terminating repeating decimals Example :

Irrational Numbers 1. Irrational numbers cannot expressed in the form of p/q Example :

(Irrational numbers are nonterminating and non-repeating decimals) Example : 3  1.732050807.....

4. Real Numbers

Conceptive Worksheet 1. 2.

Real Numbers Irrational Numbers Rational Numbers Integers

3.

Natural Numbers

4.

Formative Worksheet 1.

Find any three rational numbers between –2 and 0.

2.

1 1 Insert three rational numbers between and 3 2

3.

Examine whether

4.



22



2

is a rational or an

irrational number. If a is a positive rational number and n is a positive integer greater than 1, prove that an is a rational number.

2  1.41421356...

2. Irrational numbers, when expressed as decimals will be neither terminating nor repeating decimals.

10 7  3.3  3.5 (or) 3 2

Real numbers are the union of rational and irrational numbers.

be

Prove that between any two rational numbers a and b there exists another rational number. Insert two irrational numbers between 0.420 and 0.421. Insert a rational number and an Irrational number between 7 and 8. How many Rational and Irrational numbers can be inserted between 7 and 8? Find three rational numbers between –2 and 5

5. Even and Odd Numbers Even Numbers: The numbers that are divisible by 2 are called even numbers(E). The general notation of an even number is 2n where nZ . Example: 2, 4, 6 ............. are all even numbers. also –2, –4, –6 ............... are even numbers. E  0,  2,  4,  6..........

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8th Class Mathematics

4

Formative Worksheet

Odd Numbers: The integers that are not exactly divisible by 2 are called odd numbers. The general notation of an odd number is 2n+1 where n  Z . Example: 1, 3, 5 ................are odd numbers. Note: Every odd number is obtained by adding 1 to every even number. Example: 2 + 1 = 3, 4+1=5

5.

6. 7.

Properties of Even Numbers The sum or difference of two even numbers is even. Example: 2 + 4 = 6, 8 – 4 = 4 The square of an even number is even. Example: 62 = 36 The sum of squares of two even numbers is even. Example: 22 + 42 = 4 + 16 = 20. The product of two even numbers is even. Example: 8 × 6 = 48. The square root of an even perfect square is an even number. Example: The numbers 64, 36, 16, 4 are perfect squares and they are even numbers. (64= (8)2; 36= (6)2)

Properties of Odd Numbers

i)

ii)

iii)

The sum or difference of two odd numbers is even. Example: 3 + 7 = 10, 9 – 5 = 4 The square of an odd number is odd. Example: 52 = 25 The sum of squares of two odd numbers is even. Example: 32 + 52 = 9 + 25 = 34. The product of two odd numbers is odd. Example: 5 × 3 = 15. The square root of an odd perfect square is an odd number. Example: The numbers 81, 49, 25, 9 are perfect squares and they are odd numbers. ( 81 = (9)2,49 = (7)2). The sum or difference of an even numbers and an odd number is odd. Example: 6 + 9 = 15, 9 – 4 = 5 The product of an even number and an odd number is even. Example: 2 × 7 = 14. The sum of squares of an even number and an odd number is odd. Example: 22 + 32 = 4 + 9 = 13.

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8.

Show that (i) Any odd perfect square will be odd of the form (4k + 1). (ii) Any even perfect square will be even. If a and b are two natural numbers and if (a + b) is even ,prove that (a – b) is also even. If x, y and z are odd positive integers, then which of the following is true? 1) x2 + y2 + z2 + 12 is even 2) x2y2 + y2z2 + (z – x)2 – 45 is odd 3) x2 [y2 + z2 – (x – y)2 – (y – z)2] + 125 is even 4) x3 [y3 + z3 – (y – x)3 – (z – x)3] + 73 is even If x and y are even, then find whether x2 – y2 is even or odd?

Conceptive Worksheet 5. 6. 7. 8.

Consider the equation in positive integers: x2 + y2 = 400, x < y. Can both x and y be odd ? For what values of k and m is (19k – 3k) (18m – 11m) divisible by 112. (k, m  N) If 3 – 9 + 15 – 21 + ......... up to 19 terms = x then find whether x is even or odd ? If x and y are odd then x2 – y2 is?

6. Prime Numbers and Composite Numbers Prime Numbers A number which has exactly two different factors one and itself is called a prime number. Example: 2, 3, 5, 7, 11,13, 17, 19 etc. We can write 2 = 2 × 1.There are two factors of 2. They are 2 and 1. Now we can say that 2 is divisible by 2 and 1, i.e., 2 is divisible by itself and 1, and similarly 3 = 3×1, 5 = 5 ×1.

Composite Numbers The natural numbers other than 1 and primes are called composite numbers. Note: The Composite numbers have a minimum of three factors. Example: The factors of 4 are 1, 2, 4 and the factors of 6 are 1, 2, 3 and 6. Points to remember i) Every prime number has two factors. ii) Every composite number has at least three factors. iii) 1 is neither a prime nor a composite number. iv) 2 is the only even prime number.

Number System

5

Formative Worksheet

Types of Primes Twin Primes The two prime numbers whose difference is 2 are called Twin Primes. Example: (i) 3, 5 (ii) 5, 7 (iii) 11, 13 Prime Triplets Three consecutive odd primes are known as prime triplets. Example: 3, 5, 7 (One and only prime triplet )

9.

If F(n) represents the set of all factors of a natural number n, including 1 and itself, then write i) F(12)

10.

M  a   M  b  represents the set of all common

multiples of a and b. Then write i) M  20   M  25  ii) M  24   M  42 

Factor and Multiple If b divides a and leaves zero as remainder, then b is called a factor or divisor of a and a is called the multiple of b. Example: 5 divides 20 leaving remainder 0. Hence 5 is the factor or divisor of 20 and 20 is called the multiple of 5. Therefore, 20 = 5 × 4, where 4 is the other factor of 20. Note: Suppose a natural number n is expressed as n = 2a .3b.5c...... then the number of divisors or factors of n is given by n = (a +1) (b +1) (c +1)... Prime Factors If the factors of the given number are prime numbers, then such factors are called prime factors. Example: Consider the number 36. 36 = 4 × 9 = 2 × 2 × 3 × 3 = 22 × 32 Therefore, the prime factors of 36 are 2 and 3. Prime Factorization Expressing the given number as a product of prime factors is called prime factorization. Example: What are the prime factors of 12? It is best to start working from the smallest prime number, which is 2, so let’s check 12 ÷ 2 = 6 But 6 is not a prime number, so we need to factor it further 6÷2=3 And 3 is a prime number, so 12 = 2 × 2 × 3 As you can see, every factor is a prime number, so the answer must be right - the prime factorization of 12 is 2 × 2 × 3, which can also be written as 22 × 3 Note: F (n) indicates factors of n and M(n) indicates multiples of n.

ii) F(36)

11. lf n3 – 1 is a prime number, where n  N , then n2 + n +1 is a ________ number. 12. Find the number of prime numbers of the form

n( n  1) , where n is a natural number.. 2





2 13. Is n  n  41 a prime for n = 41?





n 14. Establish the condition for 2  1 to be a prime.

15. If a and b are twin primes and a > b, then find the value of a3 – b3.

Conceptive Worksheet 9.

If M(n) represents the set of all multiples of n, then write i) M(15)

10.

ii) M(125)

F  a   F  b  represents the set of all common

factors of a and b. Write the set i) F  28   F  40 

ii) F 12   F  36 

11. Express the number 324 as the product of the powers of prime factors. 12. Find the total number of divisors or factors of 144. 13. Find the unit’s digit of the product of all the prime numbers lying between 1 and 1000. 14. If n2 + 9n + 20 is the product of two prime numbers, for n being some integer, then find the value of n2 +1. 15. lf n3 – 1 is a prime number, where n  N, then find the nature of n2 + n + 1.

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8th Class Mathematics

6

7. Fractions

Note: Finding of fraction in between any two

a c a+c and is given by . b d b+d

p A number of the form q , where p, q are integers

fractions

and q  0 is known as a fraction.

Example: Insert a fraction between

1 3 15 , , etc. 2 5 4 Note: All fractions are rational numbers.

The fraction which comes between

1 2 and 3 5

Example:

Difference between rational number and fraction Rational numbers and fractions are of the form p . q Any integer is a rational number, since we can

integer . But in a fraction, 1 the denominator part should not be 1. express any integer as

Types of Fractions i)

Simple Fraction: A fraction in its lowest terms is known as a simple fraction.

12 5 4 , , , etc. 25 7 3 Proper Fraction: A fraction in which numerator is less than denominator is known as a proper fraction. Example:

ii)

2 15 , etc. 5 17 iii) Improper Fraction: A fraction in which numerator is greater than denominator is known as an improper fraction.

2 1 2 3   5 35 8 vi) Like Fractions: Fractions having the same denominators are called like fractions. 9 17 41 , , , etc. 14 14 14 vii) Unlike Fractions: Fractions having different denominators are called unlike fractions. Example:

9 17 41 , , , etc. 4 14 12 How to convert a given Improper Fraction into a Mixed Fraction? Example:

In general, the mixed fraction of Where

5 17 4 , , , etc. 2 15 3

p iv) Mixed Fraction: The fraction of the form n , q

p where q is a proper fraction is known as a mixed fraction.

v)

1 15 Example: 2 , 4 , etc. 3 17 Compound Fraction : A fraction whose one or both terms are fractions is called a compound fraction. Normal form of a compound fraction

a/b . c/d www.betoppers.com is

x R is : Q . y y

Q = Quotient R = Remainder

Now, let us find the mixed fraction of

17 5

5) 17 ( 3 15 2

Example:

Example:

1 and 3

i.e.,

17 2 3 5 5

2 5  3  2 17      3  5 5 5 

Fraction between any two Fractions Finding of a fraction in between any two fractions a +c a c . and is given by the form b d b+d

Equivalent Fractions Some fractions may look different, but are really the same. If we multiply numerator and denominator with same number the fraction remains unchanged.

Example:

1 1 5 5 1    2 2  5 10 2

Number System

7

Conversion of Fractions to Decimals

8. Decimals The simplified form of a fraction is called a decimal. How do we read a decimal? Consider 123.456 . Whole number part, 123, should be read as one hundred and twenty three. The decimal point ‘  ’ should be read as point. The decimal part 456 should be read as four, five, six.

Let us understand the steps by expressing

Conversion of Decimals to Fractions



Let us express 0.75 as a fraction. Step - 1: Count the numbers after the decimal point, let it be x, in 0.75 we have x = 2. Step - 2: Now multiply and divide numerator by 10x. i.e. 102 = 100.

17153 in decimal. 1000 Step - 1: Write the numerator as it is 17153 Step - 2: Count the number of zeros in the denominator. Let it be x. Here x = 3 Step - 3: From ‘right to left’ in the numerator keep the decimal point after x places. 17153  17.153 1000

Classification of Decimals

0.75  100 75  100 100 Step - 3: Reduce the above fraction to simplest 0.75 =

75 3  100 4 Note: When we multiply a decimal number with 100, the decimal point moves 2 places right. When we divide it with 100, the decimal point moves 2 places left. Example : Express 331.570063 as a fraction. form.

331.570063  

 331570063 1000000 

 we have 6 digits after the decimal point.

Decimal values of some Fractions Fraction  10 3 10 13 10 705 10

Fraction  100 3 100 13 100 705 100

Value 0.1 0.3 1.3 70.5

Fraction 1 2 2   5 2 10 1 5 5   2 5 10 3 5 15   5 5 100 27 4 108   5 4 100

Value 0.2 0.5 0.15

Value 0.01 0.03 0.13 7.05

Terminating Decimal: The decimal in which the digits end after the decimal point. Example: 0.25, 0.5, 0.75, etc. Non-terminating Decimal: The decimal in which the digits do not end after the decimal point. Example: 1.762586 ...., 0.6 , 0.27 , 0.83 , etc. Note: The non-terminating decimals can be recurring or non-recurring. The non-terminating recurring decimals are again classified into pure recurring and mixed recurring. Pure Recurring Decimals: These are the decimals in which the set of digits is repeated immediately after the decimal point. Example: 0.8,0.3,0.5,0.35etc. Mixed Recurring Decimals: These are the decimals in which the set of digits is not repeated immediately after the decimal point.

1.08

Example: 0.28,0.6573,0.0985,etc. www.betoppers.com

8th Class Mathematics

8 Conversion of a Pure Recurring Decimal into a Fraction

12.45  2 decimal places = 12.450

Write the repeated figures only once in the numerator and take as many nines in the denominator as is the number of repeating figures. 5 53 67 ; etc. Thus, 0.5  ; 0.53  ; 0.067  9 99 999 Conversion of a Mixed Recurring Decimal into a Fraction:

0.735  3 decimal places = 0.735

In the numerator, take the difference between the number formed by all the digits after decimal point (taking repeated digits only once) and that formed by the digits which are not repeated. In the denominator, take the number formed by as many nines as there are repeating digits followed by as many zeros as is the number of non-repeating digits. 16  1 15 1   Thus, 0.16  227390  22 90 22516 0.2273   . 9900 9900

Types of Decimals

Based on number of decimal places in a set of decimals, they are classified into like and unlike decimals. Like Decimals The decimal numbers that consists of same number of decimal places after decimal point are called like decimals.

6.5  1 decimal place = 6.500 145.05  2 decimal places = 145.050 to make them like decimals we need same number of decimal places

 12.450 , 6.500 , 0.735 , 145.050 is the set of like decimals.

Comparison of Decimals Let us understand the steps of comparison by the following example. Arrange 0.85 , 1.58 , 2.85 , 0.58 , 0.583 in descending order. Step – 1: Write the given decimals as the set of like decimals i.e. 0.850, 1.580, 2.850 , 0.580 , 0.583 Step – 2: Compare their integral parts i.e. 0.850, 1.580, 2.850 , 0.580 , 0.583

 0

1

2

0

0

The decimal number with greater integral part is greater.

 2.850 > 1.580 (0.850, 0.580, 0.583) Step – 3: If the integral parts are same, then compare the first decimal place. i.e. 0.850 , 0.580 , 0.583



8

5

5

For example, 3.54 , 13.68 , 0.47 are like decimals as they all have same number of decimal places i.e., two.

The number with greater first decimal place is greater.

Unlike Decimals

 0.850 > (0.580 and 0.583)

The decimal numbers that consists of different number of decimal places after decimal point are called unlike decimals.

Step – 4: If the first decimal place are same, then compare the second decimal places i.e. 0.580, 0.583

For example, 3.54 , 13.8 , 0.475 are unlike decimals as they have different number of decimal places.

 8

3.54  2 decimal places

8

But they are also equal.

13.8  1 decimal place

So, we compare the third decimal place. i.e. 0.580, 0.583

0.475  3 decimal places

 0 3  0.583 > 0.580

Let’s convert 12.45 , 6.5 , 0.735 & 145.05 into a set of like decimals.

Number with greater third decimal place is greater.

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 2.85 > 1.58 > 0.85 > 0.583 > 0.58

Number System

9

Rounding off a Decimal Various Cases Example Retain as many digits after the decimal point as are required and omit the remaining digits. In the set of omitted digits, if Let’s round off 2.6457, to two decimal places. the first digit is 5, then The first omitted digit = 5. increase the last retained So, increase the last retained digit by 1. i.e., 4 + 1 = 5. digit by 1.

required decimal number = 2.65

In the set of omitted digits, if Let’s round off 7.5689 to two decimal places. the first digit is greater than The first omitted digit = 8 5, then increase the last So, increase the last retained digit by 1. i.e., 6 + 1 = 7. retained digit by 1.

required decimal number = 7. 57 Let’s round off 5.12345 to three decimal places.

In the set of omitted digits, if

The first omitted digit = 4

the first digit is less than 5,

So, it remains unchanged.

then it remains unchanged.

required decimal number = 5.123

Formative Worksheet 5 5 to get ? 8 9 17. Arrange the following in the ascending order 16. What number must be added to

3 7 1 2006 413 , , , , 4 9 2 2007 1115 18. Evaluate: i) 35  0.07 ii) 2.5  0.0005  iii) 136.09 43.9 19. Find the products: i) 6.3204 × 100 ii) .069 × 10000 20. Express the following as fractions: i) 0.37

ii) 0.053

21.

 893  786   893  786  Simplify :  893  786 



iii) 2.536

ii) 0.1254



9. L.C.M Least Common Multiple

2

Conceptive Worksheet 16. Arrange the following in descending order : – 2006.00026, 2006.262602, –20.0626, 0.002006, 3/ 2. 17. Express the following rational number as ratio of two integers, whose decimal expansions given, i) 0.123454545… ii) 0.428571

i) 0.17

20. Find : 0.3467  0.1333 .

iii) 3.142857 2

5 7 13 16 3 , , , and in 8 12 16 29 4 ascending order of magnitude. 19. Express as fractions: 18. Arrange the fractions

iii) 0.375

Least number that is divisible by all the given numbers is called Least Common Multiple. For example, the common multiples of 7 and 2 are 14, 28, 42, etc, Among these, 14 is least and the LCM of 7 and 2.

Methods of finding LCM Multiples Listing Method In this method, LCM is found by listing several multiples of the given numbers. This method is generally used for finding LCM of numbers with less difference. Example: Let us find the L.C.M of 3 and 5. Multiples of 3 are 3, 6, 9, 12, 15, 18, ....., 45, 48, ...., 90, etc., Multiples of 5 are 5, 10, 15, 20, 25, 30, 45, ..., 90, ..., 120, etc., www.betoppers.com

8th Class Mathematics

10 The common multiples of 3 and 5 are 15, 30, 45, etc. The Least Common Multiple of 3 and 5 is 15 Prime Factorization Method In order to find the LCM of two or more given numbers we write the prime factorization of each of the given numbers. Then, the required LCM of these numbers is the product of all different prime factors of the numbers, now multiplying common prime factors having greater powers, will get L.C.M. Example:Let us find the LCM of 108, 72and 360. 2 108 2 54 3 27 3 9 3

2 72 2 36 2 18 3 9 3

2 360 2 180 2 90 3 45 3 15 5 360 = 23 × 32 × 5

108 = 22 × 33 72 = 23 × 32 Now the factors are 108 = 22 × 33 72 = 23 × 32 360 = 23 × 32 × 5 Highest common prime factors = 23  33  5 = 1080  LCM of 108, 72 and 360 = 1080 Long Division Method In order to find the LCM of any amount of numbers, we use the long this method. i) We arrange the given numbers in a line, in any order. ii) We divide by a number which divides exactly at least two of the given numbers and carry forward the numbers which are not divisible. iii) This process is repeated till no two of the given numbers are divisible by the same number. iv) The product of the divisors and the undivided numbers is the required LCM of the given numbers. Example: Let’s find the LCM of 16, 8 and 4. 2 16, 8, 4 2 8, 4, 2 2 4, 2, 1 2, 1, 1  LCM of 16, 8 and 4 = 2  2  2  2 = 16 Note: We must take prime numbers as divisors like 2, 3, 5, 7............

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Formative Worksheet 22. Find the L.C.M. of 22 × 33 × 5 × 72, 23 × 32 × 52 × 74, 2 × 3× 53 × 7 × 11. 23. Find the L.C.M. of 16, 24, 36 and 54. 24. Find the L.C.M. of 0.63, 1.05 and 2.1. 25. The traffic lights at three different road crossings change after every 48 sec. 72 sec. and 108 sec. respectively. If they all change simultaneously at 8 : 20 : 00 hours, then at what time will they again change simultaneously? 26. Four different electronic devices make a beep after every 30 minutes, 1 hour, 1½ hours and 1 hour 45 minutes respectively. All the devices beeped together at 12 noon. At what time they will again beep together. 28. Find the smallest number of five digits exactly divisible by 16, 24, 36 and 54. 29. Find the smallest four digit number which when divided by 17 and 13 leaves a remainder of 7 in each case. 30. How many numbers lying between 950 and 1050 leave a remainder of 6 when divided by 7, 8 and 14 ? 31. Find the largest number of four digits exactly divisible by 12, 15, 18 and 27.

Conceptive Worksheet 21. The LCM of two numbers is 48. The numbers are in the ratio 2 : 3. Then find the sum of the numbers. 22. Find the least number which when divided by 16, 18, 20 and 25 leaves 4 as remainder in each case, but when divided by 7 leaves no remainder. 23. When two numbers are divided by a certain divisor, the remainders are 9 and 10 respectively. When the sum of the numbers is divided by the same divisor, the remainder is 6. Then find the divisor. 24. A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and C in 198 seconds, all starting at the same point. After what time will they meet again at the starting point? 25. Find the smallest number of five digits exactly divisible by 16, 24, 36 and 54.

Number System

10 H.C.F Highest Common Factor The number that divides all the given numbers is called as a Common Factor. The greatest among the common factors of the given numbers is called Highest Common Factor (HCF). Example: The common factors for 24 and 12 are 2, 4, 6, 12. Among the factors, 12 is the highest and is the HCF of 24 and 12 . HCF is also called as: GCD – Greatest Common Divisor HCD – Highest Common Divisor GCF – Greatest Common Factor HCF – Highest Common Factor

11 Long Division Method The following steps should be used to find the HCF by long division method. i) If two numbers are given, divide the greater number by the smaller one. ii) Divide the divisor by the remainder. iii) Repeat the process of dividing the preceding divisor by the remainder last obtained, till the remainder zero is obtained. iv) The last divisor is the required HCF of the given numbers. Example: Let us find the HCF of 136, 170 and 255, using the long division method. Sol: 136) 170 (1 34) 255 (7 136

Methods of finding HCF Listing the Factors Method In this method, HCF is found by listing several factors of the given numbers. Example: Let us find the HCF of 36 and 48. Step - 1: Write down all the factors of the given numbers. 36  1, 2, 3, 4, 6, 9, 12, 18, 36. 48  1, 2, 3, 4, 6, 8, 12, 24, 48 Step - 2: Check the common factors of the given numbers. 1, 2, 3, 4, 6, 12 Step - 3: Identify the greatest among those common factors of the given numbers. The greatest among these factors is 12 and is the HCF of the 36 and 48. Prime Factorization Method We first find the prime factorization of each of the given numbers. Then the product of all common prime factors, using the least power of each common prime factor, is the HCF of the given numbers. Example: Lets find the HCF of 200 and 320

238

34) 136 (4 136

17) 34 (2 34

0

0

 HCF of 136, 170 and

255 = 17

Solving methods to solve various types of problems related to HCF

2 200

2 320

2 100

2 160

Type – 1: Find the greatest number that will exactly divide a, b and c. Approach: Required number = HCF of a, b and c. Type – 2 : Find the greatest number that will divide x, y and z leaving remainders a, b and c respectively. Approach: Required number = HCF of (x–a), (y–b) and (z–c) Type – 3: Find the greatest number that will divide a, b and c leaving the same remainder in each case. Approach: Required number = HCF of (a–b), (b–c) and (c–a) Example: Find the greatest number that will divide 148, 246 and 623 leaving remainders 4, 6 and 11 respectively. Required number = HCF of (148–4), (246–6) and (623–11) = HCF of 144, 240 and 612

2

50

2

80

144) 240 ( 1

5

25

2

40

5

2

20

2

10 5

320 = 2×2×2×2×2×2×5 200 = 2×2×2×5×5 3 2 200 = 2 × 5 320 = 26 × 5  HCF = product of lowest power prime factors

= 23 × 5 = 40

48) 612 (12

144

48

96) 144 (1

132

96

96

48) 96 (2

36) 48 (1

96

36

0

12) 36 (3 36 0

HCF of 144, 240 and 612 = 12  Required number = 12 www.betoppers.com

8th Class Mathematics

12 Co-Primes The numbers that do not have any common factor other than 1 are called co-prime or relatively prime. Example: Consider the numbers 8 and 15 Set of factors of 8 = {1, 2, 4, 8} Set of factors of 15 = {1, 3, 5, 15} The common factor of 8 and 15 is 1 only.  8, 15 are co-primes. Note: Two numbers are said to be co-prime or relatively prime, if their HCF is 1. For example, the HCF of 4, 13 is 1. Therefore, 4 and 13 are co-prime numbers.

Formative Worksheet 32. Find the HCF of 180 and 324. 33. Find the HCF of 4 × 27 × 3125, 8 × 9 × 25 × 7 and 16 × 81 × 5 × 11 × 49

11. H.C.F and L.C.M HCF and LCM of Fractions i) HCF of fractions =

HCF of numerators LCM of denominators

LCM of numerators ii) LCM of fractions = HCFof denominators

Relation between LCM and HCF Consider any two numbers a and b. Let their LCM and HCF be x and y respectively. Then ab = xy Product of any two numbers = Product of their LCM and HCF Note: Product of ‘n’ numbers = (HCF for each pair)n–1 × LCM of ‘n’ numbers.

Formative Worksheet

34. Find the HCF of 22 × 33 × 55, 23 × 32 × 52 × 7 and 24 × 34 × 5 × 72 × 11.

38. Find the LCM of

1 5  4 , , , . 3 6 9 27

35. The HCF of two numbers is 18 and the first 4 quotients obtained in the division are 2,1, 2, 2. Find the numbers.

39. Find the HCF of

9 12 18 21 , , and . 10 25 35 40

36. What is the number of number-pairs lying between 40 and 100 with their HCF as 15?

40. Find the HCF and LCM of

37. Which of the following is a pair of co-primes ? 1) (16, 62)

2) (18, 25)

3) (21, 35)

4) (23, 92)

Conceptive Worksheet 26. The HCF of 24 × 32 × 53 × 7, 23 × 33 × 52 × 72 and 3 × 5 × 7 × 11 is: 27. Three numbers are in the ratio 1 : 2 : 3 and their HCF is 12. Then find their sum. p

q

r

28. If a × b × c is a perfect cube, where a, b and c are distinct primes, then p, q, r are respectively 29. Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then find the sum of the digits in N. 30. A number has two digits whose sum is 10. If 18 is added to the number, its digits get interchanged. Find the HCF. of the number and the new obtained number.

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41. Arrange the fractions

5 4 7 , , . 6 9 18

17 31 43 59 , , , in ascending 18 36 45 60

order. 42. If the product of any two numbers is 6912 and their LCM is 288 then find their HCF. 43. Find the HCF and LCM of between p and q where p = x2 + x(a + b) + ab and q = x2 + x(b + c) + bc 44. Find the LCM and HCF of m and n, if m = 2004! and n = 2005 × 2006 × 2007 × 2008 × 2009. 45. Three wheels completing 60, 36 and 24 revolutions in a minute, start from a certain point in their circumference downwards. By what time will they come together again at the same position? 46. There are 4 numbers. The HCF of each pair is 3 and the LCM of all the four numbers is 126. What is the product of the 4 numbers?

Number System

13

Conceptive Worksheet

9.

31. The HCF and LCM of two numbers are 11 and 385 respectively. If one number lies between 75 and 125, then find that number. 32. Find the smallest fraction, which each of 33. 34.

35. 36.

6 5 19 , , 7 14 21

will divide exactly. Two numbers, both greater than 29, have HCF 29 and LCM 4147. Find the sum of the numbers . If the sum of two numbers is 55 and the HCF and LCM of these numbers are 5 and 120 respectively, then find the sum of the reciprocals of the numbers. Find the equation, whose roots are equal to the LCM and HCF of the numbers 10 and 16. The LCM of two numbers is 6 times their HCF. If the product of the numbers is 3174, then find their HCF.

Summative Worksheet 1. 2.

3. 4. 5.

6.

7.

8.

The H.C.F of

1)

2 3

2)

2 8 64 10 , , and is: 3 9 81 27 2 81

3)

160 3

4)

160 81

10. There are 312 boys and 408 girls in a school, who are to be divided into equal sections of either boys or girls alone. The maximum number of boys or girls that can be placed in a section are _______ 11. A trader has three kinds of sugar: of the first kind 368 kg, of the second kind 391 kg and of the third kind 345 kg. The least number of full casks of equal size in which this can be stored without mixing is ________ 12. Find the greatest number which, when subtracted from 3000, is exactly divisible by 7, 11 and 13. 13. Find the L.C.M of 96 and 114. 14. Find the L.C.M of 22, 33, 44 and 55.

Show that the product of three consecutive integers is divisible by ‘6’. The greatest number which will divide 25, 74 and 99, leaving remainders 1, 2 and 3 respectively, is ________ 1) 17 2) 18 3) 22 4) 24

15. Find the HCF of 370, 592, 1036

2 8 24 , and is _________ 5 7 29 The LCM of two numbers is 45. One of the numbers is 9. Find the other. A number, when divided successively by 4 and 5, leaves remainders 1 and 4 respectively. When it is successively divided by 5 and 4, the respective remainders will be 1) 1, 2 2) 2, 3 3) 3, 2 4) 4, 1 Find the least number which when divided by 20, 25, 35 and 40 leaves remainders 14, 19, 29 and 34 respectively. The sum of two numbers is 216 and their H.C.F. is 27. The numbers are: 1) 27, 189 2) 81, 189 3) 108, 108 4) 154, 162

17. The HCF and LCM of two numbers are 44 and 264 respectively. If the first number is divided by 2, the quotient is 44. The other number is ______

The LCM of

2 3 4 9 The L.C.M. of , , , is: 3 5 7 13 1) 36

2)

1 36

3)

1 1365

4)

16. Three numbers are in the ratio 1 : 2 : 3 and their H.C.F. is 12. The numbers are: 1) 4, 8, 12

2) 5, 10, 15

3) 10, 20, 30

4) 12, 24, 36

18. The LCM of two numbers is 630 and their HCF is 9. If the sum of the numbers is 153, then find their difference. 19. If a and b are two rational numbers then prove that a + b, a – b, ab are rational numbers. If b  0 , show that

a is also a rational number.. b

20. Consider the numbers : 15, 21, 28, 35, 121. The three mutually relatively prime numbers from the above numbers are 1) 15, 21, 121

2) 28, 35, 121

3) 15, 28, 121

4) 21, 35, 121

12 455 www.betoppers.com

8th Class Mathematics

14

HOTS Worksheet If n is a positive integer, then 3n (3n – 1) will be an even number for: 1) Only odd values of n 2) Only even values of n 3) For all values of n 4) Only for prime values of n 2. The last digit of 391 + 393 is 1) Even number 2) Odd number 3) Prime number 4) Composite number 3. The number of possible sets of two prime numbers such that their sum or difference will never yield a composite number is 1) 0 2) 1 3) 2 4) 5 4. If a2 – b2 is prime, then the correct option is 1) a2 – b2 = a – b 2) a2 – b2 = a + b 3) a2 – b2 = ab 4) a2 – b2 = 2 5. If p and q are prime numbers such that p < q < 70 and p + q is a factor of 243, then how many values of q are possible? 1) 0 2) 1 3) 2 4) 3 6. Three numbers which are co- primes to each other are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is 1) 75 2) 81 3) 85 4) 89 7. Three different containers contain 496 litres, 403 litres and 713 litres of mixtures of milk and water. Then, the biggest measure that can measure all the different quantities exactly is 1) 1 litre 2) 7 litres 3) 31 litres 4) 41 litres 8. The least number which is divisible by all the prime numbers between 10 and 20 is 1) 46,189 2) 6,92,835 3) 36,465 4) 62,985 9. Two numbers when divided by a certain divisor leave remainders of 12 and 17 respectively. If their sum is divided by the same divisor, the remainder is 11. Then the divisor is 1) 18 2) 22 3) 14 4) 16 10. The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is 1) 60 2) 23 3) 33 4) 37 11. Of the number of pairs which have 16 as their H.C.F. and 136 as their L.C.M. we can say that: 1) No such pair exists 2) Only one such pair exists 3) Only two such pairs exist 4) Many such pairs exist. 1.

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12. The H.C.F. and L.C.M. of two numbers are 84 and 21 respectively. If the ratio of the two numbers is 1 : 4, then find the bigger of the two numbers. 1) 12 2) 48 3) 84 4) 108 13. If p and q are relatively prime, then find the sum of H.C.F. and L.C.M. of p2 and q2. 1) (pq)2 + 1 2) p2 + q2 2 2 3) p + q + 1 4) p + q + 1 14. If (a + b) and ab are relatively prime numbers and

a +1 b +1 + is an integer, then G.C.D. of a + b +1 b a and ab is 1) 1 2) ab 3) a + b + 1 4) ab+ 1 15. If p is a factor of q, then the sum of L.C.M and H.C.F. of p and q is 1)

p 2  4 pq  q 2 2)

p 2  2 pq  q 2

3)

p 2  2 pq  q 2 4) None

16. If p is prime, then number of numbers that are less than p2 and relatively prime to it are 1) p2 – p 2) p 3) p – 1 4) 2p 17. The average score of a certain cricketer for all the innings he played is 60 runs. His highest score exceeds his lowest score by 138 runs. If these two innings are excluded, his average is reduced by 2 runs. If his lowest score is greater than 36 but less than 48, then the number of innings he played given that the number of innings which is a prime number is 1) 59 2) 67 3) 53 4) None of these 18. The number of integers from 1 to 2001 that are relatively prime to 2002 is 1) 2001 2) 720 3) 1997 4) 2000 19. A 3-digit number has only 2 factors (other than 1 and itself) both of which are co-primes. How many such numbers are possible? 1) Only 1 2) Only 2 3) Only 3 4) More than 3 20. If x2  y 2  p ( p is prime), then find x and y.

1 1  0.16134 , the value of is: 6.198 0.0006198 1) 0.016134 2) 0.16134 3) 1613.4 4) 16134

21. If

22. Find  n  2    n  1  n   n  1 n  2    n  3  ....   n  k 

Number System

15

23. If a + b = c + d , prove that a = c and b = d. 24. Prove that 2

2 is irrational. 2

25. D = a + b + c2, where c = ab and a and b are two consecutive numbers, Prove that D is odd where a  0 and b  0 .

IIT JEE Worksheet If p, q, r and s are consecutive odd numbers, then (p2 + q2 + r2 + s2) is always divisible by 1) 5 2) 7 3) 3 4) 4 2. If x3 – 1 = 1,330 and y3 – 8 = 1,720, which of the following is not false 1) y is multiple of x or x is a multiple of y 2) Both x and y are even numbers 3) Both x and y are odd numbers 4) (x + y)2 + 96 is a perfect square. 3. Which of the following numbers is exactly divisible by all the prime numbers between 30 and 40? 1) 785 2) 1,147 3) 1,295 4) 1,237 4. If p is a prime and n is a positive integer, then GCD of p and n is 1) 1 2) p 3) n 4) 1 or p 5. If a and b are twin primes and a2 – b2 = 120, then their average is 1) 18 2) 30 3) 12 4) 6 6. A is a 3-digit number below 200. It has no singledigit factor except 1. Find the number given that is not a perfect square. 1) 187 2) 143 3) Either (1) or (2) 4) Both (1) and (4) 7. Find the number n given that has the first ten natural numbers as its factors and is the smallest 5 digit number 1) 12,520 2) 10,080 3) 36,280 4) 10,040 8. The smallest number that must be multiplied by 4500 to make it a perfect cube is 1) 2 2) 3 3) 6 4) 5 9. If 2m + 5n and 3m + 4n are two distinct prime numbers, then the numbers whose G.C.D. is 50 is 1) 50m + 150n and 150m + 20n 2) 100m + 250n and 150m + 200n 3) 100m + 200n and 250m + 300n 4) 50m + 250n and 150m + 250n 10. When a number is divided, the divisor is equal to three times the remainder or equal to the square of the remainder. If ten times the divisor is more than the quotient by a value equal to the remainder, then what is the dividend ? 1) 857 2) 976 3) 786 4) Cannot be determined 1.

11. When a number is divided by 9 the remainder is 7. If twice the number is divided by 9, then the remainder is 1) 9 2) 7 3) 4 4) 5 12. Two positive integers are in the ratio of 3 : 4. Their HCF and LCM are in the ratio of 1 : 12. If the difference between their LCM and their HCF is 44, then sum of the numbers are 1) 28 2) 48 3) 56 4) 120 13. The HCF and the LCM of two numbers are 32 and 192 respectively. If the square of one of the numbers is a perfect cube, then the two numbers are 1) 64, 96 2) 32, 48 3) 128, 183 4) None of these 14. If L.C.M. of two numbers is 5 times of one of the numbers and the other number is 25, then H.C.F. of the two numbers is 1) 5 2) 25 3) 125 4) 175 15. If the sum and difference of two fractions are

56 21

16 , respectively, then L.C.M. of the two 21 fractions is and

1)

12 7

2)

20 21

3)

60 7

4)

4 21



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16

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8th Class Mathematics

Chapter -2

Learning Outcomes

Square Roots & Cube Roots

By the end of this chapter, you will understand •

Perfect square

•

Pythagorean triplet

•

Square Root

•

Cube of a number

•

Cube roots

Square Roots

Example1: Is 225 a perfect square ? Resolving 225 into prime factors,

1. Introduction You know that the area of a square = side × side Let us observe the following table. Area of the square (in cm2)

Side of a square (in cm) 4

4 × 4 = 16 = 42

7

7 × 7 = 49 = 72

9

9 × 9 = 81 = 92

11

11 × 11 = 121 = 112

20

20 × 20 = 400 = 202

x

x × x = x2

p

p × p = p2

5 3 3

45 9 3 1

5 we get, 180 = 2 × 2 × 3 × 3 × Though 2 and 3 are paired, 5 is not paired. Thus, 180 cannot be expressed as the product of pairs of equal factors. Hence, 180 is not a perfect square. Some more Examples: Perfect Non-perfect squares squares 9 = 3 × 3 = 32 5=? 2 25 = 5 × 5 = 5 21 = ? Note: Though 5 can be expressed as

5  5 it is

not a prefect square because 5 is not a natural number. Similarly, 21 is also not a perfect square.

2. Perfect Square

Properties of Squares of Numbers i)

Identifying a Perfect Square: A given number is perfect square, if it can be expressed as the product of pair of equal factors. Note: The factors can be found by prime factorisation method.

225

we get, 225 = 5 × 5 × 3 × 3 as it is expressed as the product of equal factors. Hence, it is a perfect square. Example2: Is 180 a perfect square ? Resolving 180 into prime factors,

What is special about the numbers 4, 9, 25, 64 and other such numbers ? Since, 4 can be expressed as 2 × 2 = 2 2 , 9 can be expressed as 3 × 3 = 3 2 , all such numbers can be expressed as the product of the number with itself. Such numbers like 1, 4, 9, 16, 25, ..... are known as square numbers of 1, 2, 3, 4, 5,...... respectively.

A natural number is said to be perfect square, If it is the square of some other natural number. Example: 81 = 92, 4 = 22, 36 = 62, 10000 = 1002.

5

ii)

The square of an even number is always an even number. Example: 6 is an even number and 62 = 36 which is even. 8 is an even number and 82 = 64 which is even. The square of an odd number is always an odd number.

8th Class Mathematics

18

iii)

Example: 7 is an odd number and 72 = 49 which is odd. 11 is an odd number and 112 = 121 which is odd. The square of a proper fraction is less than the proper fraction. Example:

3 4

is a proper fraction and

2

9 3 3      4  16 4 2

iv)

v)

vi)

25 5 5 5  is a proper fraction and    36 6 6 6 The square of a decimal fraction less than 1 is smaller than the decimal. Example: 0.2 < 1 and (0.2)2 = 0.04 < 0.2 0.4 < 1 and (0.4)2 = 0.16 < 0.4 A natural number ending with 2, 3, 7 or 8 is never a perfect square. Example: 52, 77, 93, 18 are not perfect squares. (or) All perfect squares end with 0, 1, 4, 5, 6 or 9 only. Example:16, 225, 1024, 900 are perfect squares. Note: Though all perfect squares ends with 0, 1, 4, 5, 6, (or) 9 all numbers ending with 0, 1, 4, 5, 6, (or) 9 need not be perfect squares. Example: 56, 10, 39, 84, 71, 95 are not perfect squares. Any natural number ending with odd number of zeros is never a perfect square. Example: 1000, 500000...... Note: Perfect squares always end with an even number of zeros but all numbers ending with even number of zeros are not perfect squares. Example: 900 is ending with 2 zeros, it is a perfect square. 800 even though ending with 2 zeros, is not a perfect square.

Interesting patterns of Square Numbers: i)

Sum of first n odd natural numbers is n2. Example: 1 = 1 = 12 [one odd number] 1+3 = 4 = 22 [sum of first two odd numbers] 1+3+5 = 9 = 32 [sum of first three odd numbers] 1 + 3 + 5 + 7 [.....] = 16 = 42 1 + 3 + 5 + 7 + 9 [.....] = 25 = 52 1 + 3 + 5 + 7 + 9 + 11 [.....] = 36 = 62

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ii)

Observe the squares of numbers 1, 11, 111...etc. They give a beautiful pattern: 12 = 1 2 11 = 121 2 111 = 12321 2 1111 = 1234231 2 11111 = 123454321 2 111111 = 12345654321 2 1111111 = 1234567654321 The square of a natural number having all digits as 1 (number of ones are £ 9) results in a palindrome. iii) Another interesting pattern: 72 = 49 2 67 = 4489 2 667 = 444889 2 6667 = 44448889 2 66667 = 4444488889 2 666667 = 444444888889 The numbers 49, 4489, 444889, . . . obtained by inserting 48 into the middle of the perceiving number are squares of integers of 7, 67, 667, 6667, . . . respectively. The fun is in being able to find out why this happens. May be it would be interesting for you to explore and think about such questions even if the answers comes some years later. iv) The difference of odd squares of successive numbers is equal to their sum. Example: 172 – 162 = 17 + 16 1012 – 1002 = 101 + 100 2352 – 2342 = 235 + 234 In general, (n + 1)2 – (n )2 = (n + 1) + (n) = 2n+1 v) If 1 is added to the product of two consecutive even natural numbers, it is equal to the square of the only odd natural number between them: 2 × 4 + 1 = 9 = 32 4 × 6 + 1 = 25 = 52 6 × 8 + 1 = 49 = 72 8 × 10 + 1 = 81 = 92 10 × 12 + 1 = 121 = 112

Square Roots & Cube Roots

19

vi) The squares of natural numbers like 11, 111, ….., have a nice pattern as shown below 121 × (1 + 2 + 1) = 484 = 222 112 × (sum of the digits in 112) = (2 ×11)2 12321 × (1 + 2 + 3 + 2 + 1) = 110889 = 3332 1112 × (sum of the digits in 1112) = (3× 111)2 1234321 × (1 + 2 + 3 + 4 + 3 + 2 + 1) = 19749136 = (4444)2 11112 × (sum of the digits in 11112) = (4 × 1111)2  Square of a number   sum of digits   ×    having all ones   in the square   number of digits  = × the number   in the number 

Note: if m > 1 and i) If m is odd, then the Pythagorean triplet is

 m2  1 m2  1  ,  m,  2 2 .  Example: If one of the numbers of a Pythagorean triplet is 3, find the triplet. Sol: As the given number 3 is odd,  m2  1 m2  1  ,  m,  2 2   i.e., m = 3 The Pythagorean triplet is

2

 m2  1 m2  1  ,  m,  2 2  

Formative Worksheet 1.

 32  1 32  1   8 10  i.e,  3, ,  i.e,  3, ,  i.e, 3,4,5 2 2   2 2  Thus, the required triplet is (3, 4, 5). Note: if m > 1 and ii) 2m is even then the Pythagorean triplet is (2m, m2 – 1, m2 + 1). Example:If one of the numbers of a Pythagorean triplet is 4, find the triplet. Sol: As the given number 4 is even i.e., 2m = 4  m = 4/2 = 2, m = 2  The Pythagorean triplet is (2m, m2 – 1, m2 + 1) i.e., (2(2), 22 – 1 , 22 + 1) i.e., (4,3,5) Thus, the required triplet is (3,4,5).

If 3a = 4b = 6c and a + b + c = 27 29 , then find a 2 + b 2 + c2

2. 3.

Find the least perfect square which is divisible by each of 21, 36 and 66. If a = 0.1039, then find the value of 4a 2  4a +1 + 3a

Conceptive Worksheet 1. 2. 3.

Find the smallest number to be subtracted from 549162 in order to make it a perfect square. Find the least number which must be added to 4931 to make it a perfect square. A General wishing to draw up his 64019 men in the form of a solid square, found that he had 10 men over. Find the number of men in the front row.

3. Pythagorean Triplet Let us observe the sum of squares of 3 and 4 i.e., 32 + 42 = 9 + 16 = 25  32 + 4 2 = 5 2 Here, the sum of squares of 3 and 4 is square of another number 5.  The sum of squares of two numbers is again square of another number.  The three numbers (3, 4, 5) are said to be Pythagorean triplets. In general, a triplet (m, n and p ) of natural numbers m, n and p is said to be a Pythagorean triplet if m2 + n2 = p2 .

Formative Worksheet 4.

If 8 is one of the numbers in a Pythagorean triplet, then find the triplet.

5.

If n3 + 3n2 + 2n  n  N  is the product of the numbers in a Pythagorean triplet, then find the triplet

Conceptive Worksheet 4.

5. 6.

In a Pythagorean triplet, the sum of the greatest number and the least number is 72 and their difference is 50, then find the other number. If the difference of two numbers in a Pythagorean triplet is 2, then find the other number. If 49 is one of the numbers in a Pythagorean triplet, then find the remaining numbers. www.betoppers.com

8th Class Mathematics

20

4. Square Root 1/2

Let us consider 81 . As 81 is raised to the power 1/2, it is said to be the square root of 81. 1

Denoted as

812

Exponential form



81 Radical form

Thus, any number a raised to the power 1/2 is said to be the square root of the given number, denoted as

a.

Finding Square Root of a Number I n the previous topics we have seen different methods of finding the squares of numbers given. Let us now have a look at the various methods for finding the square root of the given number. Prime Factorization Method: Step-1: Write the prime factorisation of the given number. Step-2: Pair the factors such that primes in each pair are equal. Step-3: Choose one prime from each pair and multiply all such primes. Step-4: The product thus obtained is the square root of the given number. Let us understand this method through the example given below. Example: Find the square root of 24336. 24336  (2  2)  (2  2)  (3  3)  (13  13)

= 2 × 2 × 3 × 13 = 156

 24336  156 Division Method: Step-1: Place a bar over every pair of digits starting from the units digit. Step-2: Find the largest number whose square is less than or equal to the number under the left most bar taking this number as the divisor and the number under the left-most bar as the dividend. Divide and get the remainder. Quotient in this step is same as the divisor. Step-3: Bring down the number under the next bar to the right of the remainder. This is the new dividend. Step-4: Double the quotient and enter it with a blank on the right for the next digit of the next possible divisor. Step-5: Find the largest possible digit to be taken beside the divisor and also to get the new digit in the quotient. Step-6: Bring down the number under the next bar to the right of the new remainder. www.betoppers.com

Step-7: Repeat steps 4, 5 and 6 till all the bars have been considered. The final quotient is the square root. Let us understand this method better through the following example. Example: Find the square root of 467856. Step-1: Placing bars over every pair of digits i.e., 4678 56 . Step-2: Finding the largest number whose square is less than or equal to the left most bar. i.e., 36 < 46  6 is the required divisor. Now divide 46 by 6. 6 6 4678 56 36 10

Step-3: Bring down the number under the next bar, beside the remainder of Step 2. 6 4678 56 36

6

10 78

Step-4: Doubling the quotient i.e., 2 × 6 = 12. Step-5: Guessing the largest possible digit to be taken beside the divisor and also to get the new digit in the quotient. The required digit is8. Now divide 1078 by 128 and get the remainder 54. 68 6

46 78 56 36

128 10 78 10 24 54

Step-6: Bringing down the number under the next bar to the right of the new remainder. 68 6 4678 56 36 128 1078 1024 54 56

Step-7: Repeating the steps 4, 5 and 6 till all the bars have been considered. The final quotient is the required square root.

Square Roots & Cube Roots

21

684 6 46 78 56 36 128 1078 1024 1364 54 56 54 56 0

 467856  684

Note: This method is more efficient with larger numbers, but it can also be used to find square root of smaller numbers i.e., 3 digit or 4 digit numbers. Assumption Method: Steps for finding the square roots up to four digit numbers, without using either factorization or division methods. Step-1: Find the largest number whose square is less than (or) equal to the number under the left most bar. This is the tens digit of the square root. Step-2: Find the units digit by squaring the relative number. Step-3: Choose the correct digit by squaring one possible square root and comparing it with the given number. Example: Find the square root of 9801. Sol: 98 01 Step-1: 9 2 = 81 is the largest square number < 98; The tens digit in the square root of 9801 is 9.  9801  9 ? ;

Step-2: 12 = 1 ; 92 = 8 1 in both the cases, the units digit is 1. Step-3: Trial 912 = 8281  9801 i.e., 1 doesn’t satisfy the units digit of square root of 9801.  9801  99 Example: Find the square roots of 144 and 6561. Sol: 144

Step-1: 1 44 12 = 1

Step-1: 65 61 82 = 64  Tens digit in square root of 6561 is 8 . 6561  8 ?

Step-2: 12 = 1 ; 92 = 8 1 In both the cases, the units digit is 1. Step-3: Trial 802 = 6400 is closer to 6561 than 902 = 8100

 6561 is closer to 80 than 90 The tens digit of square root may be 8 square root of 6561 may be 81 Verification : 81×81=6561  6561  81

Square Root of a Decimal Step-1: Place bars on the integral part of the number in the usual manner. Step-2: Place bars on the decimal part on every pair of digits beginning with the first decimal place. Step-3: Start finding the square root by the division process as usual. Step-4: Place decimal point in the quotient as soon as the integral part is over. Step-5: Stop when the remainder is zero. Step-6: The quotient at this stage is the square root. Example: Find the square root of 0.00059049 Step-1: Here, integral part is 0. So place a bar on 0. Step-2: Starting from the first decimal point, place the bars on every pair of digits. i.e., 00059049 . Step-3: Start finding the square root and place the decimal point in the quotient as soon as the integral part is over. 0.0243 2 0.00 059049 4 44 483

 Tens digit in square root of 144 is 1. 144  1 ? 2

Example: Find the square roots of 6561. Sol: 6561

2

Step-2: 2 = 4 ; 8 = 6 4 In both the cases, the units digit is 4. Step-3: Trial 122 = 144  144  12

1 90 1 76 1449 1449 0

 0.0005905  0.0243 Tip (i) : If the integral part is zero in the given decimal number, then the integral part in the square root is also zero.

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8th Class Mathematics

22 Tip (ii): If the first pair after the decimal point is 00, then the first digit after the decimal in the square root is also zero.

Square Root of a Rational Number For any two rational perfect squares m and n

 n  0 ,

For any two integral perfect squares a and b (b  0), ab  a  b . 25  36  52  6 2 = = 5 × 6 = 30 ( for any two integers a and b, (a × b)2 = a2 × b2)

Example 1:

m m  n n

Example 1: Find the square root of 21

2797 . 3364

Square root of

21

Square Root of Integers

73441 2797 2797 73441   21  3364 3364 3364 3364

(5  6) 2

Example 2: 25  36  52  62 = 5 × 6 = 30 From examples (1) & (2), we notice that

 25  36  25  36 Note: 1) a  b  a  b .

Let us first find

73441 by long division method: 271

2)

ab  a  b .

Square Roots of Imperfect Squares

5

33 64 25

Till now, we have learnt how to find square roots of numbers which are perfect squares. Now, let us find the square roots of numbers which are not perfect squares. Remember that in such cases, we add zeros after the decimal point, or after the last figure if the original number is already in decimal form. Then carry out the answer to the desired number of places. Example: Find the square root of 1869 to 2 decimal places. Sol: Since we have to find the square root of 1869 to 2 decimal places, we have to add 6 zeros after the decimal to form 3 pairs.

108

864 864 0

1869  1869.000000 . Let us now find the square root using long division method.

2 47

734 41 4 3 34 3 29

541

5 41 5 41 0

Similarly, let us find 3364 . 58

Thus, we have

73441 = 271 and

43.231

3364 = 58. 4

 21

2797 271 39  4 . 3364 58 58

Example 2: 2 2 25 52 5  a2  a   5  2      2     36 6 6  b  b   6

83 862

8643 86461

18 69 000000 16 269 249 2000 1724 27600 25929 167100 86461 80639

 1869  43.231 to 3 decimal places  1869  43.23 to 2 decimal places. www.betoppers.com

Square Roots & Cube Roots

23

Formative Worksheet 6.

18. If x  7  1  7  1  7  ...... , then find (x4 – 14x2 – x) 19. If n is a positive perfect square, then find the largest perfect square less than n 20. If a, b and c are three consecutive whole numbers, then find 1 + ac.

 225 25  16 The Value of  729  144   81 is  

(A)

1 48

(B)

5 48

(C)

5 16

(D)

1 16

21. Given, 7.

x 1  1 , then x is equal to 169 13 (A) 1 (B) 13 (C) 27 (D) 15

5 = 2.23607, then find the value of

3- 5

If 1 

.

2 + 7-3 5

22. Find the square root of 5  10  15 + 6 . 8.

What is the least number which should be subtracted from 0.000326 to make it a perfect square? (A) 0.000002 (B) 0.2078 (C) 0.02 (D) 0.04 9. The sum of 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 is (A) 64 (B) 85 (C) 81 (D) 100 10. The square root of 1471369 is (A) 1312 (B) 1211 (C) 1219 (D) 1213 11. The value of

9 1 is 16

23. If 1369  0.0615  x  37.25 , then find x. 24. A man arranges 15376 apple plants in his garden in an order,so that there are as many rows as there are apple plants in each row. Find the number of rows. 25. The product of three consecutive even numbers when divided by 8 is 720. Find the product of their square roots.

27.

2 2 1 3 (B) (C) 1 (D) 3 5 4 4 12. What is the square root of 0.0009? (A) 0.3 (B) 0.03 (C) 0.003 (D) 3 (A)

28.

13. Evaluate 175.2976 (A) 13.24 (B) 12.14 (C) 12.24 (D) 13.14 14. The value of

17 (A) 11

29.

0.289 is 0.00121 170 70 (B) (C) 11 11

30. (D)

70 110

31.

15. The value of

0.121 upto three places of

decimal is (A) 0.011 (B) 0.11 (C) 0.347 (D) 1.1 16. The least perfect square number divisible by 3, 4, 5, 6 and 8 is (A) 900 (B) 1200 (C) 2500 (D) 3600 17. If

2 x  6  x  4  5 , then find the value of x.

3 of 2872, then find x . 4 Find the smallest number by which 720 should be multiplied to get a perfect square number. i) What is the perfect square number so obtained? ii) Find the square root of this perfect square number. Find the smallest number by which 6000 should be divided to get a perfect square number. i) What is the perfect square number so obtained? ii) Find the square root of this number. What least number must be subtracted from 16160 to get a perfect square? Also find the square root of this perfect square. What least number must be added to 2945 to get a perfect square number? What is the resulting number? Find the square root of the resulting number. Find the greatest number of 5 digit which is a perfect square. Find the least number of six digits which is a perfect square.

26. If 28 x  1426 

32.

33. Find the value of

0.9 up to 3 places of decimal.

34. Find the value of 3 correct to 3 places of decimal. 35. Find the value of

3 up to four decimal places. 7

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8th Class Mathematics

24

1729 = 1728 + 1 = 123 + 13

Conceptive Worksheet 7.

1729 = 1000 + 729 = 103 + 93

If

x  x  x  .......  x x x....... , then find the value of ‘x’. If a = 0.1039, then find the value of

8.

4a 2  4 a +1 + 3a .

9.

Given,



2 = 1.414. If the value of



8 + 2 32  3 128 + 4 50 is ‘a’ and 1+ 2 1+ 2 1+ 2 1+ ... is ‘b’, then find a + b.

10. Find the greatest four-digit perfect square number. 11. Find the approximate value of 12. 13.

14.

15.

16. 17. 18.

3 12



2 21

. 2 28 98 Three-fifth of the square of a certain number is 126.16, then find the number. Find the smallest number by which 2592 be multiplied to get a perfect square number. i) What is the perfect square number so obtained? ii) What is the square root of the resulting number? Find the smallest number by which 1728 be divided to get a perfect square number. i) What is the perfect square number so obtained? ii) Find the square root of this number. Find the least number which must be added to 6203 to obtain a perfect square. Find the perfect square and its square root. Find the greatest number of six digits, which is a perfect square. Find the square root of this number. Find the square root of 0.00038809. Find the value of each of the following up to 3

1729 has since been known as the Hardy – Ramanujan Number, even though this feature of 1729 was known more than 300 years before Ramanujan. How did Ramanujan know this? Well, he loved numbers. All through his life, he experimented with numbers. He probably found numbers that were expressed as the sum of two squares and sum of two cubes also. There are many other interesting patterns of cubes. Let us learn about cubes, cube roots and many other interesting facts related to them. In the previous chapter, we have learnt about squares and square roots of the numbers. In this lesson we try to understand the cubes and cube roots.

5. Cube of a Number If a number is multiplied by itself, we say that the number is squared. In the similar manner, if a number is multiplied by itself three times, we say that the number is cubed. In other words, the square of a number when multiplied by the number itself gives the cube of the number. If n is the number, then cube of n is n × n × n (3 factors) and it is denoted by n3. Thus, the exponent of the cube of a number is 3. Number

Cube of the Number /Number Cubed

1

13 = 1

2

23 =8

3

33 = 27

4

43 = 64

5

53 =125

6

63 = 216

7

73 = 343

8

83 = 512

9

93 = 729

decimal places 7 . 19. Find the value of

4352 . 12393

Cubes & Cube Roots This is a story about one of India’s great mathematical geniuses, S. Ramanujan. Once another famous mathematician Prof. G.H. Hardy came to visit him in a taxi whose number was 1729. While talking to Ramanujan, Hardy described this number “a dull number”. Ramanujan quickly pointed out that 1729 was indeed interesting. He said,“It is the smallest number that can be expressed as a sum of two cubes in two different ways”. www.betoppers.com

The table in the last page gives the cubes of numbers from 1 to 9.

Square Roots & Cube Roots

Perfect Cube In the above table, 1, 8, 27, ..., 729 are called perfect cubes or perfect third powers of 1, 2, 3, ..., 9 respectively. A natural number is said to be a perfect cube if it is the cube of some natural number. If small numbers are given, we can identify whether it is a perfect cube or not. But if a larger number is given then it is difficult to do so. Hence, we need a method to check whether the number is a perfect cube or not. Test for a Perfect Cube Let us now have a look at the method : We know that if a prime p divides a perfect cube, then p3 also divides this perfect cube. Also in the prime factorisation of a perfect cube, every prime occurs 3 times or a multiples of 3 times. Thus, to check whether a number is a perfect cube or not, i) We first prime-factorize the given number. ii) Then group together triplets of the same prime factors. iii) If no factor is left out, the number is a perfect cube. Otherwise, it is not a perfect cube. Consider the following examples. Example 1: Is 64 a perfect cube? Let us find 64 is a perfect cube or not. Step-1: 64 = 2 × 2 × 2 × 2 × 2 × 2 (prime factorisation) Step-2: 64 = (2 × 2 × 2) × (2 × 2 × 2) (grouping together triplets of same prime factors) Step-3: Here, no prime factor is left out. So 64 is a perfect cube. Example 2: Is 392 a perfect cube ? 392 = 2 × 2 × 2 ×7 × 7 (prime factorisation) = (2 × 2 × 2) × 7 × 7 (grouping together triplets) 7 does not appear in a group of three. So 392 is not a perfect cube.

Cubes of Negative Numbers Let us consider the cubes of negative numbers: (–1) × (–1) × (–1) = (–1)3 = –13 = –1 (–3) × (–3) × (–3) = (–3)3 = –33 = –27 (–2) × (–2) × (–2) = (–2)3 = –23 = –8 (–5) × (–5) × (–5) = (–5)3 = –53 = –125 From the above examples, we can see that –1,

25 –27 , –8 and –125 are perfect cubes. Here, we can note an important idea about perfect squares and perfect cubes. We know from the previous chapter that a negative number cannot be a perfect square. But from the above examples we see that negative numbers may also be perfect cubes. That is, (–1) = ( –1 × –1 × –1) is a perfect cube. (–27) = ( –3 × –3 × –3) is a perfect cube. (–8) = ( –2 × –2 × –2) is a perfect cube. (–125) = ( –5 × –5 × –5) is a perfect cube. If smaller numbers are given, to find the cube of that number is easier as we can multiply them thrice orally. But finding cubes of two digit and three digit numbers is difficult as it involves a lot of calculation. Let us now look at the alternative method to do so.

Properties of Cubes of the Numbers Let us now understand the pattern of the units digits of the cubes of the numbers.

Units digit of the number (n)

Units digit of the cube (n3) 0, 1, 4, 5, 6, 9 0, 1, 4, 5, 6, 9 respectively 2 8 3 7 7 3 8 2 Numbers with units digits 0, 1, 4, 5, 6, 9, the units digits of their cubes are again the same digits 0, 1, 4, 5, 6, 9. Till now we have understood the idea of a cube and cube of a negative number and some other properties of cubes.

Formative Worksheet 36. When a number is divided by 25, its cube root is x. When it is multiplied by 5, then its cube root is y. If x + y = 36, then find the number. 37. Find the cube root of

b3 3b2 3b + + +1 . a3 a2 a

38. If 3,375 balls are arranged in a box in the form of a cube, then find the number of balls at one edge of the box. 39. If the cube of a number is 1331, then find the sum of factors of that number. www.betoppers.com

8th Class Mathematics

26

Step-2: Make groups in triplets of the same prime. Step-3: Find the product of the primes choosing one from each triplet. Step-4: The product from Step 3 is the required cube root of the given number. Example: Find the cube roots of 512 and 531441.

Conceptive Worksheet 20. If the product of two numbers x and y is a perfect cube, then show that 2[(x + y) 2 – (x – y2)] is a perfect cube. 21. Find the least number that 675 be multiplied to obtain a number which is a perfect cube. 22. Find the cube root of 22 × 32 × 42 × 62 × 82 × 92 × 6. 23. Find the number of perfect cube numbers in between 1 and 100. 24. The cube of a number is 8 times the cube of another number. If the sum of the cubes of the numbers is 243, then find the difference of the numbers.

6. Cube Roots

2

512

2

256

2

128

2

64

2

32

2

16

2

8

2

4 2

512 = (2 × 2 × 2) ×(2 × 2 × 2) ×(2 × 2 × 2)

Let n be an integer and a perfect cube. Then for some integer m, m3 = n in which case the number m is called the cube root of n.

3

512 = 2 × 2 × 2 = 8

Thus, the cube root of a number is that number which when multiplied by itself three times, gives the original number.



3

531441

3

177147

3

59049

3

19683

m

n = m3

3

6561

1

1

1

3

2187

2

8

2

3

729

3

27

3

3

243

4

64

4

3

81

5

125

5

3

27

6

216

6

3

9

7

343

7

8

512

8

9

729

9

3

n

10 1000 10 By looking at the above table, we observe that the cube roots of perfect cube of a number ending in 0, 1, 4, 5, 6 and 9 ends in 0, 1, 4, 5, 6 and 9, respectively. However, the cube of a number ending in 2 ends in 8 and vice versa. Similarly, the cube of a number ending in 3 or 7 ends in 7 or 3, respectively.

Finding Cube Roots of a Number Let us now have a look at various methods for finding the cube root of the given number. Prime Factorisation Method: Step-1: Express the given number as the product of primes. www.betoppers.com

3

512 = 8

3

531441 = (3 × 3 × 3) × (3 × 3 × 3) × (3 × 3 × 3) × (3 × 3 × 3) 3

531441  3  3  3  3  9  9  81



3

531441 = 81

Note: The cube root of the product of two perfect cubes is the product of their cube roots. For two perfect cubes x and y,

3

x y  3 x 3 y .

Example: i) ii)

3

9261  21  3  7  3 27  3 343  3 27  343 3

91125  45  5  9

 3 125  3 729  3 125  729

iii)

3

551368  82  2  41

 3 8  3 68921  3 8  68921

Square Roots & Cube Roots

27

Unit Digit Method: In this method, we can find cube root of a perfect cube by using units digit. Let us understand the steps to be used by finding the cube root of 2197. Step-1: Look at the digit in the units place of the perfect cube and determine the digit in the units place of the cube root. Units digit in 2197 is 7. Therefore, units digit in its cube root is 3. Step-2: Strike out from the right, last three (i.e., units, tens and hundreds) digits of the number. On striking out units, tens and hundreds digits

Let us observe the following examples: Example 1:

iii)

in 636 0 5 6 , the left out number is 636. The largest single digit number whose cube is less than 636 is 8 (  83 = 512 < 636)

3

3

1  3 512   1  8   8



3

512   8



3

3

125  3 1  125

1  3 125   1  5   5 

Example 3:

3

x×y=

3

3

m 

3

1  3 m   1 3 m   3 m

3

m   3 m Cube root of a negative perfect cube is negative.

Cube Roots of Rational Numbers The cube root of the quotient of two perfect cubes is the quotient of their cube roots. For any two perfect cubes x and y, y  0 , 3

1.

2.

x  y

3 3

x is a rational number. Example of y

Find the cube roots of 3

343 3 343   125 3 125

9261 = 3 27 × 343 =

3

ii)

3

91125 = 3 125 × 729 = 3 125 × 3 729

iii)

3

551368 = 3 8 × 68921 = 3 8 × 3 68921

3

343 . 125

777 5 5 5



7 5

2197 . 1331

2197  3 2197  3 13  13  13 13  3  3  1331 11 1331 11  11  11

Cube Root of Product of Integers Observe the following examples : 1.

3

125  64 

3

5  5  5  4  4  4  5  4  20 –––––– (1)

125  3 64  3 5  5  5  3 4  4  4 = 5 × 4 = 20 –––––– (2) From (1) and (2), we have 3

27 × 3 343

Cube Roots of Negative Numbers

3

Find the cube roots of 3

Examples: 3

125   5

 3 1  3 343   1  7   7 From the above examples, we can conclude that

x× 3 y

i)

3

343  3 1  343

3

636056 = 86. We have seen various methods of finding the cube roots of numbers. Let us now have a look at the method of finding cube roots of negative numbers. Cube Roots of Positive Numbers Let us now have a look at the method of finding cube roots of positive numbers. The cube root of the product of two perfect cubes is the product of their cube roots. Note: For two perfect cubes x and y,



512  3 1  512



Example 2:

in 21 9 7 ,the left out number is 2. Note: If nothing is left, we stop and the digit in step 1 is the cube root. Step-3: Consider the number left out from step 2. Find the largest single digit number whose cube is less than or equal to the left out number. This is the tens digit of the cube root. The largest single digit number whose cube is less than 2 is 1 (  13 = 1 < 2)

 3 2197 = 13 Let us see one more example. Find the cube root of 636056. i) Units digit in 636056 is 6.  units digit in its cube root is 6. ii) On striking out units, tens and hundreds digits

3

2.

3

125  64  3 125  3 64 .

3

216  ( 343)  3 6  6  6  (7)  ( 7)  ( 7)

= 6 × (–7) = –42 ------- (1) 3

216  3 ( 343)

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8th Class Mathematics

28

Conceptive Worksheet

 3 6  6  6  3 ( 7)  (7)  (7)

= 6 × (–7) = –42 From (1) and (2), we have 3

3

---------- (2)

of

216  ( 314)  216  (343) 3

From the above examples, we can conclude that for any two integers a and b,

3

ab  3 a  3 b .

40. What least number by which 13720 must be divided so that the quotient is a percfect cube? (A) 2 (B) 3 (C) 5 (D) 6

5832 =? (A) 22 (B) 18

(A) 

3

(C) 16

3

N

3

N

3

N .

29. a, b, c, d, e are the digits in a 5 digit number abcde, where a = c – 7, b = c – 5, d = c – 6,

1. (C) 

3 4

(D) 

12 17

3

3

156  x   12, then the value of x is

2.

3.

4. 5.

8  3 27  3 343

46. The vlaue of

 2

2

3

is

(A) 7 (B) –8 (C) 8 (D) –5 3 47. Calculate the value of (-0.4) (A) 0.640 (B) 0.064 (C) –0.064 (D) –0.640 48. Find the cube root of 49. If x =

3

a,y=

where a = 3

 2001

3

1728 2744

b,z=

4096 , b =

x+ y+z

3

6.

7.

c,

64 , c =

729 , then find

8.

.

50. If two perfect cubes differ by 271 and their cube roots are consecutive natural numbers, then find the sum of the two perfect cubes. 3

51. If 681472 = 2a × 11b , where a  x , b  3 y , x = 729 and y = 27 then find the value of 3

( a  b )3  3 ( a  b )3 .

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abcde .

Find the square of

3 ii) 0.17 7 Using prime factorization method, find which of the following are perfect square numbers i) 3528 ii) 11025. Using prime factorization method, find the square root of each of the following numbers: i) 7056 ii) 30625 Find the smallest number by which 588 be multiplied to get a perfect square number. Find the smallest number by which 7776 be divided to get a perfect square number. i) What is the resulting number? ii) What is the square root of the number so obtained? Find the square root of each of the following by division method. i) 4401604 ii) 9653449 Find the least number which must be subtracted from 7581 to obtain a perfect square. Find this perfect square and its square root. Find the least number which must be added to 506900 to make it a perfect square. Find this perfect square and its square root. Find the square root of i) 5.4289 ii) 176.252176 i)

(A) 1570 (B) 1572 (C) 1560 (D) 1512 3

3

Summative Worksheet

1728 2744

6 6 (B)  11 7

e = c – 4, c = 8, then find

(D) 14

0.000064 equals (A) 0.04 (B) 0.4 (C) 0.004 (D) 0.02 44. 3 144  3 12 equals (A) 12 (B) 14 (C)13 (D) 6 45.

28. If N > 1, then find

3

42. Evaluate:

43.

512  3 3.375 . (A) 12 (B) 9.5 (C) 8 (D) 1.5 3 26. Given that x  6 , then x = (A) 216 (B) 18 (C) –18 (D) –216 3

27. The vlaue of (27×2744)1/3 is (A) –40 (B) 42 (C) –22 (D) –32

Formative Worksheet 41.

25. Given that 512=83 and 3.375=1.53, find the value

9.

10. Find the value of up to 3 decimal places 0.4 . 11. Find the value of each of the following, correct to 2 places of decimal

0.121 .

12. Find the square root of 42

583 . 1369

Square Roots & Cube Roots 13. Find the value of

29

3 , correct to 2 places of 8

decimal. 14. Find the cubes of the following :

3 5 Show that 648 is not a perfect cube. Find the smallest number by which 11979 must be multiplied so that the product is a perfect cube. Find the smallest number by which 8788 must be divided so that the quotient is a perfect cube. Find the cube root of : i) 30

15. 16. 17. 18.

ii) 1.2

iii) 1

508 i) 4 ii) 0.000110592 1331 19. Find the least number of four digits which is a perfect square. 20. Find the cube root of 4.096.

HOTS Worksheet 1.

If x =

2 2 2 2.... and y =

4.

2  1.414 , then find the nearest square root of

2 1

. 2 1 11. Find the value of: 3 5 7 9 11 13 + 2 2+ 2 2+ 2 2+ 2 2+ 2 2 2 1 .2 2 .3 3 .4 4 .5 5 .6 6 .7 2

+

15 17 19 + 2 2+ 2 2 2 2 7 .8 8 .9 9 .10

12. Consider the following statements : i) If a is a natural number, then

a is a rational

number. ii) If p and q are natural numbers, then p 2  2q 2 . iii) If p and q are perfect squares, then

p is a q

rational number. iv) The square root of a prime number is not a rational number.

3 3 3 3.... ,

then the true statements in the following are I) x + y  5 II) x2 + y2 = 6xy 2 2 III) x y = 6xy 1) I only 2) III only 3) I and III 4) None of these 2. 3.

10. If

Find the last digit in the expansion of 999 . If a, b and c are the numbers of a Pythagorean triplet and a < b < c, then find ab. Find the number of pairs (x, y) of positive integers

Which of the above statement/s is/are correct? 1) (i) and (ii)

2) (ii) and (iii)

3) (iii) only

4) (iii) and (iv)

13. If

x+a + y+a =

 x + y + 2a  , then find the

value of a. 14. If a number is both perfect square and perfect cube, then all the exponents in its prime factorization are 1) Multiples of 5 2) Multiples of 7 3) Multiples of 4

4) Multiples of 6

2

such that 5.

x  y  1980 .

If m is any positive integer, then find the possible value of    m  m  m  ........  m  m  m........   

6.

7.

If ‘x’ is a positive integer, then x  x among which of the following cannot be possible 1) 20 2) 30 3) 60 4) 90 If ‘a’ and ‘b’ are the perfect squares between 50

a +b (up to two decimals). a b Find the smallest positive integer n such that and 100, then find

8.

15. If n is a perfect cube, then the correct statement is 1) n is odd

2) n is even

3

3) n is a perfect square 4) n is a perfect cube 16. If all the exponents in prime factorisation of a number are multiples of 6, then the cube root of the number must be 1) Irrational

2) A perfect square

3) A perfect cube

4) Both B and C

17. From a two-digit number N, subtract the number with the digits reversed such that the result is a positive perfect cube, then find all such two digit numbers. 18. If a, b differ by ‘1’, then show that a2 + ab + b2 is difference of the cubes of a, b.

n  n  1 < 0.01. 9.

If x is a positive integer, then x2 + x +1 among which of the following cannot be equal to 1) 85 2) 631 3) 261 4) 1296

19. 8 3 a m + 27 3 b m + 64 3 c m = p, where a = 10, b = 20, c = 30, m = 3, then find the number added to make it a perfect cube. www.betoppers.com

30 20. If a = b + c = c + d = d + e = 24, where b = 12, then find the value of 3

 a ×b× c× d × e 

a+b+c+d+e

2. 3.

15. If 1  1) 1

4.

Evaluate : 3

3) 8

2) 5

3) 6

80  6 5 ? 1) 13.41 2) 20.46 3) 21.66 4) 22.35 .081 .484 is equal to : .0064  6.25 1) 0.9 2) 0.99 3) 9 18. The value of

If x * y  x  y  xy , the value of 6*24 is :

6.

1) 41 2) 42 3) 43 4) 44 What is the square root of 0.16? 1) 0.004 2) 0.04 3) 0.4 4) 4

7.

17.

1 )

0.00004761 equals : 1) 0.00069 2) 0.0069 3) 0.0609 4) 0.069

8.

1.52  0.0225  ? 1) 0.0375 2) 0.3375 3) 3.275 4) 32.75

9.

1.5625  ? 1) 1.05 2) 1.25

3) 1.45

1 in the equation 2  ? ? 32

1 4) None of these 2 12. What should come in place of both the question 2) 7

1) 12

2) 14

?

128 3) 144



3  729 , then the value of n is : 1) 6 2) 8 3) 10 4) 12

13. If

n

1) 0.02

1) 0.6 22.

0.16 is : 0.4 2) 0.2 3) 0.63

If

1  0.01

is close to : 1  0.1 2) 1.1 3) 1.6 4) 1.7

5  2.236 , then the value of

24.

 3 2  4 3 6     ? 6 2 8  12   6 3

1) 25.

3 2

2)

3 2

3+ 6 5 3  2 12  32 + 50 1)3

2) 3 2

3) 6

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4) None of these

5 10   125 is equal to : 2 5 1) 5.59 2) 7.826 3) 8.944 4) 10.062 23. The least perfect square, which is divisible by each of 21, 36 and 66, is : 1) 213444 2) 214344 3) 214434 4) 231444

162 . ? 4) 196

is :

20. The value of

3) 7

marks in the equation

2

4 a 2  4 a  1  3a is : 1) 0.1039 2) 0.2078 3) 1.1039 4) 2.1039

4) 1.55

4

2

1) 0.1 2) 10 3) 102 4) 103 19. If a = 0.1039, then the value of

21. The value of

52 169  10. If , the value of x is : x 289 1) 52 2) 58 3) 62 4) 68 11. Which number can replace both the question marks

1) 1

2

4) 99

 0.03   0.21   0.065 2 2 2  0.003   0.021   0.0065

4) 6.4

5.

55 x 1 , then the value of x is : 729 27 2) 3 3) 5 4) 7

of

4) 10

41  21  19  9

4) 1444

16. If 3 5  125  17.88 , then what will be the value

10  25  108  154  225 is :

2) 6

3 of 2872 4 2) 676 3) 1296

28 ?  1426  1) 576

53824 = ? 1) 202 2) 232 3) 242 4) 332 The square root of 64009 is : 1) 253 2) 347 3) 363 4) 803 The value of

1) 4

14.

.

IIT JEE Worksheet 1.

8th Class Mathematics

3) 5 3

4) 1

=? 4) None of these

By the end of this chapter, you will understand •

Ratio

•

Proportion

•

Properties of Ratio and Proportions

•

Variations

1. Introduction Ratios are one of those ideas, that go too far back to be sure where they originated. It is rather profitless to speculate the domain in which the concept of ratio, first appeared. The idea that one tribe is twice as large as another and the idea that one leather strap is only half as long as another, involve the notion of ratio. Both are, such as would develop early in the history of the race, and yet, one has to do with ratio of numbers and the other with the ratio of geometric magnitudes. Indeed, when we come to the Greek writers, we find Nicomachus, including ratio in his arithmetic, Eudoxus in his geometry, and then of Smyrna in his chapter on music. Daily in the morning after you brush your teeth and freshen up, you normally have either coffee or tea. These are made with different contents in them. What are these contents in terms of their quantities, mathematically called ? Could you guess ? If not, don’t worry. Please read about the same hereunder, for knowing about them and their importance. The importance of Ratio and Proportion Who doesn’t like Hyderabadi Biryani? Can your mother prepare delicious and mouth- watering Biryani, only with Basmathi Rice? Obviously, no. As, she has to add different ingredients to make it tasty and unforgettable. These ingredients should be added in the required quantities (or) proportions and cooked. Then only the tasty Biryani can be enjoyed.

Chapter -3

Learning Outcomes

Ratio Proportions & Variations

2. Ratio Suppose your age is 13 years and your father’s age is 39 years. then,

Your age 13 1   . Your father's age 39 3

This is written as, Your age : Father’s age = 1 : 3. This can also be written as 1 : 3. The symbol “:” stands for “is to”. Similarly, if there are 30 pizzas and 10 burgers in a

no. of pizzas 30 3 box, no. of burgers  10  1 . We say that the pizzas and burgers in the box are in the ratio 3 : 1. Therefore, the ratio is the fraction between two similar quantities, which has same units.

Definition If a and b are two non-zero quantities of the same kind and in the same units, then the

a is called the ratio between a and b, b written as a : b (read as a is to b). Example: The cost of an object A is Rs. 20 and the cost of object B is Rs. 5, then the ratio of cost of object A to object B is : fraction

cost of A 20 4   (or) 4 : 1 cost of B 5 1

Points to remember i) ii)

Before finding the ratio, make sure that the quantities given are in the same units. In the ratio a : b, a is the Antecedent and b is the Consequent.

8th Class Mathematics

32 iii) iv)

Since the ratio a:b is a fraction, all the rules of operations on fractions are also applied on ratios. Always, the ratio should be in the lowest terms. A ratio a : b is said to be in the lowest terms, if the H.C.F of a and b is 1.

Formative Worksheet 1.

2.

3.

4. 5. 6. 7.

8.

Suppose the age of Hary potter is 15 years and the age of Albus Dumbledore (Philosopher) is 75 years. How are x and y related age wise? Laurel and Hardy are the most famous comedian pair of Hollywood, Laurel is carrying a load of 3kg and Hardy being fat, carrying 30,000g. What is the ratio of the loads carried by Laurel and Hardy? An office opens at 10.00 A.M. and closes at 6.00 P.M., with lunch interval of 30 minutes. What is the ratio of lunch interval to the total period in offices ? If (2a + 3b) : (3a + 5b) = 18 : 29, then find a : b. If x : y = 2 : 3, then find the value of (3x + 2y) : (2x + 5y). If (6a2 – ab) : (2ab – b2) = 6 : 1, then find a : b. Two numbers are in the ratio of 2 : 3 and the differences of their squares is 320. Find the numbers. The ratio of two numbers is 3 : 4 and their sum is 420. Find the greater of the two numbers.

a b 3a  3b   , then find the value of ‘k’. 7 3 k 10. What number must be added to each term of the ratio 4 : 5, so that it becomes 5 : 6 ? 9.

If

1 1 1 : : and 2 3 4 its perimeter is 104cm. Find the length of the longest side. 12. If a + b + c = 0, then prove that 11. The sides of a triangle are in the ratio

 b  c   c  a   a  b   a    b    c         b c   a  b  c  c  a  a  b   9 13. Divide Rs.610 among Sarita nand Rajan in the ratio

1 5 3 :1 . 4 6 14. Divide Rs. 612 into three parts such that the first

3 part is of the second part and the ratio between 4 second and third parts is 5 : 4. 15. If A : B = 8 : 15, B : C = 5 : 8 and C : D = 4 : 5, then find A : D. www.betoppers.com

Conceptive Worksheet 1. 2.

If 2A = 3B = 4C, then find A : B : C. If 2A = 3B and 4B = 5C, then find A : C.

3.

If A 

4. 5.

B C and B  , then find A : B : C. 3 2 If x : y = 5 : 2, then find the value of 8x + 9y: 8x + 2y If (4x2 – 3y2) : (2x2 + 5y2) = 12 : 19, then find x: y.

4 x  3 y 11  , then find x : y. 6 x  5 y 17 7. If x : y = 3:5, then find the ratio (8x + 3y) : (5x + 2y). 8. Two numbers are in the ratio of 8 : 5 and the difference of their squares is 156. Find the numbers. 9. Find the ratio a : b : c ; if a : b = 2 : 3; b : c = 4 : 5. Also find the ratio a : c. 10. The lengths of sides of a right triangle are in the ratio 3 : 4. If the perimeter of the triangle is 24cm, then find the lengths of the sides of the triangle. 11. The work done by (a + 7) workmen in (a – 8) days and the work done by (a – 8) workmen in (a + 9) days are in the ratio 5 : 6. Find the value of a. 12. If a : b = 3 : 7; b : c = 6 : 11, then find a : b : c.

6.

If

Inequalities in Ratio If a : b is the given ratio, then a is antecedent and b is consequent. i) The ratio a : b is said to be a ratio of greater inequality, if a > b. Example: Consider the ratio 6 : 5. As 6 > 5, the ratio

6  1 . Hence, it is the ratio of 5

greater inequality. ii) The ratio a : b is said to be a ratio of lesser inequality, if a < b. Example: Consider the ratio 4 : 5. As 4 < 5, the ratio

4  1 . Hence, it is the ratio of 5

lesser inequality. iii)The ratio a : b is said to be a ratio of equality, if a = b. The ratio a : b is always 1 : 1

Comparison of Ratios To compare two ratios we follow either of the two procedures given below: i) Convert the consequent of each ratio to the same value and compare the antecedent.

Ratio, Proportions and Variations

2 4 Example: 2 : 3  ; 4 : 5  3 5

33 1)

L.C.M. of denominators 3 and 5 is 15.

2 2  5 10 4 4  3 12   ;   3 3  5 15 5 5  3 15 12 10   4:5  2:3 15 15 ii) Convert the antecedent of each ratio to the same value and compare the consequent.  The L.C.M. of numerators 2 and 4 is 4 Example:

a× c× e× g.... . b× d × f × h....

Example: The compound ratio of ratios 2 : 3, 5 : 6 and 1 : 7 is

2 2  2 4 4 4 1 4   ;   3 3 2 6 5 5 1 5

4 4 4 2  i.e.,   4 : 5  2 : 3 5 6 5 3

2)

Formative Worksheet 16. Show that the ratio of less inequality is increased, by adding the same quantity to both of its terms. 17. When the same quantity is subtracted from both the same terms of the given ratio, compare the initial and final ratio of greater inequality ratio. 18. Divide Rs 2420 among A, B, C in the ratio

3)

a: b.

Example: The sub-duplicate ratio of 25 : 49 is

3 2 1 19. If a : b  1 : 2 and b : c  4 : 3 , find: 4 3 4 and

2  5  1 10  3  6  7 126 Duplicate Ratio The duplicate ratio of a : b is the ratio of the squares of a and b i.e., a2 : b2. Example: The duplicate ratio of 2 : 3 is 22 : 32 = 4 : 9 Sub - Duplicate Ratio The sub-duplicate ratio of a given ratio is equal to the ratio of square roots of the antecedent and the consequent of the given ratio. The sub-duplicate ratio of a : b is

1 1 5 1 :1 : . 3 2 6

i) a : b : c

Compound Ratio The compound ratio of two ratios a : b and c : d is defined as ac : bd. In other words, compound ratio is the ratio of the product of antecedents to the product of consequents of the given ratios. Similarly, compound ratio of a : b, c : d, e : f, g : h.... is a × c × e × g × ... : b × d × f × h × .... or

4)

ii) a : c

Conceptive Worksheet 13. Show that the ratio of greater inequality will diminish, when we add the same quantity, to both of its terms. 14. Which is greater, 4 : 5 or 7 : 8? 15. Which ratio is greater 7:3, 2:9, 4:5 16. Write in descending order 4:5, 6:7

Types of Ratios 1) Compound Ratio 2) Duplicate Ratio 3) Triplicate Ratio 4) Sub-duplicate Ratio 5) Sub-triplicate Ratio 6) Reciprocal Ratio

5)

25 : 49  5 : 7 . Triplicate Ratio The triplicate ratio of a given ratio is the ratio of the cubes of the antecedent and the consequent of the given ratio. The triplicate ratio of a : b is a3 : b3 Example: The triplicate ratio of 2 : 3 is 23 : 33 = 8 : 27. Sub - Triplicate Ratio The sub-triplicate ratio of a given ratio is the ratio of the cube roots of the antecedent and the consequent of the given ratio. The sub-triplicate ratio of a : b is 1

1

a : 3 b or a 3 : b 3 . Example: The sub-triplicate ratio of 27 : 125 is 3

3

27 : 3 125  3: 5 . www.betoppers.com

8th Class Mathematics

34 6)

Reciprocal Ratio

1 1 The reciprocal ratio of a : b is : , which is a b same as b : a. Example: The reciprocal ratio of 1 1 : 16 : 25 is or 25 : 16. 16 25 Commensurable Ratios If the ratio of any two quantities can be expressed exactly by the ratio of two integers, the quantities are said to be commensurable. Otherwise, they are incommensurable. Example: i) 3 : 5 is commensurable ratio. ii) 2 :1 and ratios.

3 : 5 are incommensurable

ii) x : y , y : z and z : p iii)2x : 3y, a : 2b, 3p : 4q

a : b2

22. Find the duplicate ratio of the following:

 p  q

ii) 3 : 5

7

3

: p  q

7

3

ii) a

1

3

2

b 3 :a

2

3

b

1

3

27. Find the sub-triplicate ratio of the sub-duplicate ratio of 4096a6 : 729b12. 28. Check whether the ratio of 1.25 litres to 500ml is commensurable.

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21. 22. 23. 24.

1 9

ii)

a b : 3 5

a c  . b d By cross multiplication we have, ad = bc i.e., the product of extremes = the product of means. Conversely, if ad = bc  (The product of means = The product of Since a : b = c : d, we have

23. If (x + 8) : 2 (x + 16) is the duplicate ratio of 5 : 8, then find x. 24. Find the sub-duplicate ratio of i) 16p4 x6 : 9q6y4 ii) 9p2 : 16q4 iii) 729 x12 : 4096x6 25. If (2a – x):(b – 2x) is the sub-duplicate ratio of a : b, then show that x2 = ab. 26. Find the triplicate ratio of, i)

20.

3: 6 . If (x + 2) : (3x + 4) is the duplicate ratio of 6 : 10, then find x. If p : q with p  q is the duplicate ratio of p + r : q + r, then show that r2 = pq. Find the sub-duplicate ratio of i) 9x2a4 : 25y6 ii) 64a6 : 729x4 Find the sub triplicates ratio of i) 64a3 : 27b6 ii) a2 : b2 Find the reciprocal ratio of

Four quantities a, b, c, d are said to be in proportion, if a : b = c : d. The term ‘a’ and ‘d’ are called extremes (end terms) and ‘b’ and ‘c’ are called means (middle terms). If the four quantities a, b, c, d are in proportion, then the product of the extremes is equal to the product of the means. is represented as a : b :: c : d

2 :1,3: 5, 20 : 9

i) 2p : 4q

19. Find the duplicate ratio of

3. Proportion

20. Find the compound ratio of the following:

21. Find the duplicate ratio of

17. Find the compound ratio of 2 : 3 and 3 : 7 18. Find the compound ratio of (p + x) : (p – x), (p2 + x2) : (p + x)2 and (p2 – x2)2 : (p4 – x4).

i) 4 :

Formative Worksheet i)

Conceptive Worksheet

a c  b d i.e., a, b, c and d in proportion. extremes), then

Continued Proportion: a b  . b c Similarly, a, b, c, d are said to be in continued proportion. a, b, c are in continued proportion if

a b c   a, b, c, d are called proportionals, b c d where, a = First proportional, b = Second proportional, c = Third proportional, d = Fourth proportional. In

Ratio, Proportions and Variations

35

Mean Proportional:

1 1 2   , then prove that a d c

If a, b, c are in continued proportion, a : b = b : c

39. If a + c = 2b and

a b =  b 2  ac  b  ac b c Here b is called the mean proportional of a and c.

a : b = c : d. 40. Three numbers are in continued proportion of which the middle one is 16 and the sum of the other two is 130. Find the numbers.

i.e.,

Conceptive Worksheet

Formative Worksheet 29. i) Are 40, 30, 60, 45 in proportion? ii) Are the numbers 5, 6, 20, 18 in proportion? 30. If 2, 4, 6 and x are in proportion, then find the value of x. 31. If x, 2x,y and y2 are in proportion, then find the values of x and y. 32. If a : b = c : d and e : f = g : h, then show that ae : bf = cg : dh. 33. Find the mean proportional between 7 and 63. 34. Find the third proportional of 8 and 24. 35. Find the third proportional to (p 2 –q 2 ) and (p + q). 36. Find the fourth proportional of 1,2,3. 37 If a, b, c, d are in continued proportion, then prove that, 2a + 3d : 3a – 4d = 2a3 + 3b3 : 3a3 – 4b3. 38. Find the four proportionals, such that the sum of the extremes is 21, the sum of the means is 19 and the sum of the squares of all the four numbers is 442.

25. Find the mean proportional between 13 and 637. 26. Find the third proportional of 16 and 4. 27. Find third proportional to the numbers i) 1.2,1.8 ii) 225,75. 28. Find fourth proportional to the numbers 6,8 and 9. 29. Find the mean proportional of 16 and 4, if the three are in continued proportion. 30. If 16, 8 are in continued proportion, then find the third proportional of 16 and 8.

q+r p+q ;r  and q is the mean 2 2 proportional between ‘p’ and ‘r’, then prove that

31. If

p

1 1 2   . p r q

32. Find the mean proportional between

x2 p and 2 . qy 4 pq

4. Properties of Ratio and Proportions S.No

Property Name

1 Invertendo

Property If a : b = c : d, then b : a = d : c.

Example 21 : 14 = 15 : 10 14 : 21 = 10 : 15

Proof a c a c = (given), 1   1  b d b d (Unity is divided by given ratio’s) As

 If a : b = c : d, then a : c = b : d.

2

21 : 14 = 15 : 10 21 : 15 = 14 : 10

Alternendo

b d =  b: a=d :c a c

a c = (given) b d  ad = bc (By cross-multiplication). Dividing both sides by cd, we have As we have

ad bc a b =  =  a : c= b : d cd cd c d 3

Componendo

If a : b = c : d, then a+ b c+ d = b d

21 : 14 = 15 : 10 21  14 35 5 i)   14 14 2 15  10 25 5 ii)   10 10 2 From the above, the result follows.

We know that a : b = c : d a b a c =  1  1 c d b d (Adding 1 to both sides)

i.e.,

a+ b c+ d = ––––––– (A) b d a+b :b=c +d :d This is called componendo. 

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8th Class Mathematics

36 Property Name

S.No 4

Dividendo

Property

Example

Proof

If a : b = c : d, then a - b c- d = b d

Let the ratios be 21 : 14 and 15 : 10 21  14 7 1 i)   14 14 2 15  10 5 1 ii)   10 10 2 From the above, the result follows.

If a : b = c : d Let the ratios be 2 : 8, 4 : 16 , then By componendo and Dividendo, a+ b c+ d 2  8 10 5  = i)   a - b c- d 2  8 6 3

5

Componendo Dividendo

4 16 20 5   4 16 12 3 From the above, the result follows. Let the ratios be 21 : 14 and 15 : 10 21 21 3 i)    3 :1 21  14 7 1 15 15 3 ii)    3 :1 15  10 5 1 From the above, the result follows. ii)

If a : b = c : d, a: a–b = c : c – d.

6

Convertendo

Formative Worksheet

a c a c =  1  1 . b d b d (Subtracting 1 from both sides)

ii) We know that

a - b c- d = ––––––––– (B) b d a –b: b=c –d: d From the above, dividing i) by ii), we get a+ b c+ d a+ b c+ d  b = d  = a- b c -d a- b c - d b d The above two results are called componendo and dividendo. If a : b = c : d then a : (a – b) = c : (c – d) a c a : b = c : d  = ________ (1) b d a c a - b c- d ____ or  1  1 or = (2) b d b d Dividing (1) by the corresponding si des a c a c b d of (2) = or = a-b c -d a -b c -d b d  a :  a - b = c c - d 

Conceptive Worksheet

x  x  2 x  2x  3  2x 2x  3 42. If (3a + 6b + c + 2d) : (3a – 6b + c – 2d) = (3a + 6b – c – 2d) : (3a – 6b – c +2d), then prove that a,b,c,d are in proportion.

33. Show that a, b, c, d are in proportion, if 6a + 7b : 6c + 7d = 6a – 7b : 6c – 7d. 34. What quantity must be added to the terms of the ratio (p + q) : (p – q) to make it equal to (p + q)2 : (p – q)2?

9 x  7 y 9 a  7b 43. If  , then show that x : y = a: b. 9 x  7 y 9 a  7b 44. Prove that (2p + q + 4r + 2s) : (2p – q + 4r – 2s) : : (2p + q – 4r – 2s) : (2p – q – 4r + 2s), if p : q = r : s. 45. Solve the equation using componendo and dividendo

35. If a 

2

2

41. Solve the equation

a  1  a  1 4a  1  2 a  1  a 1 46. If

x

4 6 2 3

, then find the value of

x2 2 x2 3  x2 2 x2 3

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2 xy

, then find the value of

x+ y  a+ x a+ y      a-x a- y 

p q r 2 p  3q  5r    , then find ‘k’. 2 3 4 k 37. If (4p + 9q) : (4r + 9s) = (4p – 9q) : (4r – 9s), then prove that p : q = r : s. 36. If

K-Method: It is a method introduced to solve some types of problems related to ratio and proportions. In this method we equate ratios to ‘k’ and start solving the problem as follows.

Ratio, Proportions and Variations

37

Formative Worksheet

5. Variation

a c 47. If = , then prove that (a2 + ac + c2 ): b d (a2– ac + c2 ) = (b2 + bd + d2) : (b2– bd + d2). 48. If p, q, r, s are in proportion, then show that  1 1   1 1   p - q  p - r        . pqr  p s q r

49. If

x y z = = , then prove that p q r

x 2  p2 y 2  q2 z 2  r 2   x p yq zr 2

 x  y  z   p  q  r  x  y  z  p  q  r 50. If

2

a c e   , then prove that each of these ratios b d f 1

If the two quantities A and B are related to each other the information how A changes with the change in B is conveyed by the term variation.

Direct Variation One quantity A is said to vary directly as another B, when the two quantities depend upon each other in such a manner that if B is changed, A is changed in the same ratio. It’s written as A  B and A = k(B) where k is a constant. For instance: If a train moving at a uniform rate travels 40 kilometers in 60 minutes, it will travel 20 kilometres in 30 minutes, 80 kilometres in 120 minutes, and so on; the distance in each case being increased or diminished in the same ratio as the time. this is expressed by saying that when the velocity is constant the distance is proportional to the time. or more briefly, the distance varies as the time.

Inverse Variation

 k a n  k 2c n  k3e n  n is equal to  1 n where k1, k2, k3 n n   k1b  k2 d  k3 f 

One quantity A is said to vary inversely as another B when A varies directly as the reciprocal of B;

are constants.

written as A 

51. If

x y z   , then prove that a b c

x3 y 3 z 3 3xyz    . a 3 b3 c 3 abc

Conceptive Worksheet 38. If p : q = r : s, prove that p + q : r + s =

p2 + q 2 : r 2 + s 2

a c e 39. If = = , then prove that b d f a+c - e a+c+e = . b+ d - f b+ d + f 40. If a : b = c : d, then prove that ab + cd : ab – cd = a2 + c2 : a2 – c2. 41. The ratio of the sum of money in three bags A, B and C is 4 : 3 : 2. If Rs. 50 is added to each of the bags, the ratio becomes 14 : 13 : 12. Find the sum of money in each of the bags. 42. If a, b, c, d are in continued proportion, then prove that a : d : :(ma3 + nb3 – rc3) : (mb3 + nc3 – rd3).

43. If

x y  , then (x + 5) : (y + 8) is equal to: 5 8

1 and 1 (Where k is a A  k B B

constant). For instance: If 6 men do a certain work in 8 hours, 12 men would do the same work in 4 hours, 2 men in 24 hours and so on. Thus, it indicates when the number of men is increased, the time is proportionately decreased and vice versa.

Formative Worksheet 52. If x varies as y and x = 8 when y = 15, find x, when y = 10. 53. If the square of x varies as cube of y and x = 3 when y = 4, find the value of y, when x 

1 . 3

54. P varies directly as Q and inversely as R, also

P

2 3 9 when Q  and R  ; find Q when 3 7 14

P  48 and R  75 . 55. If x varies as y, prove that x 2 + y 2 varies x2 – y2.

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8th Class Mathematics

38

Conceptive Worksheet 44. If x  y and y = 7 when x = 18. Find x when y = 21. 45. If x  y , and y = 3 when x = 2, find y when x = 18. 46. If x 

1 and y = 4 where x = 15, find y where y

12. If A : B be in the duplicate ratio of A + x : B + x, prove that x2 = AB. 13. If (p+ q + r) (p – q + r) = p2 + q2 + r2, then show that p, q, r are in continued proportion. 14.. If p 

4ab , then find the value of a +b

p  2a p  2b  . p  2 a p  2b

x = 6.

1 47. If y  and y = 1 when x = 1, find x when y = 5 x 48. If  3a  7b    3a  13b  , and when a = 5, b = 3 find the equation between a & b.

1 1 and y  prove that z  x z y 50. If the square root of a varies as the cube root of b, and if a = 4 when b = 8, Find equation between a and b. 49. If x 

Summative Worksheet 1. 2. 3.

4.

5. 6.

7.

a b c a +b + c   , then find . 3 4 7 c What number must be added to each term of the ratio 7:13 so that the ratio becomes 2:3 ? The ratio between two numbers is 3:4 and their L.C.M. is 180. Find the smallest of the two numbers. Two trains travel, respectively 448 km in 8 hours and 504 km in 12 hours. Find the ratio of their speeds. If

2 p  3q 2r  3q p r  = , then prove that . 2 p  3q 2r  3q q s The length of a steel tape for measurement of buildings is 10m and its width is 2.4cm. What is the ratio of its length to its width ? The ratio between two numbers is 6 : 13 and their L.C.M. is 312. Find the numbers.

If

1 1 If a : b  6 : 4 ;b : c = 8.5 : 4.5,then find a:c 2 2 9. If (3a + 3) : (9a + 7) is the duplicate ratio of 3 : 5 then find a. 10. Two numbers are in the ratio of 5 : 8. If 9 be added to each they are in the ratio of 8 : 11. Find the numbers. 11. (a + 4), (a + 12), (a – 1), (a + 5) are in proportion. Find the value of ‘a’. 8.

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15. If a 

y 2  xy  y 2  xy y 2  xy  y 2  xy

, then show that

xa2 – 2ya + x = 0. 16. If

x y z = = , then prove that q+r - p r+ p - q p+q - r

x + y + z x y + z+ y  z + x+ z  x + y = p+q+r 2  px + qy + rz  17. Using the properties of proportion solve a:

a  7  a 1 2  . a  7  a 1 1

18. Find the third proportional to 5  2 3 and

37  20 3 . 19. If

x+ y y+ z z+ x = = , then prove that px + qy py + qz pz + qx

each ratio is equal to

2 , provided p+q

x + y + z  0. 20. If a, b, c are three proportionals, then show that (b2 + bc + c2) (ac – bc + c2) = b4 + ac3 + c4. 21. If a : b = c : d = e : f, then prove that (5a + 7c + 3e) : (5b + 7d + 3f) = c : d. 22. Find three numbers in the ratio of 3:2:5 such that the sum of their squares is equal to 1862. 23. If A : B = 3 : 4, B : C = 8 : 9 and C : D = 15 : 16, then find A : D. 24. 30 men do a piece of work in 27 days. In what time 18 men can do another piece of work 3 times as great?

p q r = = , then show that b-c c - a a -b p + q + r = 0.

25. If

Ratio, Proportions and Variations

HOTS Worksheet 1.

The ratio of the number of apples to that of mangoes in a basket is 5 : 4. If the number of apples is 210, then find the number of mangoes in the basket.

2.

There are 120 coins in a purse containing one rupee coins and ten paise coins. If the value of the coins in the purse is Rs.57, then find the number of one rupee coins in the purse.

3.

In a factory, the ratio of male workers to female workers was 5:3. If the number of female workers was less by 40, what was the total number of workers in the factory ?

4.

If (2x + 3y) : (3x + 4y) = 13 : 18, then find x : y.

5.

If (p – x) : (q – x) is the duplicate ratio of p : q, then 1 1 1 show that = + . x p q

39 12. If

a 3b  2c 2e  3ae 2 f ace  b 4  2d 2 f  3bf 3 bdf 13. If

If

7.

ay - bx cx - az bz - cy If , then show that = = c b a

1

14. If

15. If

1

 a  1 3   a  1 3 y 1 1  a  1 3   a  1 3

, then find the value of

x y = p  qb+ rc - pa  q  rc+ pa - qb 

z r  pa + qb - rc  , then prove that p q = x  by + cz + ax  y  cz + ax - by  =

r z  ax+ by - cz 

Find the compound ratio of the three ratios

IIT JEE Worksheet

2a : 3b, 6ab : 5c2, c : a 9.

x y z = = . p q r

y3 – 3ay2 + 3y – a.

x y z = = . a b c 8:

q+r - p r + p - q p+q - r = = , then prove that y+ z - x z+ x - y x+ y - z

each ratio is equal to

p q r = = , then prove that ap + bq + b-c c-a a-b cr = 0 and p + q + r = 0.

6.

a c e = = then show that b d f

Five numbers are in continued proportion. The product of the first and the fifth numbers is 144 and the sum of the second and the third numbers is 30. Find the numbers.

10. If p, q, r, s are in continued proportion, then prove that i) p : q + s = r3 : r2s + s3

1.

If a : b = c : d and b : x = d : y, then show that a : x = c : y.

2.

Find the mean proportional of 16 and 4.

3.

Find the third proportional of 16 and 8.

4.

If a, b, c, d are in continued proportion, then prove that 2a + 3d : 3a – 4d = 2a3 + 3b3 : 3a3 – 4b3

ii) (p2 + q2 + r2) (q2 + r2 + s2) = (pq + qr + rs)2 5. p q r = = 11. If and q+r - p r+ p - q p+q - r

p  q  r  0 , then prove that p = q = r.

Find the mean proportional between 6  3 3 and

84 3 . 6.

(x + 4), (x + 12), (x – 1), (x + 5) are in proportion. Find the value of x.

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8th Class Mathematics

40 7.

If p + r = 2q and

1 1 2 + = , then prove that q s r

p : q = r : s. 8.

If b is the mean proportional between a and c, then prove that

9.

a2  b2  c2  b4 . 2 2 2 a b c

What must be added to the numbers 6, 10, 14, 22, so that they become proportional ?

10. Find the mean proportional between and



27  18



27  18





11. Find the third proportional to (a2 – b2) and (a + b) 12. Find the third proportional of 27 and 9. 13. Three numbers are in continued proportion, of which the middle one is 16 and the sum of the other two is 130. Find the numbers. 14. The ratio of the third proportional to 12 and 30 and the mean proportional of 9 and 25 is ––––– 1) 2 : 1

2) 5 : 1

3) 7 : 15

4) 9 : 14

15. If (a + b + c) (a – b + c) = a2 + b2 + c2, then show that a, b, c are in continued proportion.



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    

1.

Comparing Quantities

By the end of this chapter, you will understand  Compound Interest Recalling Ratios and Percentages  Formula for Compound Interest Finding the Increase or Decrease Per cent  Rate Compounded Annually or Half Yearly Finding Discounts (Semi Annually) Prices Related to Buying and Selling (Profit and Loss)  Applications of Compound Interest Formula Sales Tax/Value Added Tax

Recalling Ratios and Percentages

Since contains only apples and oranges, So, percentage of apples + percentage of oranges = 100 or percentage of apples + 20 = 100 or percentage of apples = 100 – 20 = 80 Thus the basket has 20% oranges and 80% apples.

We know, ratio means comparing two quantities. A basket has two types of fruits, say, 20 apples and 5 oranges.

Formative Worksheet 1.

2. Then, the ratio of the number of oranges to the number of apples = 5 : 20. The comparison can be done by using fractions 5 1 1  The number of oranges are th the as, 20 4 4 number of apples. In terms of ratio, this is 1 : 4, read as, “1 is to 4” OR 20 4  Number of apples to number of oranges  5 1 which means, the number of apples are 4 times the number of oranges. This comparison can also be done using percentages. There are 5 oranges out of 25 fruits. So percentage of orange is 5 4 20    20% 25 4 100

3. 4.

5. 6.

1. 2.

OR



5 100  20 25

Find the ratio of the following. (a) Speed of a cycle 15 km per hour to the speed of scooter 30 km per hour. (b) 5 m to 10 km (c) 50 paise to Rs 5 Convert the following ratios to percentages. (a) 3 : 4 (b) 2 : 3 72% of 25 students are good in mathematics. How many are not good in mathematics? A football team won 10 matches out of the total number of matches they played. If their win percentage was 40, then how many matches did they play in all? If Chameli had Rs 600 left after spending 75% of her money, how much did she have in the beginning? If 60% people in a city like cricket, 30% like football and the remaining like other games, then what per cent of the people like other games? If the total number of people are 50 lakh, find the exact number who like each type of game.

Conceptive Worksheet

[Denominator made 100]

By unitary method Out f 25 fruits number of oranges are 5. So out of 100 fruits, number of oranges

Chapter - 4

Learning Outcomes

3.

4.

There were 25 chocolates in a bag, out of which Rohit ate 7 chocolates. What percentage of chocolates was eaten by Rohit? A shopkeeper has a stock of 200 cricket bats of three brands – A, B, and C. Out of these, 65 bats are of brand A, 90 bats are of brand B, and the remaining bats are of brand C. Find the percentage of bats of each brand with the shopkeeper. There are 500 students in a school. Out of these, 325 students are day scholars while the remaining students are hostlers. Find the percentage of the students who are hostlers? Vicky spends Rs 4000 and saves Rs 1000 every month. What percentage of his monthly income constitutes his monthly savings?

8th Class Mathematics

42 5.

a. b. c.

2.

A picnic is being planned in a school for Class VII. Girls are 60% of the total number of students and are 18 in number. The picnic site is 55 km from the school and the transport company is charging at the rate of Rs12 per km. The total cost of refreshments will be Rs 4280. Can you tell. The ratio of the number of girls to the number of boys in the class? The cost per head if two teachers are also going with the class? If their first stop is at a place 22 km from the school, what per cent of the total distance of 55 km is this? What per cent of the distance is left to be covered?

Then let us find the price of scooter. Price of scooter = Rs 34000 Reduction

 Rs

= Rs 34000 – Rs 1700 = Rs 32300

3.

Finding Discounts Discount is a reduction given on the Marked Price (MP) of the article. This is generally given to attract customers to buy goods or to promote sales of the goods. You can find the discount by subtracting its sale price from its marked price. So, Discount = Marked price – Sale price Let us consider an example. Example : An item marked at Rs 840 is sold for Rs 714. What is the discount and discount %? Solution: Discount = Marked Price – Sale Price = Rs 840 – Rs 714 = Rs 126 Since discount is on marked price, we will have to use marked price as the base. On marked price of Rs 840, the discount is Rs 126. On MP of Rs 100, how much will the discount be?

We often come across such information in our daily life as. (i) 25% off on marked prices (ii) 10% hike in the price of petrol Let us consider an example. Example :The price of a scooter was Rs 34,000 last year. It has increased by 20% this year. What is the price now? Solution: First find the increase in the price, which is 20% of Rs. 34,000, and then find the new price.

New price

5  34000  Rs1700 100

New price = Old price – Reduction

Finding the Increase or Decrease Percent

20% of Rs 34000 = Rs

= 5% of Rs 34000

20 ×34000 100

Discount 

=Rs 6800 = Old price + Increase

126  100  15% 840

We can also find discount when discount % is given.

= Rs 34,000 + Rs 6,800

Estimation in percentages

= Rs 40,800

OR

Unitary method 20% increase means, Rs 100 increased to Rs 120. So, Rs 34,000 will increase to? Increased price = Rs120 ×34000 100

(i) (ii)

i.e., Rs

= Rs 40,800 Similarly, a percentage decrease in price would imply finding the actual decrease followed by its subtraction the from original price. Suppose in order to increase its sale, the price of scooter was decreased by 5%. www.betoppers.com

Your bill in a shop is Rs 577.80 and the shopkeeper gives a discount of 15%. How would you estimate the amount to be paid? Round off the bill to the nearest tens of Rs 577.80, i.e., to Rs 580. Find 10% of this,

10  5801  Rs 58. 100 1  58  Rs 29. 2

(iii)

Take half of this, i.e.,

(iv)

Add the amounts in (ii) and (iii) to get Rs 87. You could therefore reduce your bill amount by

Comparing Quantities

43

Rs 87 or by about Rs 85, which will be Rs 495 approximately. 1. Try estimating 20% of the same bill amount. 2. Try finding 15% of Rs 375.

4.

Bill No S.No.

300 30 100%  %  10% 3000 3 P P%   100 CP

This sales tax is charged by the government on the sale of an item. It is collected by the shopkeeper from the customer and given to the government. This is, therefore, always on the selling price of an item and is added to the value of the bill. These days however, the prices include the tax known as Value Added Tax (VAT).

Formative Worksheet 7.

A man got a 10% increase in his salary. If his new salary is Rs 1,54,000, find his original salary.

8.

On Sunday 845 people went to the Zoo. On Monday only 169 people went. What is the per cent decrease in the people visiting the Zoo on Monday?

9.

A shopkeeper buys 80 articles for Rs 2,400 and sells them for a profit of 16%. Find the selling price of one article.

10.

The cost of an article was Rs 15,500. Rs 450 were spent on its repairs. If it is sold for a profit of 15%, find the selling price of the article.

11.

A VCR and TV were bought for Rs 8,000 each. The shopkeeper made a loss of 4% on the VCR and a profit of 8% on the TV. Find the gain or loss percent on the whole transaction.

12.

During a sale, a shop offered a discount of 10% on the marked prices of all the items. What would a customer have to pay for a pair of jeans marked at Rs 1450 and two shirts marked at Rs 850 each?

13.

A milkman sold two of his buffaloes for Rs 20,000 each. On one he made a gain of 5% and on the other a loss of 10%. Find his overall gain or loss. (Hint: Find CP of each)

14.

The price of a TV is Rs 13,000. The sales tax charged on it is at the rate of 12%. Find the amount that Vinod will have to pay if he buys it.

15.

Arun bought a pair of skates at a sale where the discount given was 20%. If the amount he pays is Rs 1,600, find the marked price.

16.

I purchased a hair-dryer for Rs 5,400 including 8% VAT. Find the price before VAT was added.

Profit =

5.

Sales Tax/Value Added Tax The teacher showed the class a bill in which the following heads were written.

Amount

ST means Sales Tax, which we pay when we buy items.

2  100  25% only.. 8

Let us consider an example. Example :Sohan bought a second hand refrigerator for Rs 2,500, then spent Rs 500 on its repairs and sold it for Rs 3,300. Find his loss or gain per cent. Solution: Cost Price (CP) = Rs 2500 + Rs 500 (overhead expenses are added to give CP) = Rs 3000 Sale Price (SP) = Rs 3300 As SP > CP, he made a profit = Rs 3300 – Rs 3000 = Rs 300 His profit on Rs 3,000, is Rs 300. How much would be his profit on Rs 100?

Rate

Total

For the school fair (mela) I am going to put a stall of lucky dips. I will charge Rs 10 for one lucky dip but I will buy items which are worth Rs 5. So you are making a profit of 100%. No, I will spend Rs 3 on paper to wrap the gift and tape. So my expenditure is Rs 8. This gives me a profit of Rs 2, which is,

Finding cost price/selling price, profit %/loss%

Menu Quantity Bill amount + ST (5%)

Prices Related to Buying and Selling (Profit and Loss)

Sometimes when an article is bought, some additional expenses are made while buying or before selling it. These expenses have to be included in the cost price. These expenses are sometimes referred to as overhead charges. These may include expenses like amount spent on repairs, labour charges, transportation etc.

Item

Date

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8th Class Mathematics

44

Conceptive Worksheet 6.

The length and breadth of a rectangle are respectively increased by 20% and 25%. What is the percentage change in the area of the rectangle?

7.

An election was contested by only two persons “ Meenakshi and Khushboo. Once the results came in, it was found that Meenakshi got 20% more votes than Khushboo. There were a total of 111100 valid votes polled in the election. What was the difference between the numbers of votes received by the two contestants?

8.

A clothes merchant sells cloth pieces such that the cost price of 15 cloth pieces equals the selling price of 10 cloth pieces. What is the profit percent made by the merchant on each cloth piece?

9.

Earlier, Meenakshi spent 75% of her salary. Recently, she got a raise of 25% in her salary. By what percent must her expenditure increase so that her savings remain the same?

10.

A shopkeeper sold a shirt for Rs 450, thereby suffering a loss of 10%. How much did the shirt cost the shopkeeper?

11.

An item marked at Rs 840 is sold for Rs 714. What is the discount and discount %?

12.

The list price of a frock is Rs 220. A discount of 20% is announced on sales. What is the amount of discount on it and its sale price.

13.

A shopkeeper purchased 200 bulbs for Rs 10 each. However 5 bulbs were fused and had to be thrown away. The remaining were sold at Rs 12 each. Find the gain or loss %.

14.

Meenu bought two fans for Rs 1200 each. She sold one at a loss of 5% and the other at a profit of 10%. Find the selling price of each. Also find out the total profit or loss.

15.

The cost of a pair of roller skates at a shop was Rs 450. The sales tax charged was 5%. Find the bill amount.

16.

Waheeda bought an air cooler for Rs 3300 including a tax of 10%. Find the price of the air cooler before VAT was added.

6.

Compound Interest We might have come across statements like “one year interest for FD (fixed deposit) in the bank @ 9% per annum” or ‘Savings account with interest @ 5% per annum’.

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1.

Interest is the extra money paid by institutions like banks or post offices on money deposited (kept) with them. Interest is also paid by people when they borrow money. We already know how to calculate Simple Interest. We have some money in the bank. Every year some interest is added to it, which is shown in the passbook. This interest is not the same, each year it increases. Normally, the interest paid or charged is never simple. The interest is calculated on the amount of the previous year. This is known as interest compounded or Compound Interest (C.I.). Let us take an example and find the interest year by year. Each year our sum or principal changes. Calculating Compound Interest A sum of Rs 20,000 is borrowed by Heena for 2 years at an interest of 8% compounded annually. Find the Compound Interest (C.I.) and the amount she has to pay at the end of 2 years. Aslam asked the teacher whether this means that they should find the interest year by year. The teacher said ‘yes’, and asked him to use the following steps : Find the Simple Interest (S.I.) for one year. Let the principal for the first year be P1. Here, P1 = Rs 20,000 SI1 = SI at 8% p.a. 20000  8  Rs 1600 100 Then find the amount which will be paid or received. This becomes principal for the next year. Amount at the end of 1st year = P1 + SI1 = Rs 20000 + Rs 1600 = Rs 21600 = P2 (Principal for 2nd year) Again find the interest on this sum for another year.

for 1st year = Rs 2.

3.

SI2 = SI at 8% p.a.for 2nd year = Rs 4.

21600  8 100

= Rs 1728 Find the amount which has to be paid or received at the end of second year. Amount at the end of 2nd year = P2 + SI2 = Rs 21600 + Rs 1728 = Rs 23328 Total interest given = Rs 1600 + Rs 1728

Comparing Quantities

45 5000  5  5  1   100  100  R  R  or  P1 1    100  100

= Rs 3328 Reeta asked whether the amount would be different for simple interest. The teacher told her to find the interest for two years and see for herself. SI for 2 years = Rs

 Rs

20000  8  2  Rs 3200 100



Reeta said that when compound interest was used Heena would pay Rs 128 more.

7.

5  5000  5  5   A2  Rs 5000 1    Rs 1   100  100   100  A2 = P2 + SI2 5  5    Rs 5000 1   1    100  100  R  R  R    P1  1    P1 1   100  100   100 

Formula for Compound Interest Suppose P 1 is the sum on which interest is compounded annually at a rate of R% per annum. Let P1 = Rs 5000 and R = 5% per annum. Then by the steps mentioned above

1.

SI1  Rs or

P1×R×1 100

so, A1  Rs 5000  or

2

5    Rs 5000  1    P3  100  R  R    P1 1   1    100   100 

5000  5  1 100

SI1  Rs

2

R    P1 1    100  Proceeding in this way the amount at the end of n years will be

5000  5  1 100

A1  P1  SI1  P1 

P1R 100

R   A n  P1 1    100 

5    Rs 5000  1    P2  100 

5  5 1  SI2  Rs5000 1    100  100 P2  R  1 or SI2  100

n

n

R   A  P 1  Or, we can say   100  Using this we get only the formula for the amount to be paid at the end of n years, and not the formula for compound interest. Aruna at once said that we know that CI = A – P, so we can easily find the compound interest too.

R   or  P1  1    P2  100 

2.

P1R  R  1   100  100 

8.

Rate Compounded Annually or Half Yearly (Semi Annually)

Time period and rate when interest not compounded annually The time period after which the interest is added each time to form a new principal is called the conversion period. When the interest is compounded half yearly, there are two conversion periods in a year each after 6months. In such situations, the half yearly rate will be half of the annual rate. What will happen if interest is compounded quarterly? In this case, there are 4 conversion periods in a year and the quarterly rate will be one-fourth of the annual rate.

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8th Class Mathematics

46

P = Rs 100 at 10% per annum compounded annually

P = Rs 100 at 10% per annum compounded half yearly

The time period takes is 1 year

The time period is 6 months or ½ year

100  10 1 I  Rs  Rs 10 100 A = Rs 100 + Rs 10 = Rs 110

100×10× I  Rs

100

Rate becomes half

1 2  Rs.5

A = Rs 100 + Rs 5 = Rs 105 Now for next 6 months the P = Rs 105 105  10  So, I  Rs

1 2  Rs 5.25

100 and A = Rs 105 + Rs 5.25 = Rs 110.25

9.

(i) (ii) (iii)

Applications of Compound Interest Formula There are some situations where we could use the formula for calculation of amount in CI. Here are a few. Increase (or decrease) in population. The growth of a bacteria if the rate of growth is known. The value of an item, if its price increases or decreases in the intermediate years. Let us consider an example. Example The population of a city was 20,000 in the year 1997. It incr eased at the rate of 5% p.a. Find the population at the end of the year 2000. Solution There is 5% increase in population every year, so every new year has new population. Thus, we can say it is increasing in compounded form. Population in the beginning of 1998 = 20000 (we treat this as the principal for the 1st year)

5  20000  1000 Increase at 5%  100 Population in 1999 = 20000 + 1000 = 21000 [Treat as the Principal for the 2nd year.] Increase at 5% 

5  21000  1050 100

Population in 2000 = 21000+1050 = 22050 [Treat as the Principal for the 3rd year.]

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

Increase at 5%

5  22050 = 1102.5 100

At the end of 2000 the population = 22050 + 1102.5 = 23152.5 or, Population at the end of 2000 5   = 20000  1    100   20000 

3

21 21 21   20 20 20

= 23152.5 So, the estimated population = 23153.

Formative Worksheet 17.

Calculate the amount and compound interest on

(a)

Rs 10,800 for 3 years at 12 % per annum compounded annually.

(b)

Rs 18,000 for 2

1 2

1 years at 10% per annum 2

compounded annually. (c) (d)

(e)

Rs 62,500 for 1

1 years at 8% per annum 2

compounded half yearly. Rs 8,000 for 1 year at 9% per annum compounded half yearly. (You could use the year by year calculation using SI formula to verify). Rs 10,000 for 1 year at 8% per annum compounded half yearly.

Comparing Quantities 18.

19.

20.

21.

22.

Kamala borrowed Rs 26,400 from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan? (Hint: Find A for 2 years with interest is compounded yearly and then find SI on the 2nd

interest. Which of the following statements is

4 year amount for years). 12

(B) Sumit earns Rs 7200 more than Sarita in the

Fabina borrows Rs 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much? I borrowed Rs 12,000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay? Vasudevan invested Rs 60,000 at an interest rate of 12% per annum compounded half yearly. What amount would he get (i) after 6 months? (ii) after 1 year? Arif took a loan of Rs 80,000 from a bank. If the rate of interest is 10% per annum, find the difference in amounts he would be paying after

(C) Sarita earns Rs 5616 more than Sumit in the

1

23.

24.

correct? (A) Sumit earns Rs 5616 more than Sarita in the given period given period

1 years if the interest is 2

(i) compounded annually. (ii) compounded half yearly. Maria invested Rs 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find (i) The amount credited against her name at the end of the second year. (ii) The interest for the 3rd year. Find the amount and the compound interest on Rs 10,000 for 1

25.

47

1 years at 10% per annum, 2

compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually? Find the amount which Ram will get on Rs 4096,

1 2

if he gave it for 18 months at 12 % per annum,

given period (D) Sarita earns Rs 7200 more than Sumit in the given period

Conceptive Worksheet 17.

18.

19.

20.

21.

22.

23.

interest on this sum and the amount to be paid at

Sumit invested Rs 125000 in a bank for three years

the end of 2 years.

at the rate of 12% per annum compounded annually. On the other hand, his friend Sarita invested the same sum in another bank at the same rate for the same duration, but on simple

A sum of Rs 10,000 is borrowed at a rate of interest 15% per annum for 2 years. Find the simple

interest being compounded half yearly. 26.

The population of a place increased to 54,000 in 2003 at a rate of 5% per annum (i) find the population in 2001. (ii) what would be its population in 2005? In a Laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5, 06,000. A scooter was bought at Rs 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year. The simple interest on a certain sum invested for one year at the rate of 8% p.a. isRs 640. What is the compound interest on the same sum invested for 2 years at the same rate compounded annually? In what time period will an amount 1331 become  times the original sum, when the 1000 rate of interest is 10% per annum compounded annually? If the simple interest on a certain sum for 2 years is Rs 2000 and the compound interest on the same sum for the same time period at the same rate of interest compounded annually is Rs 2160, then what is the rate of interest per annum?

24 25.

Find CI on Rs 12600 for 2 years at 10% per annum compounded annually. What amount is to be repaid on a loan of Rs 12000 1 for 1 years at 10% per annum compounded half 2 yearly. www.betoppers.com

8th Class Mathematics

48 26.

Find CI paid when a sum of Rs 10,000 is invested for 1 year and 3 months at 8

27.

28.

1 % per annum 2

compounded annually A TV was bought at a price of Rs 21,000. After one year the value of the TV was depreciated by 5% (Depreciation means reduction of value due to use and age of the item). Find the value of the TV after one year. Deepak borrowed Rs 16000 from a bank as loan at a certain rate, which is compounded annually. He repaid the loan after 2 years by paying Rs. 1640 as compound interest. At what rate did Deepak borrow the loan?

9.

10.

Summative Worksheet 1.

2.

3.

4.

5.

6.

7.

The marked price of an article is Rs 1850. After announcing a discount of 20% on sales, the shopkeeper can earn a profit of 15% on the sale of the article. What is the approximate cost price of the article? (A) Rs 1390 (B) Rs 1310 (C) Rs 1287 (D) Rs 1205 The value of a washing machine worth Rs 16000 is depreciated by 5% every year. What will be the value of the machine after two years? (A) Rs 13800 (B) Rs 14440 (C) Rs 15280 (D) Rs 15700 The population of a district in the year 2007 was 625000. If it increased at the rate of 8% per annum, then what will be the population of the district at the end of the year 2009? (A) 780000 (B) 729000 (C) 820000 (D) 839000 Raima bought a gold ornament worth Rs 20000. The value of gold is appreciated by 5% every year. What will be the value of the ornament after 3 years? (A) Rs. 22150.50 (B) Rs. 23152.50 (C) Rs. 24265 (D) Rs. 25154 Yash Builder Group bought land worth Rs 1562500 in a town. The value of the land appreciated at the rate of 4% per annum. What will be the cost of the land after two years? (A) Rs 1590000 (B) Rs 1690000 (C) Rs 1725800 (D) Rs 1825800 What price should a shopkeeper mark on a jacket that costs him Rs 1600 to earn a profit of 20% even after allowing a discount of 20%? (A) Rs 2200 (B) Rs 2340 (C) Rs 2400 (D) Rs 2580 A shopkeeper bought a ceiling fan at Rs1800. He marked it at Rs 2400. How much discount percent should be allowed by the shopkeeper on the ceiling fan to earn a profit of 20%? (A) 10% (B) 15% (C) 20% (D) 5%

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1 years 2 when Rs 75000 are invested at the rate of 10% per annum compounded annually? (A) Rs 80217 (B) Rs 82347.50 (C) Rs 90256 (D) Rs 95287.50 What is the compound interest received on Rs 20000 invested for 1 year at the rate of 6% per annum compounded half yearly? (A) Rs 1218 (B) Rs 1312 (C) Rs 1405 (D) Rs 1560 Manish borrowed Rs 62500 from his friend at the rate of 8% per annum compounded annually. After some time, he repaid Rs 72900 to his friend. For how many years did Manish borrow the money? (A) 1 (B) 2 (C) 3 (D) 4 The compound interest on a certain sum invested for two years at the rate of 10% p.a. compounded annually is Rs 4200. What is the simple interest on the same sum invested for the same time period at the same rate? (A) Rs 4400 (B) Rs 4000 (C) Rs 3890 (D) Rs 3000 Somnath bought a shirt for Rs 702 including sales tax. The marked price of the shirt is Rs 650. How much sales tax is imposed on the shirt? (A) 6% (B) 8% (C) 12% (D) 14% A television set costs Rs 16060 including 10% value added tax. What is the cost of the television set without tax? (A) Rs 14200 (B) Rs 14600 (C) Rs 15400 (D) Rs 15800 In a shop, 14% sales tax is imposed on the purchase of every article. Rajni wants to buy a sweater whose list price is Rs 1200. How much money does she need to spend to buy the sweater? (A) Rs 1298 (B) Rs 1368 (C) Rs 1396 (D) Rs 1426 What is the amount received after 2

8.

11.

12.

13.

14.

HOTS Worksheet 1.

2.

Rupert wants to purchase an ice hockey stick that is put up on sale at a departmental store. The marked price of the hockey stick is Rs. 25. The store offers a discount of 20% on all the items. The store manager is friend of Rupert and he gives Rupert an additional discount of 20% on the hockey stick. What is the selling price of the hockey stick? (A) Rs.10 (B) Rs.14 (C) Rs.16 (D) Rs.20 A sum of money doubles itself at compound interest in 6 years. In how many years will it become sixteen times itself? (A) 12 years (B) 16 years (C) 20 years (D) 24 years

Comparing Quantities 3.

4.

5.

49

A footwear shop is offering a discount of 10% on all footwear. Also, an extra discount of 5% on the discounted price is being offered for purchases above Rs.100. The effective discount on a purchase of Rs.150 is (A) 13.5 % (B) 14.0% (C) 14.5% (D) 15.0% Sammy invests some money in a bank. The bank pays him interest at the rate of 5% per year compounded annually. After 2 years, the value of his investment increased to Rs.84 100. What was the original value that he invested? (A) Rs.4 000 (B) Rs.8 000 (C) Rs.40 000 (D) Rs.80 000 If Rs.1 000 is invested at 9% compounded semiannually, then the expression that gives the value of the investment after 15 years is (A) 1000(1 + 0.09)30 (B) 1000(1 + 0.09)15  1  0.09  (C) 1000    2 

6.

7.

8.

9.

10.

30

 0.09  (D) 1000  1   2  

5 6

(B)

6 5

(C)

4 5

(D)

5 4

12.

13.

30

A tennis racket costs Rs.71.50, after a sales tax of 10% is levied on it. Before the sales tax is levied, the tennis racket costs (A) Rs.61 (B) Rs.63 (C) Rs.65 (D) Rs.67 Nancy decided to buy a car priced at Rs.24 000. The salesman then told her that she would have to pay an additional 7.5% of the amount as sales tax. The total amount to be paid for the car is (A) Rs.1 800 (B) Rs.6 000 (C) Rs.25 800 (D) Rs.42 000 An amount ofRs.20 000 is being compounded at 20% p.a. The rate of interest is charged half yearly. The amount after 2 years will be (A) Rs.29 282 (B) Rs.29 412 (C) Rs.30 512 (D) Rs.31 615 David purchased a car 3 years ago for Rs. 200 000. Its value depreciated each year by 25% p.a. The present value of the car is (A) 61 215 (B) 84 375 (C) 65 325 (D) 41 345 Johny and Tony invested same amounts in a bank at the rate 20% p.a. compound interest. Johny took the amount after 4 years while Tony took the amount after 5 years. The ratio of the amount received by Johny and Tony was (A)

11.

14.

15.

16.

17.

18.

Sabrina wants to save a total of Rs.10 000 in two years in order to buy a diamond necklace. She finds an account that pays 4% interest, quarterly. How much should Sabrina put into her account, so that she will have Rs.10 000 at the end of two years? (A) Rs.7 306.90 (B) Rs.7 407.54 (C) Rs.7 837.63 (D) Rs.7 943.31 Theamount to be returned against a loan of amount ‘A’ in given by the expression A ×  (1.1 250 881)t, where t is  time in years.  What in the annual rate of interest if the interest is compounded quarterly? (A) 7% (B) 9% (C) 11% (D) 12% Sam is supposed to return an amount of Rs.101 22.55 for a loan he took 3 years ago. The interest was compounded half yearly at the rate of 8%. What amount did he take on loan? (A) Rs.8 000 (B) Rs.8 025 (C) Rs.8 000 (D) Rs.8 062 Bill takes a lone of Rs.10 000 on a rate of 2% compound interest per quarter for 5 years. The amount he will return is (A) Rs.146 93.28 (B) Rs.147 32.44 (C) Rs.14 859.47 (D) Rs.14 978.85 A sum is borrowed from a bank at a rate of 10% compounded half-yearly. The effective rate of interest is (A) 10% (B) 10.25% (C) 11% (D) 11.25% At a factory, the value of a machine decreases every year at the rate of 20% of its value at the beginning of that year. If the present value of the machine is Rs.512, its value after 3 years will be (A) Rs.500 (B) Rs.900 (C) Rs.1 000 (D) Rs.2 500 Mr. Anthony invested money in two schemes A and B offering compound interest at 8% per annum and 9% per annum, respectively. The total interest accrued through the two schemes together in two years was Rs.4 818.30 and the total amount invested was Rs.27 000. The amount invested in scheme A was (A) Rs.11 000 (B) Rs.12 000 (C) Rs.13 000 (D) Rs.14 000 Peter had two sons. He divided Rs.1 301 between his sons Sam and Patrick such that the amount of Sam after 7 years was equal to the amount of Patrick after 9 years at 4% per annum compound interest. Sam’s and Patrick’s share, respectively was (A) Rs.676 and Rs.625 (B) Rs.625 and Rs.676 (C) Rs.766 and Rs.535 (D) Rs.535 and Rs.766

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8th Class Mathematics

50

IIT JEE Worksheet I.

Single Correct Answer Type

1.

Mahender sold a cell phone at a loss of 8%. Had he sold it at Rs 378 less, the loss would have been 15%. At what price did Mahender buy the cell phone? (A) Rs 4800 (B) Rs 5400 (C) Rs 6200 (D) Rs 7500 Kunal bought a bag of 5 kg basmati rice at Rs 117 at a discount of 10% from a grocery shop. What price was marked on the bag of 5 kg basmati rice? (A) Rs 127 (B) Rs 130 (C) Rs 133 (D) Rs 140 During a festive season, a microwave oven whose list price is Rs 3250 is available at Rs 2470 after a discount in a particular shop. What is the discount percent offered by the shopkeeper on the microwave oven? (A) 15% (B) 20% (C) 24% (D) 30% In a book-fair, 16% discount is offered on each book. If the marked price of a book is Rs 250, then what is the selling price of the book? (A) Rs 210 (B) Rs 226 (C) Rs 230 (D) Rs 236 The cost of a flat was Rs 1800000 last year. Now, the cost of the same flat is Rs 1926000. What is the percentage increase in the cost of the flat? (A) 5% (B) 6% (C) 7% (D) 8%

2.

3.

4.

5.

6.

7.

8.

9.

1 1 If the sales tax be reduced from 4 %  to  4 % , 2 3 then the money saved by a person who purchases an article which had an initial cost price of Rs.9 000, is (A) Rs.10 (B) Rs.15 (C) Rs.18 (D) Rs.20 A certain sum of money doubles itself at compound interest in 15 years. How long will the sum take to become eight times? (A) 20 years (B) 25 years (C) 40 years (D) 45 years If the compound interest on a certain sum at 16 2 3 % for 3 years is Rs.1 270, then the simple interest on the same sum at the same rate and same time period will be (A) Rs.1 000 (B) Rs.1 020 (C) Rs.1 050 (D) Rs.1 080 Robert and Luke are two brothers. Their father divides Rs.7 569 between them such that Robert’s share at the end of 9 years is equal to Luke’s share at the end of 11 years, compounded annually at the rate 5%. Luke’s share was (A) Rs.3 950 (B) Rs.3 960 (C) Rs.3 969 (D) Rs.3 990

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10.

Sandy and Lara decided to borrow same amount of money from a friend. Sandy borrowed it at compound interest and Lara on simple interest. The difference between the compound interest and simple interest on that amount at 12% per annum, came out to be Rs.90 at the end of two years. The value of that amount at the end of 3 years will be (A) Rs.9 234.80 (B) Rs.9 113.70 (C) Rs.8 780.80 (D) Rs.8 380.80

II.

Multi Correct Answer Type

11.

The selling price of 12 articles is equal to the cost price of 15 articles. The gain percent is

12.

13.

1 12 (A) 6 % (B) 22 % 3 4 (C) 25% (D) 80% A discount series of 10%, 20% and 40% is equal to a single discount of (approx.) (A) 50% (B) 56.8% (C) 57% (D) 70.28% Kabir buys an article with 25% discount on its marked price. He makes a profit of 10% by selling it at Rs. 660. The marked price is (A) Rs.

660  40 33

660  33 40 An umbrella marked at Rs.80 is sold for Rs.68. The rate of discount is (D) Rs.

(C) Rs. 800 14.

12 11 % (C) 17 % (D) 20% 4 17 A retailer buys a sewing machine at a discount of 15% and sells it for Rs. 1955. Thus, he makes a profit of 15%. The discount is (A) Rs. 270 (B) Rs. 3× 9× 10 (C) Rs.300 (D) Rs. 3×102 A Sum of money at S.I. doubles in 7 years, It will become four times in (A) 168 month (B) 252 month (C) 28 years (D) 21 years (A) 15%

15.

16.

(B) Rs.700

(B) 12

Comparing Quantities

III.

Paragraph Type

I.

P% of x=

17.

18.

19.

20.

II. 21.

51 22.

p x 100 Find 135% of 80 cm (A) 10800 m (B) 10.8 cm (C) 1.08 m (D) 108 m Find a if 8.4% of a is 42 (A) 500 (B) 50 (C) 5 (D)5000 What percent of 24 m is 6m (A) 60% (B) 25% (C) 5% (D) 6.25% Ajit lost 20% of his money after which he was left with Rs. 6400. The amount he had in the begining with him is (A) Rs. 8000 (B) Rs. 9000 (C) Rs. 7000 (D) Rs. 1000 Increase =Increase % of original value and Decrease = Decrease% of original value The salary of a bank clerk was increased by 7%. If his present salary is Rs. 8025, his salary before increament is (A) Rs. 7500 (B) Rs.8500 (C) Rs. 7000 (D) Rs.8000

23.

24.

IV.

Integer Type

25.

72% of 25 students are good at Mathematics. How many are not good at it? If 50% of (x – y) = 30% of (x + y), then x = ky, find k? A shopkeeper buys 80 articles for unit digit of 2400 and sells them for a profit of 16%. Find the unit digit of selling price of one article. The value of a machine depreciates every year by 15%. The value after a year if its present value is Rs. 50,000 will be 425×10n. Find n. A Person sells an article for Rs. 550, gaining 1/10 of its C.P. If gain % is 10n. Find n.

26. 27.

28.

29.

V.

Matrix Matching (Match the following)

30.

Convert the following ratios into percents. Column I Ratio (A) 2: 5 (B) 2: 7 (C) 13: 20 (D) 13: 75 Column I

31.

The value of a machine depreciates every year by 5%. If the present value os Rs. 40,000, the value after a year would be (A) Rs. 380 (B) Rs. 30,000 (C) Rs. 3800 (D) Rs. 38,000 The population of a town increase by 10% annually. If the present population is 60,000, the population after a year is (A) 6600 (B) 66,000 (C) 660 (D) 60, 000 The value of a machine depreciates 10% every year. If its present value is Rs. 38700, the value a year ago is (A) Rs.4300 (B) Rs.430 (C) Rs.43000 (D) Rs. 40300

Column II Percent (p) 28.55% (q) 17.33% (r) 65% (s) 40% Column II

1 (A) The number whose 6 % is 2 is 4 (B) Marked price of a book is Rs. 30. It is sold at a discount of 15%. The discount allowd on the book in Rs. is (C) Rohan purchased a house for Rs. 6500 and sold it on the same day for Rs. 7150. His gain in Rs. is (D)A shopkeeper purchased goods worth Rs. 8000. He sold them at a profit of 7%. His selling price in Rs. is

(p) 8560 (q) 650

(r) 32

(s) 4.50

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52

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8th Class Mathematics

Chapter -45

Exponents and Radicals

Learning Outcomes

By the end of this chapter, you will understand •

Laws of Exponents

•

Numbers with non-integer exponents

•

Rational Exponent

•

Radical and Radicand

1. Introduction

ii) Cube Root of 27 is 3 i.e., ( 33 = 27)

Exponent If a certain number a is multiplied m times in succession, then the continued product so obtained is called the m th power of a and is written as am (read as, a to the power m). Thus, am = a × a × a × a........ to m factors. Here, a is called the base of am and m is called the index or exponent of am . Examples: i) x5 = x . x . x. x . x ii) (–3)6 = (–3) (–3) (–3) (–3) (–3) (–3) Note: In particular, a2 is called the square of a (or, a to the power 2) and a3 is called the cube of a (or a to the power3)

iii)Sixth Root of 64 is 2 i.e.,

Again, (–5)2 = 25  25  5 Therefore, it is evident that both 5 and (–5) are square roots of 25. Hence, by the Square root of a real positive number x we mean  x (i.e.,  x or  x ) Note: i) If x > 0 and n is any positive integer, then

 x n

2

12

written as a  x or a  x

then

x does not exist in the set of real numbers.

n

x.

In particular, if a2 = x, then a is called the second root or square root of x and is

x is positive.

If x < 0 and n is any odd integer, then n x is negative. If x < 0 and n is any positive even integer,

ii)

x ) is such a number

whose nth power is equal to x i.e., i)

n

iii) n

n

Formative Worksheet 1.

or simply

Identify base and index of the following. 2

 9  i) 4 ii) (–2) iii)   iv) (–1)2009  25  Express the following in the exponential notation. i) 192 ii) 729 Find the values of the following i) 27 – 72 ii) (2.5)3 – (1.5)3 Which is greater in each of the following? i) 212, 28 ii) 510, 315 Write the following in the index form 4

ii)

a x. If a3 = x, then a is called the third root or cube root of x and is written as a =

i) Square Root of 25 is 5 i.e., (5 = 25) 2

2. 3.

x or a  x1 3 Examples: 3

64  2

Since 52 = 25  25  5

x or a  x1 n .

Clearly, nth root of x (i.e.,

6

(2 = 64) Example:

If a and x are two real numbers and n is a positive integer such that a n = x, then a is called the n th root of x and is written as a n

27 = 3

6

Root

=

3

4.

25  5 5.

i)

5

32 1

4

ii)

3

a 6 y 3

8th Class Mathematics

54 Case – 1 :If m > n

Conceptive Worksheet 1.

Identify base and index of the following. Also find their values. 4

3.

 2 i) 33 ii) (–2)6 iii)   iv) (–1)100 5 Express the following in the exponential notation. i) 288 ii) 1296 Find the values of the following.

4. 5.

3 3 i)      ii) (3.2)2 – (1.6)2 2 4 Which is greater in each of the following ? 78, 96 Write the following in the index form.

2.

4

i)

4

4

2

16a b

ii)

3

3

3

27a b

2

2. Laws of Exponents I)

(Fundamental Index Law) For multiplying the power of same base, powers are added. Proof: a m  a n   a  a  a  ........to m factors    a  a  a  a...... to n factors 

 am  a n  a m n Examples: n

m+n



 23  24  27 ii) (x + y)2 × (x + y)3 = (x + y)2 + 3

 a  a  a   (x + y)2 × (x + y)3 = (x + y)5 m+n

Divison Property Property: + a m ÷ an = am n (m,n  z ) For dividing the powers of same base, we subtract the indices. Proof:

Case –2 : If m < n am 1  n a a  a  a  ......  n  m  factors am 1   nm  n a a



32 35

3 3 1 1   3 3  3  3  3  3 3 3 3 3

III) Power of a Power Property: n a m = a mn (m,n  z+)

 

n

m m m m Proof:  a   a  a  a  ...... to n factors

m n

a 

 a mn

a  a  a  ...... to m factors am a a  n = a  a  a  ..... to n factors a

3 i)  5  = 53 × 2 = 56 (  (am)n = (amn)) 4

2 ii)  y  = y2 × 4 = y8 4

  y 2   y8

IV) Power of a Product Property: m (m,n  z+) a m .b m =  ab  Proof: a m  a m   a  a  a  ..... to m factors    b  b  b  ..... to m factors 

=  ab    ab    ab   ...... to m facors

n

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35  am  5 –2 3   a mn  = 3 = 3  2 n 3  a 

2

i) 23 × 24 = 23 + 4  a  a  a

m

5 2 Example: 3  3 

Examples: m

II)

am  a mn n a

 a m m  m ....... to n factors

  a  a  a  a..... to  m  n  facors 

n



2 5 Example: 3  3 

Multiplication Property: a m × a n = a m+n (m,n  z+)

m

am  a  a  a  a......  m  n  factors an

 a m  b m   ab 

m

Exponents & Radiclas

55 Case-3 : Either m is a negative number

Example: x 3 . y 3   x  x  x    y  y  y 

Meaning of a–m , where m is a Positive Real Number and a  0

=  xy    xy    xy  = (xy)3

Since the fundamental index law is true for all indices, hence, a–m .am = a–m+m = a° = 1

3

 x3 . y 3   xy 

V) Power of a Divison Property: m am  a  =  (m,n  z+) bm  b  Proof: =

1 m 1  a-m = m and -m  a a a –m Thus a is the reciprocal of am.

a m a  a  a  .... to m factors  b m b  a  b ..... to m factors

Case-4: Either m = 

a a a    ..... to m factors b b b

positive integers.

Meaning of a–p/q, where p and q are Positive Integers

m

am  a   m   b b all the above laws are defind for m, n  z+ only..

3. Numbers with Exponents

p where, p and q are q

We have, a–p/q =



1 q

.

ap

If a, m, n are three real numbers and am = an , then m = n (a  0, 1,  ) Proof : Since a m = a n and a  0, 1,  hence,

Numbers with Non-integer Exponents What if m is not a positive integer? If m is not a positive integer, then there exist four cases. Case-1: either m = 0

am  1 or am–n = a° an m – n = 0  m = n proved

Meaning of a° (a  0)

am a° = m  1 ( a  0) a Note: a° has no meaning when a = 0 i.e., 0° has no meaning.

a

p q

Equations and Identities involving Indices

Non-Integer

We know the Fundamental Index law being true for all indices hence, a°.am = a° + m = am Now Dividing both sides by am we get,

1

Formative Worksheet 2n  4  2.2n  2 3 . 2.2n  3

6.

Simplify :

7.

9n  35  (27) 3 Simplify : . 3  (81) 4 x7 / 2. y3

8. p Case-2 : Either m = where p and q are positive q 9. integers.

Simplify:

x5 / 2 . y

.

Simplify : x 1 y  y 1 z  z 1 x . n 1

n 1

2 4 Meaning of a p/q , where p and q are 10. Simplify: n n 1  n 1 n 1 . (2 ) (2 ) Positive Integer Since q is a positive integer, hence, q p/q p/q p/q p/q a =a .a .a ......... to q factors

  =a

p/q+ p/q+ p/q...to q factors

q p a = a Thus, ap/q is the qth root of ap . p/q

=a

p/q .q

=a

11. Find the value of

p

12. Find the value of

(0.6)0  (0.1) 1 1

3

 3   3   1   3      2  2  3 

1

.

9993  3336  1117 . 32  94  115

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8th Class Mathematics

56 1 3

1 4   3  4 16 4  1  3  13. Find the value of 81  2     . 6  27     

 xb  14. Simplify:  c  x 

b+c-a

 xc   a  x 

c+a-b

 xa   b  x 

30. If x, y, z are not equal and if x 2  yz y 2  zx z 2  xy   , then prove that (a a b c + b + c) (x + y + z) = ax + by + cz.

a+b-c

.

1 1 + . n-m 1+ x 1+ x m-n

15. Simplify:

x

a

(2 ) .8  32 = 2–4, then find 3b – 5a. 23b.4 22. If a + b + c = 0, then show that 1 1 1 + + c -a +1 = 1 . a -b b -c x + x x + x +1 x + x +1

23. If 2x = 3y = 12z then prove that xy = z(x + 2y). 1 1 1 24. Find the value of a + b + c if x a = y b = z c and

xyz = 1. 1

1 1 25. Prove that 1  1  (1  n)    n .

2

26. If x 3  y 3  z  0 , then prove that (x + y2 + z3)3 = 27xy2z3. 27. If x2 = y + z, y2 = z + x and z2 = x + y, then find 1

1  x

7.

Simplify:

8.

Simplify:

9.

Simplify: (8 x 3  27 a 3 ) 3 .

1

a 3 b 4 c 3  3 a 2 b 4 c 1 . 4

x 2 y 5  x 4 y 3 .

1

 m  14 m  25 . 5.5 10. Simplify:  m  5. 5 

m   .   1

2

a

21. If

1

Simplify : a 2 b  3 ab 3 .

2

if a  0, b  0, ab  1, ab  1 . Solve 2x + 5 = 4x + 2. If 4x – 4x – 1 = 24, then find x. Find x and y if 2x+3 = 4y–2 and 3x–2 = 93y–2x. If amn = am.an, then find the value of m(n – 2) + n(m – 2).  a 2

6.

2

y

1  1   a +  . a   b b     16. Simplify:  1  x  1  y .  b +  . b   a  a 

17. 18. 19. 20.

Conceptive Worksheet

.

28. If a + b + c = 0, then find a 2  b2  c 2 . (a  b)2  (b  c ) 2  (c  a ) 2

2 1  9 2 11. Find the value of    3(8) 3 (4) 0    . 4  16  12. Find the value of 2

3

(0.000008) 3  (0.0081) 4  10 2 1

.

1

(0.0256) 2  (0.125) 3  (0.3)2 4

4

 1   1 3    4   4   3 13. Find the value of  1 2  1 2 . 3    4   4   3  xa  14. Simplify:  x b   

a+b

 xb   c  x 

b+c

 xc   a  x 

c+a

.

15. Simplify: 1 1 + 1 + x a - b + x a - c 1+ xb - c + xb - a . 1 + 1+ xc - a + xc - b

16. Simplify:

29. a, b, c are real numbers such that a + b + c = 12,

1 1 1 and    15 , then find a b c a b c a c b + + + + + . b c a c b a

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 xa   xb 

   

 xc   xa 

a 2  ab  b 2

   

 xb     xc   

c 2  ca  a 2

b2  cb  c 2

.

Exponents & Radiclas 17. If

5

 3  9

18. If 22x–1=

2

 3x  3  3 , then find x.

x 3

, then find the value of x.

1  2 2 , then find the value of 3x + 4y. 19. If (64)x = (256) y 20. If

 4 3

2 x

1 2

= 32, then find the value of x.

21. If ax = m, ay = n and a2 = (mynx)z, then show that xyz = 1. 22. If a x = b y = c z and b 2 = ac, then prove that

1 1 2   x z y.

1 1 1 24. If 2x = 3y = 6–z, then prove that x  y  z  0 . 25. If x = ya, y = zb and z = xc, find the value of abc. 1 1 3 26. Prove that 1  1  (1  x 3 ) 1   x .  

27. If a + b + c = 0, then

a

2

1 .  b2  c 2

a b bc ca ,y= and z = , then find the a+b b+c c+a

p , where q  0 q -

n

p q

and a is a positive real number, then a = a =

1 p

aq

-

Consider, x

p q

p

1  -m 1  1 q  p  x  m  =   x   x xq 

p

p

x q is the reciprocal of x q .

r Thus, if x = (r, s > 0), then s

23. If a = xy p –1, b = xy q –1 and c = xy r–1, then show that aq – r. br – p. cp – q = 1.

28. If x =

III) Negative Rational Exponent If n is a negative rational, i.e. n = 

1 8

57

-m

r   s

-

p q

p

 s q   r

m

r s       where m  Q s   r IV) Also, all the laws of indices applicable for integral index are also applicable for rational index. i.e., if a is a positive real number and m, n are rational numbers, then, n

m mn ii)  a   a

i) a m × a n =a m+n

m

m

am a iv)    m b b

m m

iii)  ab   a b

5. Radical and Radicand 1

1 x 1 y 1 z value of 1  x .1  y .1  z . 29. If a = xm+n.yp, b = xn+p.ym, c = xp+m.yn, then prove that am–n .bn–p .cp–m =1. 30. If ax = bc, by = ca and cz = ab then prove that x + y + z – xyz = –2.

We know that if y > 0 and y q  x , then we can write it as x = q y 1

Here,

q

y is called ‘Radical form’ of y q

Radical

4. Rational Exponent I)

Rational Number The fraction p/q where p and q are integers, q  0 is called a rational number..

II)

Index

q

Positive Rational Exponent Let a be a positive real number and n a positve fraction equal to p/q, i.e.,n = p/q, where p and q are positive integers, the equation xq = ap has one and only positive solution for x, given by

y

Radicand

Radical sign

p q

x = a q = a p = q th root of ap. i) In

q

y,

is called Radical Sign www.betoppers.com

8th Class Mathematics

58 q

ii)

Summative Worksheet

y is called a Radical

iii) q is called Index of the radical iv) y is called the Radicand Note : Index of a radical is always positive

3

1.

Important Results i)

n

n

iii) v)

n

n

a b

=n

a b

am = a



1 2 x

n

ii)

an = a

iv)

ab = n a n b

m n

a = mn a

If

3.

Simplify :

5. 6.

Simplify

7.

 x b  bc  x c  ca  x a  ab Simplify  c    a    b  . x  x  x 

Formative Worksheet 2 / 3    p  32. Express   q      

   

xy



1/ 7

in radical form.

, then find the value of x.

y a x yz a az   zx . y az ax a 3 5  a 2b   ab 1     . Simplify :   a3b4   a 3b 2      n 3 Simplify : ( x  y )3  ( x + y )n .

4. 31. Write the radical form of (x1/p)1/q.

p q   q  p

2.

m n

3/ 2

 1 4 Find the value of   .  81 

 

2n.6m  1.10m - n.15m+ n  2 4m.32m  n.25m  1 1

35

33. If

3

(60) 128. (108)

2

4 243.135 a, b and c.

= 2a.3b.5c, find the values of

2

 125     34. Simplify :  64 

35. If x

x x



 x x



1

m

 3 216   4  256  264  625 1

3

4

0

8.

x

, then find the value of x.

Conceptive Worksheet 2

 4 3

2x

1 2

a a 1 a b  b If a = b , show that    a . b b

a

1 1  m p n p m  n n  m 1 x +x 1 x +x 1 .  1 pm pn 1 x +x

1  3 25 1. 33. Prove that  p   q 4 2 3 2   x   11. Simplify  1 34. If x = 10 (512)3 .100 (512)3 .1000 (512)3 ....., then find q  x  p the value of x.    3

2  1  3 2  1 , then prove that

x3 + 3x = 2.

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n

 1  1  p   . p   q  q  m n Simplify  1  1 . q  . q      p  p 

 32 .

2

35. If x =

1

10. Show that

  1  12  8 31. Simplify and express  4    in radical form.   x  

32. Find the value of x, if

9.



1  p-q q 1 p 1   q p .x .x  .   

2  n  9n32  3 2   27 n   12. If m – n = 1, then . 3 m 3 23

Exponents & Radiclas

59

2 x  1.4 x  1 13. Solve : = 64. 81  x

1 1 1 12. If (4.2)x = (0.42)y = 100, then prove that x  y  2 .

14. Solve : 5x.2x+1= 200 15. Solve : 22x+1 – 33.2x–1 + 4 = 0 16. If xp = yq = (xy)pq, then prove that p + q = 1

13. If ax – 1 = bc, by – 1 = ca, cz – 1 = ab, then prove that xy + yz + zx = xyz.

1 2 17. If x = 2  2 3  2 3 then prove that x3 – 6x2 + 6x – 2 =0 18. If 2x = 4y = 8z and 5x = p(y + z), then find p.

14. If 64a =

1  2 2 then prove that 3a + 4b = 0. 256b 15. If x2 = y + z, y2 = z + x and z2 = x + y, then prove

1 1 1 that x  1  y  1  z  1 = 1.

z (2 x+ y ) . xy

19. If 3x = 5y = 75z, then find

IIT JEE Worksheet

20. If a = 2x, b = 4y, c = 8z and ayz. bzx. cxy = 46, then prove that xyz =2.

HOTS Worksheet 2n 2

n2

n 1

1.

If (a )  (a ) , then prove that

2.

Solve 22 x 1.23 y 1  8 ; 2 x  2.2 y  2  16 .

n2  2 .

1

a 

a b

3

1 3

  a 

3.

If x =

4.

show that x3+3bx–2a=0. If 9x – 10.3x + 9 = 0, then find x.

2

a b

3

1 3

,

then

5. 6.

7.

8.



1 1 5

Find the value of [(10)150 ÷ (10)146]. If (18)3.5  (27)3.5 × 63.5=2x , then find the value of x.

4. 5.

 256 

 32 



.

x3

a b If      , then find the value of x . b a If m and n are whole numbers such that mn = 121, then find the value of (m – 1)n+1 .

Find the value of

1 1 a

 n-m 



1 1  a

m-n 

Find the value of  b+c-a 

 c+a-b 

 a+b-c 

 xb   xc   xa  . a  . b  .  c x  x  x  If 3(x–y) = 27 and 3(x + y) = 243, then find the x value. n

 243 5  32 n 1 9n  3n 1

? 1

1 34   1  3 3  Find the value of 5  8  27   .    



a, b. 9.

3  4

2. 3.

 216 

a

a

1

Find the value of

a 1  a b 6. If a = b ,    a b and a = 2b then find the b 7. value of b. If x and y are real numbers such that (x2 – 3y2) = 2 and (3x2 – y2) = 11, then find x2 + y2 8. n2 2 m m n If 2  (2 ) , then find m in terms of n. 9.  2   a 1     a 2  b    If = 1 and a + b = 5, then find 10. 2 ( a  b )  2 ab

b



1.

x 1

2

2  3

x n If x x x  x n x , then find x.

 

10. If a + b + c = 0 , then prove that a 2 – bc = – (ab + bc + ca).

1 1 2   , then prove that a2 + b2 = b+c c+a a+b 2c2 .

11. If

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60

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8th Class Mathematics

By the end of this chapter, you will understand • • • • • • 1.

Basic terms • Polynomial • Degree of Polynomial Classification of Polynomials Arithmetic operations of polynomials Factorization

Remainder Theorem Factor Theorem

Introduction

2. Introduction

In mathematics, a polynomial is a finite length expression constructed from variables (also known as indeterminates) and constants, using the operations of addition, subtraction, multiplication, and constant non-negative whole number exponents.

i)

Example: x2 – 4x + 7  a polynomial Polynomials are one of the most important concepts in algebra and throughout mathematics and science. They are used to define polynomial functions, which appear in settings ranging from basic chemistry and physics to economics, and are also used in calculus. History About 4000 years ago, Babylonian scholars solved quadratic equations and system of equations. But, the literal notation which we use today in algebra was not known to them. The first abridged notation for unknown quantities are found in the writings of the ancient Greek mathematician, Diophantus. Neither the Babylonians nor the Greeks considered negative numbers. The Chinese were acquainted with negative numbers and irrational numbers. The Hindu scholars made extensive use of abridged notation for unknown quantities and their powers. The founder of algebra as a special branch of learning was Central Asian scholar Mohammed of Korezme.

Chapter -56

Polynomials

Learning Outcomes

Variable A symbol which takes various values is known as a variable or literal. Any letter can be used to denote a variable. Example: x, y, q, etc. ii) Constant A symbol having a fixed numerical value is called a constant. Example: 3, 25, 8, etc. iii) Coefficient In the product of a variable and a constant, each is called the coefficient of the other. Example: 3x. Here, 3 is the coefficient of x and also x is called the coefficient of 3. iv) Term Numbers and variables are multiplied to form a product which is known as the Term. Also, a constant or a variable is also known as a term. Example: 3x, 4, y, 5y, etc. v) Expression or Algebraic Expression A combination of constants and variables by the four basic mathematical operations (+, –, ×, ¸) is called an Expression or an Algebraic Expression. Example - 1: 1

3x2 + 4x – 3, 4 x 3  3 x  5 , 4x–2 + 3x + 5 Example - 2: In the expression 3x2 + 4x – 3 , x is the variable, 3, 4 are coefficients, –3 is the constant and the total number of terms in the expression are three.

Important Points The powers of variables of an algebraic expression can take any real number. Every term is an expression. Example: 3x = 3x – 0x5 + 0x7 Every constant is a term as well as an expression.

8th Class Mathematics

62 Example: 2 = 2 × x0 2 = 2 × x 0 + 0 × x 2 – 0 × x3 Every variable is a term as well as an expression. Example: y=y×1 y = y + 0 × y3 – 0 × y4

3. Polynomial An expression of the form p(x) = a0 + a1x + a2x2 + ……. + anxn where powers of variable are non–negative integers, and coefficients are any real numbers (a0, a1, a2,……..an Î R ). is called a polynomial in x of nth degree over real numbers. Example:(x4 – 3x2 + 7), (4x3 + 3x), . . . . etc. (x4 – 3x2 + 7), (4x3 + 3x), . . . . etc. Note: Thus, an expression with non-negative, non-fraction powers and real coefficients is called a polynomial. General form of a Polynomial Polynomials with their standard or general forms and their respective degrees are given as follows: S.No

General form of a Polynomial

Example

Degree of a Polynomial

1 2

a ax + b

3 4x + 5

0 1

3 4

ax2 + bx + c ax 3 + bx 2 + cx + d

3x2 – 5x + 7 x2 – 6x2 + x + 1

2 3

5

ax 4 + bx 3 + cx 2 + dx + e

x 2 – 3x 3 + 2x2 + x + 2

4

---

-------

-------

---

n–1

n

n+1

n

n –1

ax + bx

+ cx

n –2

+ – – –+ px + q

4. Degree of a Polynomial i)

Degree of a polynomial in one variable In the case of a polynomial in one variable, the highest power of the variable is called the degree of the polynomial. Example:

7 is a polynomial in x of degree3. 8 ii) 3x5 – 7 is a polynomial in x of degree 5. Degree of a Polynomial in two or more Variables In the case of a polynomial in more than one variable, the sum of the powers of the variables in each term is taken up and the highest sum so obtained is called the degree of the polynomial. Example: i) 3x2 – 5x3y2 + 9y is a polynomial of degree 5 in x and y. ii) 7x2yz3 + 4xy2z4 – 2x2y is a polynomial of degree 7 in x, y and z. i) 4x3 – 7x +

ii)

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n

24x + 42x

+ – – 4x +2

iii) Zero of a Polynomial A real number a is a zero of a polynomial P(x) if P(a) = 0. Example: Let P(x) = x2 – 5x + 6 Putting x = 2 Þ P(2) = 22 – 5 (2) + 6 = 4 – 10 + 6 = 0 Putting x = 3 Þ P(3) = 32 – 5 (3) + 6 = 9 – 15 + 6 = 0.  2, 3 are the zeros of the polynomial, P(x) = x2 – 5x + 6 Note: 1) The number of zeros of a polynomial is equal to the degree of the polynomial. 2) Every linear polynomial in one variable has a unique zero..

Polynomials

63

5. Classification of a Polynomial i)

Based on different criteria, we have different types of classifications as: Classification based on Number of Terms S.No Given expressions No. of terms

1

7x, –15x4, 24

1

Monomial

2

7x + 15;

2

Binomial

3

6x2 – 7x + 15;

3

Trinomial

4

Polynomial

3

4 ii)

Name

ax + b – 3y2 + 4y – 12

2

3

2

x + 6x + 4x + 15; 2y + 8y – 3y – 22

Classification based on Types of Coefficients: S.No

Given expressions

Types of coefficients

Name

1

3x 2  2x  3

Real numbers

Polynomial over real numbers.

2

5x – 6x + 3

Integers

Polynomial over integers

3

3 3 4 2 x  x  2x  1 4 3

Rational numbers

Polynomial over rational numbers.

2

iii) Classification based on Degree:

S.No

Expression

Degree

Name

1

3 7 5  7x,  x 2 2

1

Linear polynomial

2

2x – 3x + 7

2

Quadratic polynomial

3

3x – 4x + 5x

3

Cubic polynomial

4

Bi-Quadratic

2

2

4

4 5

3

3

2

4x 3x + 5x + 2

5, 7, 9

Like Terms and Unlike Terms

i)

Before we see the arithmetic operations related to polynomials, let us know about like terms and unlike terms in a polynomial. Like Terms: Terms with the same variables and which have the same exponents are called like terms. Example: 3x3, –7x3,

0 Constant polynomial ii)

Unlike Terms: Terms with different variables and which have different exponents are called unlike terms. Example:

4 2 xy . . . . etc. are unlike terms. 2 Now, let us know about the arithmetic operations of polynomials. x2y, – 5x3,

1 3 x . . . . . . etc. are like terms. 5 www.betoppers.com

8th Class Mathematics

64

Formative Worksheet 1.

ii ) P  x  = 2x +1, x =

Which of the following expressions are polynomials and which are not? i) 3x3 –

4 5 x+ 3 2

2 3 ii) 3x 3  5x  4x +

8.

iii) 5 y 2  3 y 3  17

9.

iv) 7x3 – 3x2 + 5 x  11 1

v) 5x 2 y 2 – 3xy3 + 19x2y3 vi) 3x2y2 – 5xy3 + 12y4

5 vii) 2x – 3x + – 8 x Write the following in their standard form : i) 4u2 + u – 3u6 + 2 ii) y2 – y4 + 3y – 2y3 + 5 iii) 3x5 + 4x7 + 2x2 + 9x + 2 iv) – x3 + x6 + x4 – x5 + x – x2 + 1 v) 14x4 – 6x2 + 4x + 2x3 + 2 vi) 3ax2 – ax3 + x3 + 2ax – 3x4 + 4 vii) x6 + x2 + x4 – x3 – x + x5 – 1 viii) 4x4 – 5x7 + 2x8 – x5 + x2 + 9x Note : Polynomials are said to be in the standard form when their terms are arranged in the descending order of the powers of the variable. Identify the following as monomial, binomial or trinomial expressions: i) 7u6 ii) y2 – 2y + 6 iii) x2 4

2.

3.

3

iv) x6 – 4.

5.

3 x v) x3 – 3x + 2 vi) u2 + 2u Write down the degree of each of the following polynomials: i) – 498 ii) 3 – y2 + y7 – 14y4 iii) 4x2y3 – 3x3y + 14x2y2 – 8 iv) x3y2z – 3x2y2z2 + 5xy4z – y7 + x3y5 + 17z5 Write the following monomials in the ascending order according to their degrees: 1 11 x , – 5x7, 4, 5x10, 2x3, x, 9x5, 3x2, a 5 b) 6x15, 4x2, 9, x, 5x6, 7x9, 3x3 Which of the following ‘x’ values is a zero for the given polynomials?

Conceptive Worksheet 1.

i ) P  x  = 2x + , x = www.betoppers.com

4 5

Which of the following are not polynomials? (i) 2 – 3x

(ii)

2y3  3y (iii) x 2  x x

7 2 x 9 (v) u 1/ 2  3u  2 6 Which of the following expressions are not polynomials? 2 (iv) x 

2.

(i) 3x3 – 2x + 1 (ii) x 2  x x  2 (iii) x4 + 2x (iv) 4x2 + 3x – 5 3.

4.

5.

(v)

2x  3 Find the degrees of the following polynomials: i) x2 + 4x2 + 4x3 + x ii) t4 – 3t3 – 6t + 2 iii) 7x8 + 8x12 + x5 + 9x2 + 4x + 3 iv) 12x11 + 15x9 + 2x3 + 11x4 + 3x Write the following polynomials in standard form (i) y2 – 2y4 + 3y – y3 + 4 (ii) 5x – 4x3 + x4 + 4 (iii) x5 + x2 – 3x +2x4 (iv) 2t – t4 + 6 – 3t3 (v) 6u + u4 + 2 – 3u2 Write the following monomials in the ascending order of their degrees.

1 11 x , – 5x7, 4, 5x10, 2x3, x, 9x5, 3x2 5 b) 6x15, 4x2, 9, x, 5x6, 7x9, 3x3 Find the values of the following monomials when x = 2, 3, – 1.5 a)

a) 6.

-1 3 If the values of the polynomials ax3 + 3x2 – 13 and 2x3 – 5x + a are equal at x = 2, find the value of a. If –2 is a zero of the polynomial ax3 + bx2 + x – b and the value of the polynomial is 4 at x = 2, then find the values of a and b. Write the simplified forms of the following in ascending and descending orders. i) 2x2 + 2 – 5x + 7x2 – 20x + 5 + 18x – 6x2 ii) 4 + xy2 + 2.7x2y + 5x3y +xy2 – x2y – 1.4x3y iii) P  x  = 3x +1, x =

7. 5 2

1 2

6.

1 3 x (iv) 2x3 2 Find the zero of the polynomial P(x) = 2x + 5. (i) 3x2

7.

(ii) –1.2x2

(iii)

Polynomials

6. Arithmetic Polynomials

65

Operations

of

The following are the basic arithmetic or Mathematical operations which can be performed on a polynomial: i. Addition ii. Subtraction iii. Multiplication iv. Division

Addition of Polynomials Two or more polynomials can be added in two ways. They are: i. Horizontal method ii. Vertical method i) Horizontal Method To add any two polynomials by this method, the following steps are to be followed: Let’s add p(x) = 3x3 + 4x4 + 2x2 + x + 2 and q(x) = 7x2 – 5x4 + 2x + x3 - 3 Step - I : Arrange the terms of the given polynomials according to the descending order or the ascending order of their powers. p(x) = 4x4 + 3x3 + 2x2 + x + 2, q(x) = – 5x4 + x3 + 7x2 + 2x - 3 Step - II : Now, group the terms retaining their signs. p(x) + q(x) = (4x4 + 3x3 + 2x2 + x + 2) + (– 5x4 + x3 + 7x2 + 2x - 3 ) = (4x4 – 5x4 ) + (3x3 + x3) + (2x2 + 7x2) + (x + 2x) + (2 – 3) Step - III : Add the coefficients of like terms. (4x4 – 5x4 ) + (3x3 + x3) + (2x2 + 7x2) + (x + 2x) + (2 – 3) = (– x4 ) + (4x3) + (9x2) + (3x) + (– 1)  P(x) + q(x) = – x4 + 4x3 + 9x2 + 3x - 1 ii) Vertical Method To add any two polynomials by this method, following steps are to be followed: Let’s add p(x) = 3x3 + 4x4 + 2x2 + x + 2 and q(x) = 7x2 – 5x4 + 2x + x3 - 3 . Step - I : Arrange the terms of the given polynomials according to the descending order or the ascending order of their powers. p(x) = 4x4 + 3x3 + 2x2 + x + 2, q(x) = – 5x4 + x3 + 7x2 + 2x - 3

Step - II : Now, place p(x) and q(x) one below the other such that like terms must be one below the other, Add the coefficients of like terms.

 P(x) + q(x) = – x4 + 4x3 + 9x2 + 3x - 1 Subtraction of Polynomials Similar to addition of polynomials, we can subtract polynomials, using two methods. They are: i. Horizontal method ii. Vertical method Note : In subtraction, the signs of a polynomial to be subtracted are changed and added to the other polynomial i) Horizontal Method To subtract any two polynomials by this method, the following steps to be followed: Let’s subtract q(x) = x2 + 2x4 + 2x + 3x3 - 6 from p(x) = x3 + 4x4 + 2x2 + x + 1. Step - I : Arrange the terms of the given polynomials according to the descending order or the ascending order of their powers. p(x) = 4x4 + x3 + 2x2 + x + 1, q(x) = 2x4 + 3x3 + x2 + 2x - 6 Step - II : Now, group the terms retaining the signs of the first polynomial and changing the signs of the second polynomial. p(x) – q(x) = (4x4 + x3 + 2x2 + x + 1) – (2x4 + 3x3 + x2 + 2x - 6) = 4x4 + x3 + 2x2 + x + 1 – 2x4 – 3x3 – x2 – 2x + 6 = (4x4 – 2x4 ) + (x3 – 3x3) + (2x2 – x2) + (x – 2x) + (1 + 6) Step - III : Add the coefficients of like terms. (2x4 ) + (– 2x3) + (x2) + (– x) + (7)  P(x) – q(x) = 2x4 – 2x3 + x2 – x + 7 ii) Vertical Method To subtract any two polynomials by this method, the following steps are to be followed: Let’s subtract q(x) = x2 + 2x4 + 2x + 3x3 - 6 from p(x) = x3 + 4x4 + 2x2 + x + 1 Step - I : Arrange the terms of the given polynomials according to the descending order or the ascending order of their powers. p(x) = 4x4 + x3 + 2x2 + x + 1, q(x) = 2x4 + 3x3 + x2 + 2x - 6 Step - II : Now, place p(x) and q(x) one below the other such that like terms must be one below the other; change the signs of the polynomial to be www.betoppers.com

8th Class Mathematics

66 subtracted, Add the coefficients of like terms.

 P(x) – q(x) = 2x4 – 2x3 + x2 – x + 7 Multiplication of Polynomials Two or more polynomials can be multiplied in two ways. They are: i. Horizontal method ii. Vertical method i) Horizontal Method To multiply any two polynomials by this method, following steps are to be followed: Let’s multiply p(x) = x2 + x + 1 from q(x) = 2x - 3 Step - I : Multiply each term of p(x) by each term of q(x). p(x) × q(x) = (x2 + x + 1) × (2x - 3) = x2(2x - 3) + x(2x - 3) + 1(2x - 3) Step - II : Now, group the like terms, retaining their signs and add the coefficients of like terms x2(2x - 3) + x(2x - 3) + 1(2x - 3) = 2x3 + (- 3x2 + 2x2 ) + (- 3x + 2x) - 3 = 2x3 + (- x2) + (- x) - 3 = 2x3 - x2 - x - 3  P(x) × q(x) = 2x3 – x2 – x – 3 ii) Vertical method To multiply any two polynomials by this method, following steps to be followed: Let’s multiply p(x) = x2 + x + 1 from q(x) = 2x - 3 Step 1 : Place the given two polynomials one below the other. x2 + x + 1 2x - 3 Step 2 : Every term of first expression should be multiplied by first term of second expression.

Step 3 : Every term of first expression should be multiplied by second term of second expression.

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Step 4 : Continue until all the terms of second expression are multiplied, Add the products.

 P(x) × q(x) = 2x3 – x2 – x – 3 Example: Let us find the product of 3x2 – 2x + 1 and 2x2 – 3x + 4, using vertical method

1.

2.

3.

4.

In this method, we write the multiplicand and the multiplier in descending powers of x. Arrange the terms one under the other, and multiply the multiplicand by every term of the multiplier, and add. Multiplication - Important Identities Let’s see some identities and their proof: (a + b)2 = a2 + 2ab + b2 Proof: (a + b)2 = (a + b)(a + b) = a(a + b) + b(a + b) = a2 + ab + ab + b2 = a2 + 2ab + b2 (a - b)2 = a2 - 2ab + b2 Proof: (a - b)2 = (a - b)(a - b) = a(a - b) - b(a - b) = a2 - ab - ab + b2 = a2 - 2ab + b2 (a + b)(a – b) = a2 – b2 Proof: (a + b)(a - b) = a(a - b) + b(a - b) = a2 - ab + ba - b2 = a 2 - b2 (a + b)3 = a3 + b3 + 3ab(a + b) Proof: (a + b)3 = (a + b)(a + b)2 = (a + b) (a2 + 2ab + b2) = a(a2 + 2ab + b2) + b(a2 + 2ab + b2) = a3 + 2a2b + ab2 + ba2 + 2ab2 + b3 = a3 + 3a2b + 3ab2 + b3 = a3 + b3 + 3ab(a + b)

Polynomials (a – b)3 = a3 – b3 – 3ab (a – b) Proof: (a - b)3 = (a - b)(a - b)2 = (a - b) (a2 - 2ab + b2) = a(a2 - 2ab + b2) - b(a2 - 2ab + b2) = a3 - 2a2b + ab2 - ba2 + 2ab2 - b3 = a3 - 3a2b + 3ab2 - b3 = a3 - b3 - 3ab(a - b) 6. a3 – b3 = (a – b) (a2 + ab + b2) Proof: Taking RHS (a – b) (a2 + ab + b2) = a(a2 + ab + b2) - b(a2 + ab + b2) = a3 + a2b + ab2 - ba2 ? ab2 ? b3 = a3 - b3 = LHS 7. a3 + b3 = (a + b) (a2 – ab + b2) Proof: Taking RHS (a + b) (a2 - ab + b2) = a(a2 - ab + b2) + b(a2 - ab + b2) = a3 - a2b + ab2 + ba2 ? ab2 + b3 = a3 + b3 = LHS 8. a3 – b3 = (a – b) (a2 + ab + b2) Proof: Taking RHS (a – b) (a2 + ab + b2) = a(a2 + ab + b2) - b(a2 + ab + b2) = a3 + a2b + ab2 - ba2 ? ab2 ? b3 = a3 - b3 = LHS 9. (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca 10. a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca) 11. If a + b + c = 0, then a3 + b3 + c3 = 3abc Important Points The following are some points to be noted in multiplying polynomials: i) The general rule is that each term in the first factor has to multiply each term in the other factor. ii) The number of products you get has to be the number of terms in the first factor times the number of terms in the second factor. Example: A binomial times a binomial gives four products, while a binomial times a trinomial gives six products. iii) Be very careful and methodical to avoid missing any terms. iv) After the multiplication is over, you can try to collect like terms and write them side by side and simplify the result. Multiplication of polynomials can be done in two ways: Here, p(x) is called the multiplicand and q(x) the multiplier. 5.

67

Formative Worksheet 10. If A = 3x3 – x + 4x2 – 1 B = 5 – 3x2 + 4x3 + 4x C = 4 – 5x + x2 – 2x3 find: (i) A + B ; B + A (ii) B + C ; C + B (iii) (A + B) + C; A + (B + C); (A + C) + B 11. Simplify (i) (3x3 – 4x2 + 5) – (2x3 – 2x2 + 3) (ii) (3x3 – 2x2 – 3x – 3) – (x3 – 2x2 + 3x – 4) (iii) (7x5 – 6x4 + 5x3 – 4x2 + 3x – 2) – (2x5 – 2x + x3 – x2 + x – 5) 12. If A = 5x3 – 3x2 + 2x – 5 B = 3x3 – x2 – x + 2 C = x3 – 5x + 4 Find (i) A – B ; B – A (ii) (A – B) – C; A – (B – C) From the results obtained state if subtraction of polynomials is (iii) Commutative (iv) Associative 13. (a) What should be added to 2x2 + 6x – 5 to get 3x2 – 2x + 6? (b) What must be subtracted from a2 + b2 + c2 – 3abc to get 2a2 – b2 – 3c2 + abc? 14. Multiply the following a) x2 + y2 + z2 and xy – yz + zx + 3 b) x2 + 3x + 1 and x2 + 3x + 2 c) 4p2 + 2p – 5 by 9p2 – 4p + 2 d) y2 + 2y + 2 by 6y2 + 3 e) m3 + 4m2 + 2m + 1 by 4m3 + m2 + 5m + 9 15. Multiply 2x2 – 3x + 4 by 3x2 – 2x + 1 16. If p(x) = x2 – 2x + 1 and q(x) = x3 – 3x2 + 2x – 1, find p(x) × q(x) and find the degree of p(x) × q(x). 17. Write the following products using appropriate identity. 1  i)  x   x 

2

ii) (n – 1) (n + 1) (n2 + 1) 2

1  iii)  2x   x  18. Use the appropriate identity to find the value of 2

 1 i)  50   2 ii) 107 × 93 = (100 + 7) (100 – 7) = …….… iii) (96)2 = (100 – 4)2 = ............

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8th Class Mathematics

68 19. If x 

1  5, find the x

2 value x 

1 1 and x4  4 2 x x

1 20. Using a2 + b2 + c2 – ab – ca = {(a – b)2 + 2 2 2 (b – c) + (c – a) } Prove that if a2 + b2 + c2 = ab + bc + ca, then a = b = c. 21. If a + b = c, show that a3 + b3 + 3abc = c3

i)

ii)

iii)

Conceptive Worksheet 8.

9.

10. 11.

12. 13.

14. 15. 16.

Simplify the sums given below. (i) (–3x2) + (6x2) – (0.5x2) + (1.5x2) (ii) (–3x) + (–4x) – (–4.5x) + (2.5x) (iii) (3x) + (–4x) – (–3x) + (–7x) (iv) (–5x2) + (5.2x2) + (1.5x2) – (0.7x2) Find the sum of the following: (a) P(x) = x2–2x + 1 and Q(x) = x3–3x2–2x –1 (b) P(x) = 4t3 – 3t2 + 2, Q(x) = t4 – 2t3 + 6 and R(x) = t3 + 4t2 – 4 What sould be added to 2x2 + 6x – 5 to get 3x2 – 2x + 6? a) By how much is a2 – 2ab – b2 more than 2a2 + 2ab – 2b2? b) By how much is 2(a2 + b2) less than a2 + 2ab + b2 ? Multiply (y2 –5y + 2) by (3y2 + 2y – 5). If A = 3x2 – 4x + 5, B = x2 – 2x – 1, C = 3x – 4, find (i) A × B, B × A (ii) C × A, A × C (iii) B × C, C × B State whether the multiplication of polynomials is commutative. If a + b = 8, ab = 15, find the value of a3 + b3. If a + b + c = 0, show that a3 + b3 – 3abc = – c3. If x + y +z = 0, show that x3 + y3 + z3 = 3xyz.

1 1  3 find the value of x 3  3 . x x 18. If a + b + c = 5, a2 + b2 + c2 = 29, find the value of ab + bc + ca. 17. If x 

Division of Polynomials Two polynomials can be divided in two ways, they are: i. Long division method or direct division method ii. Division using identities i) Long Division Following are the steps followed in long division method Let’s understand this method by the division of (– 2x2 + 3x3 + x – 3) by (x2 + 2x) www.betoppers.com

iv)

v)

Arrange the terms of the dividend and the divisor in descending order of their degrees. (3x3 – 2x2 + x – 3) ? Dividend (x2 + 2x) ? Divisor Divide the first term of the dividend by the first term of the divisor to obtain the first term of the quotient. x2 + 2x ) 3x3 – 2x2 + x – 3 ( 3x Multiply all the terms of the divisor by the first term of the quotient and subtract the result from the dividend. x2 + 2x ) 3x3 – 2x2 + x – 3 ( 3x 3x3 + 6x2 – – ___________ – 8x2 + x Consider the remainder as a new dividend and proceed as before.

Repeat this process till we obtain a remainder which is either ‘0’ or a polynomial of degree less than that of the divisor.

Some points to be noted i) If any terms are missing in the dividend or divisor, leave spaces and treat them as terms with coefficient zero. Example: If we are dividing a polynomial like (x2 + 1) by (x + 4) In dividend (x2 + 1), x term is missing. So, we can write this as (x2 + 0x + 1) ii) The degree of the quotient is equal to the degree of the dividend minus the degree of the divisor.

Division - By using Identities The division of polynomials is also represented as p( x) , where, p(x) and q(x) are two polynomials q( x) in ‘x’.

Polynomials

69

The division can also be done by using following identities: 1. (a + b)2 = a2 + 2ab + b2 2. (a – b)2 = a2 – 2ab + b2 3. (a + b) (a – b) = a2 – b2 4. (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca 5. (a + b)3 = a3 + b3 + 3ab(a + b) = a2 + 3a2b + 3ab2 + b3 6. (a – b)3 = a3 – b3 – 3ab (a – b) = a3 – 3a2b + 3ab2 – b3 7. a3 + b3 = (a + b) (a2 – ab + b2) 8. a3 – b3 = (a – b) (a2 + ab + b2) 9. a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca) 10. If a + b + c = 0, then a3 + b3 + c3 = 3abc

Examples of Division Example -1: Divide a4 - b4 by a + b 4

Given

a b a b

2

4



 a2   b2 

2

a b Which is in the form of identity, a2 - b2 = (a + b)(a – b) where a = a2, b = b2  RHS  2



(a 2  b 2 )( a 2  b2 ) a b

2

(a  b )(a  b)(a  b) a b

Example -3: Divide (x3 + x2 + 2x + 8) by (x + 2) x3  x 2  2 x  8 x2 By rearranging the terms and simplifying by applying the identity a3 + b3 = (a + b) (a2 – ab + b2)

Given

x 3  x 2  2 x  8 ( x 3  8)  ( x 2  2 x )  x2 x2 

( x 3  23 )  x ( x  2) x2

( x  2)( x 2  x (2)  22 )  x ( x  2) x2 By taking (x + 2) common, we get 



( x  2) ( x 2  x(2)  22 )  x  x2

= x2 - 2x + 22 + x

= x2 - x + 4 Example - 4: Divide (x2 + y2 + 2(xy + yz + zx)) by (x + y + 2z) Given

x 2  y 2  2( xy  yz  zx) x  y  2z x 2  y 2  2 xy  2 yz  2 zx x  y  2z

(a4  b4 )   ( a 2  b 2 )(a  b) ab



(a4  b4 ) a b are (a2 + b2) and (a + b ) Which is in the form of identity, a 2 – b 2 = (a + b)(a – b) Example -2: Divide (x3 - 64) by (x ? 4) Given

Taking out 2z common in the last two terms, we get:

The factors of

3

3

x 3  64  x    4   x4 x4 Which is in the form of identity, a3 – b3 = (a – b) (a2 + ab + b2) where a = x, b = 4  RHS  

( x  4)( x 2  ( x)4  42 ) = x2 + 4x + 16 x4

R.H .S . 

x 2  y 2  2 xy  2 z ( y  x) x  y  2z

Applying the identity (a + b)2 = a2 + 2ab + b2, we get:

R.H .S . 

( x  y ) 2  2 z ( y  x) x  y  2z

Taking out (x + y) common, we get: R.H .S . 



( x  y) ( x  y)  2 z  = (x + y) x  y  2z

x 2  y 2  2( xy  yz  zx)  x y x  y  2z

( x 3  64) = x2 + 4x + 16 x4

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8th Class Mathematics

70

Formative Worksheet 22. Divide (x2 + x2 – 2x – 5) by (x + 2) Divide (x2 + x2 – 2x – 5) by (x + 2) Here both dividend and divisor are in descending powers of x. In case they are not so, write them in descending powers of x. 23. Divide : x2 + 8y3 by x + 2y 24. Divide : x7 – 64x by x(x2 – 2x + 4) 25. Divide : x7y– xy7 by (x + y) (x2 – xy + y2) 26. Divide : 4(p – q)2 – 12 (p – q) (p + q) + 9(p + q)2 by p + 5q 27. Divide : p4 – 4q4 by p –

2q

28. Divide : 8p3 – 36p2q + 54 pq2 – 27q3 by 4p2 – 12pq + 9q2

Conceptive Worksheet 19. Go through the division and verify your answer by using division algorithm. (x3 – 5x2 + 11x – 10)  (x – 2) 20. Go through the division and verify your answer by using division algorithm. (8x4 – 8x3 + 15x – 10x2 +2)  (4x2 + 2x – 3) 21. Divide : x4 + x3 – 9x2 – 3x + 5 by x2 + 4x + 2. 22. Divide a3 + a2 + 2a + 8 by a + 2 23. Divide : 45x2 – 25y2 + 30y – 9 by 2x + 5y – 3 24. Divide x6 – 729 by x2 – 9 25. Divide 4x2 – 17xy + 4y2 by x – 4y 26. Divide x7 + xy6 by x2 + y2

7. Factorization When an algebraic expression can be expressed as the product of two or more expressions, then each of these expressions is called a factor of the given expression Example: We know that: (x2 - 36) = (x + 6)(x - 6) Clearly, (x + 6) as well as (x - 6) is the factor of (x2 -36). Thus, the process of writing an algebraic expression as the product of two or more expressions is called factorization.

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i) ii)

Methods of Factorization There are four methods of factorizing any polynomial. They are: i) Factorization by taking out common factor ii) Factorization by regrouping terms iii) Factorization by using identities iv) Method of splitting middle term Factorization by taking out Common Factor In this method of factorizing a polynomial, we have two cases, they are: When each term of the given polynomial contains the same monomial factor. When a polynomial is a multiplier of each term of the given olynomial. Case - I: When each term of the given polynomial contains the same monomial factor. Let us understand the method or procedure with an example Example: Factorize 6a2 – 8ab + 4a Step - I: Find highest factor that divides all the terms in the given expression. In 6a2 – 8ab + 4a, clearly ‘2a’ is the highest factor that divides all the terms. Step - II: Take the highest common factor as common from all the terms. i.e. by taking ‘2a’ common from all the terms, we get: = (2a)(3a – 4b + 2) Therefore, the two factors of the given expression are (2a) and (3a – 4b + 2) Factorization by taking out Common Factor - Example Let us factorize 6xy2 – 9x2y + 24xy by taking out the common factor method: If we observe 6xy2 – 9x2y + 24xy, we get ‘3x’ is the highest common factor that divides all the terms of the given expression. i.e. we can take ‘3x’ common term from all the terms of the given expression. = (3x)(2y2 + 3xy – 8y) Therefore, the two factors of 6xy2 - 9x2y + 24xy are (3x) and (2y2 + 3xy – 8y) Case - II: When a polynomial is a multiplier of each term of the given polynomial. In this case, we take out the common multiplier and use the distributive law. Let’s understand by factorizing 3x(a + 2b) – 2y(a + 2b): In this expression, we have (a + 2b) as common multiplier.

Polynomials

71

So, we can take out (a + 2b) common:  (a + 2b) (3x – 2y)  The two factors of the given expression are (a + 2b) and (3x – 2y). Factorization by Regrouping terms Sometimes in a given expression, it is not possible to take out a common factor directly. In such cases, we have to make a suitable arrangement of terms and group them in a such manner as to have a common polynomial. Example: Let’s understand by factorizing x2 – y + xy – x In this expression, we cannot take out any common term directly. So, x2 + xy – y – x  x(x + y) – 1(x + y)  The two factors of the given expression are (x – 1) and (x + y). Factorization by Regrouping terms Example Let’s factorize x3 - x2 – xy + x + y – 1 x3 - x2 – xy + x + y – 1 can be written as: (x3 - x2) – (xy – y) + (x – 1)

S.No 1 2 3

x2(x - 1) – y(x – 1) + 1(x – 1) By taking out (x - 1) common, we get: (x -1) (x2 - y + 1)  The two factors of the given expression are (x – 1) and (x2 – y + 1). Factorization using Identities Before we see the examples, let us have a look at identities we learnt: 1. (a + b)2 = a2 + 2ab + b2 2. (a – b)2 = a2 – 2ab + b2 3. (a + b) (a – b) = a2 – b2 4. (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca 5. (a + b)3 = a3 + b3 + 3ab(a + b) = a2 + 3a2b + 3ab2 + b3 6. (a – b)3 = a3 – b3 – 3ab (a – b) = a3 – 3a2b + 3ab2 – b3 7. a3 + b3 = (a + b) (a2 – ab + b2) 8. a3 – b3 = (a – b) (a2 + ab + b2) 9. a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca) 10. If a + b + c = 0, then a3 + b3 + c3 = 3abc 11. (x + a)(x + b) = x2 + (a + b) + b2 The factors of the given polynomial, when it is in the form of identities, are given below:

Form of polynomials

Factors

2

2

(a + b), (a + b)

2

2

(a  b), (a  b)

a + 2ab + b a – 2ab + b 2

a –b

2

2

(a + b), (a  b) (x + a), (x + b)

2

4

x + (a + b) + b

5

a + b + 3ab(a + b) or a + 3a b + 3ab + b

6

a – b – 3ab (a – b) or a – 3a b + 3ab – b

3

3

2

2

2

3

(a + b), (a + b), (a + b)

3

3

3

2

2

3

2

2

(a  b), (a  b), (a  b) (a + b + c), (a + b + c)

3

3

3

3

3

3

2

7

a + b + c + 2ab + 2bc + 2ca

8

a +b

9

a –b

10

a + b + c  3abc

2

2

2

2

(a + b), (a – ab + b ) (a  b), (a + ab + b ) (a + b + c), 2 2 2 (a + b + c – ab – bc – ca)

3

Factorization using Identities - Examples i) Let us factorize 4x2 + 12x + 9 using an identity: The given 4x2 + 12x + 9 can be written as (2x)2 + 2 × (2x) × (3) + (3)2 i.e, it is in the form of a2 + 2ab + b2, where a = 2x, b = 3.  4x2 + 12x + 9 = (2x + 3)2  The two factors of 4x2 + 12x + 9 are (2x +3), (2x +3)

ii)

Let us factorize 1 - 8x + 16x2 using an identity: By arranging in decreasing order, we get; 16x2 - 8x + 1 It can be written as (4x)2 - 2 × (1) × (4x) + (1)2 i.e, it is in the form of a2 - 2ab + b2, where a = 4x, b = 1.  1 - 8x + 16x2 = (4x - 1)2  The two factors of 4x2 + 12x + 9 are (4x – 1), (4x – 1)

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8th Class Mathematics

72 iii)

iv)

Let us factorize 4x2 - 81y2, using an identity: The given 4x2 - 81y2 can be written as (2x)2 - (9y)2 Clearly, it is in the form of a2 - b2, where a = 2x, b = 9y. We know that (a2 ? b2) = (a + b)(a ? b)  4x2 - 81y2 = (2x + 9y)(2x - 9y)  The two factors of 4x2 - 81y2 are (2x + 9y), (2x - 9y) Let us factorize x3y3 + 729 using an identity: The given x3y3 + 729 can be written as (xy)3 + (9)3 Clearly, it is in the form of a 3 + b 3 , where a = xy, b = 9. We know that (a3 + b3) = (a + b)(a2 ? ab + b2)  x3y3 + 729 = (xy + 9)(x2 - 9xy + 81)  The two factors of x3y3 + 729 are (xy + 9), (x2 - 9xy + 81) Factorization using Identities – Special case From the above examples, it is clear that the given polynomial can be expressed in the form of some known identity and its factors are found directly. In some cases, the given polynomial cannot be expressed and factorized using identities directly. We need to add and subtract some quantity or take some factor outside the brackets required. Regroup the terms so that the polynomial is expressed in the form of an identity and find the factors A special case - An example Let’s factorize a4 + a2b2 + b4. Given a4 + a2b2 + b4 Here, the first and last terms are perfect squares in the form of a2 and b2 Whereas the middle term is not exactly in the form of 2ab. By adding and subtracting a2b2 (the result remains same), we get, a 4 + a 2 b2 + b 4 = a 4 + a 2 b2 + b 4 – a 2 b2 + a 2 b2 = a4 + 2a2b2 + b4 – a2b2  [(a2)2 + 2 × a2 × b2 + (b2)2]= (ab)2  (a2 + b2)2 – (ab)2 Now, this is in the form of a2 – b2  (a2 + b2)2 – (ab)2 = (a2 + b2 + ab) (a2 + b2 – ab) (a2 – b2) = (a + b) (a – b))  The two factors of a4 + a2b2 + b4 are (a2 + b2 + ab) and (a2 + b2 – ab)

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1.

2.

3.

Factorization - Method of Splitting the Middle Term This method is basically used to factorize trinomials only. Let’s see the steps are to be followed by factorizing 2x2 + 7x + 5: We need to multiply x2 with constant term and let P be the product obtained. By multiplying x2 and constant term, we get:  2x2 × 5 = 10x2  P = 10x2 Split the middle term into two terms, such that their addition or subtraction results in middle term and their product is equal to P. The middle term 7x must be split such that their addition or subtraction results in 7x and their product is equal to P. 5x + 2x = 7x (5x) × (2x) = 10x2 = P Express the resultant polynomial into two terms by taking common term. We can write 2x2 + 7x + 5 as: 2x2 + 5x + 2x + 5 For taking common, let’s interchange 5x and 2x:  2x2 + 2x + 5x + 5 Now, by taking 2x common from (2x2 + 2x) and 5 from (5x + 5), we get: 2x(x + 1) + 5(x + 1)  The two factors of 2x2 + 7x + 5 are: (2x + 5) and (x + 1) Let us factorize x2 + 8x + 15 by splitting the middle term method: By multiplying x2 and constant term, we get:  x2 × 15 = 15x2  P = 15x2 The middle term 8x must be split such that their addition or subtraction results in 8x and their product is equal to P. 3x + 5x = 8x (3x) × (5x) = 15x2 = P  x2 + 8x + 15 = x2 + 3x + 5x + 15 Now, by taking x common from (x2 + 3x) and 5 from (5x + 15), we get: = x(x + 3) + 5(x + 3)  The two factors of x2 + 8x + 15 are: (x + 3) and (x + 5)

Polynomials S.No

73 About Polynomial

An identity to use 2

2

1

If the polynomial can be reduced to (a + b) = a + 2ab+ b 2 and (a – b) 2 = a2 – three terms, two of which are squares. 2ab + b 2 may be useful.

2

If the polynomial can be reduced to an expression, three of whose terms are squares.

(a + b + c) = a + b + c + 2ab + 2bc + 2ca may be useful.

3

If the polynomial can be reduced to a form of difference of two squares.

a2 – b 2 = (a + b) (a – b) may be useful.

4

3 3 3 If the polynomial can be reduced to four terms such that i) two terms are (a + b) = a + b + 3ab (a + b) (or) the cubes of some expressions 3 3 3 2 2 ii) The remaining two terms are the (a – b) = a – b – 3a b + 3ab may be useful. multiples of 3.

5

If the polynomial is of the for (x + bx (x + a) (x + b) = x + + c) and you can find two constants p & q such that b = p + q, adn c = pq

6

If the polynomial can be reduced to a + b = (a+b) (a –ab+b ) form of sum of difference of two a3 – b 3 =(a–b)(a2 + ab+ b 2) cubes.

7

If the polynomial can be reduced to a + b + c – 3abc = (a + b + c) 2 2 2 four terms such that the first three (a + b + c – ab – bc –ca) terms are cubes of some expressions and the remaining term is –3 times the product of the three expressions.

2

2

2

2

2

3

3

3

3

Formative Worksheet 29. Factorize the following polynomials. i) 36y3z + 48y2z2 ii) 15y2z3 – 20y3z4 + 35y2z2 iii) 14m5n4p2 – 42m7n3p7 – 70 m6n4p3 30. Factorize : (i) 6(2a + 3b)2 – 8(2a + 3b) (ii) x(x – y)3 + 3x2y (x – y) 31. Factorize : (i) x3 – x2 – xy + x + y – 1, (ii) ax – (ax + by)2 + a2x + aby + by 32. Factorize : (x2 + 2x)2 – 3(x2 + 2x) – y(x2 + 2x) + 3y. 33. Factorize : x2 – z2 – 2xy + 2yz 34. a4 + 4(a – 1)2 – 4 (a3 – a2) 35. Factorize the following polynomials i) x2 + 6x + 9 ii) 1 – 8x + 16x2 iii) 4x2 – 81y2 iv) x3 y3 + 729

2

36. 37. 38.

39. 40. 41. 42. 43. 44.

2

(a + b)x + b

2

2

3

v) x3 – 6x2 + 12x – 8 vi) x3 + y3 – z3 + 3xyz vii) n4 + 4 Factorize : (2x + 3y)2 + 2(2x + 3y) (x + y) + (x + y)2 Factorize : a2 b2 + c2 d2 – a2 c2 – b2 d2 Factorize : (i) 9 – a6 + 2a3 b3 – b6 (ii) x16 – y16 + x8 + y8 (iii) (p + q)2 – (a – b)2 + p + q – a + b Factorize : (x + y)3 + z3 Factorize : x2 + 7x + 12 Factorize : 7x2 – 8x – 12 Factorize x4 – 2x3 + 2x2 – 2x + 1. Factorize a3 – 2a2 – 5a + 6. 1 Factorize : x 2  2  3 x

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8th Class Mathematics

74

Remainder = – 29. Let’s verify the Division Algorithm : [ (x – 3) × (x2 – 6x – 8)] + (– 29) = x3 – 6x2 – 8x – 3x2 + 18x + 24 – 29 = x3 – 9x2 + 10x – 5 = Dividend  Dividend = [(Divisor) × (Quotient)] + remainder.

Conceptive Worksheet 27. Factorize : (i) 3x (a + 2b) – 2y (a + 2b) (ii) 2a (a2 + 1) – 3 (a2 + 1) 28. Factorize the following expressions : (i) a2 – b + ab – a (ii) xy – ab + bx – ay (iii) 6ab – b2 + 12ac – 2bc (iv) a (a + b – c) – bc (v) a2 x2 + (ax2 + 1) x + a (vi) 3ax – 6ay – 8by + 4bx 29. Factorize : (i) a2 + 2a + ab + 2b (ii) x2 – xz + xy – xz

9. Remainder Theorem

General form of Division Algorithm Let us consider two polynomials of p(x) and g(x) such that g(x) ¹ 0, where p(x) is the dividend and g(x) is the divisor. Let q(x) be the quotient and r(x) be the remainder. Then, by division algorithm, we have: p(x) =[g(x) × q(x)] + r(x) An example Let’s divide x3 – 9x2 + 10x – 5 by x – 3

Divisor = x – 3 Dividend = x3 – 9x2 + 10x – 5 Quotient = x2 – 6x – 8

i) ii) iii)

If polynomial f(x) is divided by a linear polynomial (x – a), then the remainder is f(a). Let us find the remainder when 27x3 – 9x2 + 11x – 4 is divided by x – 2. Given that f(x) = 27x3 – 9x2 + 11x – 4 is divided by x – 2. Then by remainder theorem, the remainder is f(2).  f(2) = 27(2)3 – 9(2)2 + 11(2) – 4 = 27 × 8 – 9 × 4 + 22 – 4 = 216 – 36 + 22 – 4 Therefore, the remainder = 198. Conditions for remainder theorem: In the polynomial, power ‘x’ should be a positive integer. Degree of remainder(R) < Degree of divisor g(x). This method is not applicable if the divisor is a nonlinear polynomial, i.e., more than one degree polynomial. Note: If a polynomial f(x) is divided by (x – a), then the remainder is given by equating x – a to zero. That is, x – a = 0 or x = a. That means, we have to substitute a for x in f(x). Remainder on division of f(x) by (x – a), (x + a), (ax – b) and (ax + b) are shown in the table below:

Note:This method is not applicable if the divisor is a non-linear polynomial, i.e., it must be a one degree polynomial.

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Polynomials

75

Remainder theorem – Examples Let’s find the remainder when f(x) = x3 – 2x2 + 3x + 2 is divided by i) x – 1 i) Given f(x) = x3 – 2x2 + 3x + 2 Divisor = (x – 1) If we equate (x – 1) to zero, we get x = 1. By substituting x = 1 in f(x), we get, f(1) = (1)3 – 2(1)2 + 3 (1) + 2 =1–2+3+2 =6–2 = 4 (remainder) ii) x + 2 ii) Given, divisor = (x + 2) If we equate (x + 2) to zero, we get x = – 2 By substituting x = – 2 in f(x), we get, f(2) = (–2)3 – 2(–2)2 + 3(-2) + 2 =–8–8+6+2 = – 16 + 8 = – 8 (remainder) iii) x – 1/2 iii) Given, divisor = (x – 1/2) If we equate (x – 1/2) to zero, we get x = 1/2 By substituting x = 1/2 in f(x) 2

2

1 1 1 1 f       2    3   2 2 2 2       2



1 1 3  2    2 4 4 2



1  2  6  8 13  (remainder ) 4 4

10. Factor Theorem Let f(x) = 3x2 – 7x + 4 be a polynomial. Let’s find a factor of a f(x) without actual division. By trial and error method, substituting x = 1 in f(x). Þ f(1) = 3(1)2 – 7(1) + 4 =3–7+4 =–4+4=0 i.e., remainder = 0 We can say that, (x – 1) is a factor of f(x). What is this method called ? If a polynomial f(x) is divided by a linear polynomial (x – a) and the remainder is zero (f(a) = 0), then (x – a) is said to be a factor of f(x). Let’s check if x2 – 5x + 6 is a factor of 2x4 – 17x3 + 49x2 – 52x + 12 or not. x2 – 5x + 6 = (x – 2) (x – 3)

If (x – 2) is a factor, then f(2) should be zero. f(2) = 2(2)4 – 17(2)3 + 49(2)2 – 52(2) + 12 = 32 – 136 + 196 – 104 + 12 = 0 \ (x – 2) is a factor. Similarly, if (x – 3) is a factor, then f(3) should be zero. f(3) = 2(3)4 – 17(3)3 + 49(3)2 – 52(3) + 12 = 162 – 459 + 441 – 156 + 12 = 0 Therefore, f(x – 3) is a factor. Hence (x – 2)(x – 3) = x2 – 5x + 6 is a factor of the given expression. Some Interesting points 1. The degree of the remainder is always less than the degree of the divisor. 2. The dividend should be of the degree 1. 3. If the degree of the divisor is greater than the degree of the dividend, then the quotient is zero. 4. If, for an expression f(x), f(0) = 0, then x is a factor of f(x) 5. If, for an expression f(x), f(a) = f(b) = 0, then x – a and x – b are factors of f(x)

Formative Worksheet 45. Divide : y3 + y2 – 2y + 1 by y – a and verify the division algorithm property. 46. Find the remainder when p(x) = 2x3 – 5x2 + 3x + 7 is divided by i) 2x – 1 ii) x – 2 iii) x + 2 47. If g(a) = a6 – 3a4 + 24a2 – 3, find g(–1), g(–2) and g(–3). 48. The polynomial P(x) = kx3 + 9x2 + 4x – 8, when divided by x + 3, leaves a remainder – 20. Find the value of k. 49. If one factor of (x + 1)7 + (2x + k)3 is (x + 2), find the value of k. 50. The polynomials kx3 + 3x2 – 3 and 2x3 – 5x + k, when divided by x – 4, leave the same remainder in each case. Find the value of k.

Conceptive Worksheet 39. Find the quotient when x4 – a4 is divided by x – a. 40. If x3 + 3x2 + 3x + 1 is divided by x +  , find the remainder. 41. Find the remainder, when 2x3 – x2 + 3x – 2 is divided by 2x – 1. 42. Find the value of k if kx2 – 3x – 2 is divisible by x – 2 43. Find the remainder when x4 + 15x3 + 6x2 – 12x + 3 is divided by x + 2. 44. What must be added to x3 + 3x2 – 2x + 4, so that it may be exactly divisble by 2x2 + 3x – 6? www.betoppers.com

8th Class Mathematics

76 45. What must be i) subtracted from x3 – 2x2 + 7x – 9 so that it may be exactly divisible by x2 –5x + 3 ? ii) added to x3 – 2x2 + 7x – 9 so that it may be exactly divisible by x2 – 5x + 3?

9. 10.

Summative Worksheet 1.

8.

(a) Write the following polynomials in standard form. (i) y2 – 2y4 + 3y – y3 + 4 (ii) 5x – 4x3 + x4 + 4

3 x + 2x4 (iv) 2t – t4 + 6 – 3t3 (v) 6u + u4 + 2 – 3u2 (b) In the following, identify monomials, binomials and trinomials : (i) x (ii) m2 + 2m (iii) 3xy + y + 2x

11. 12.

(iii) x5 + x2 –

5 3 6 (v) t  t  6 3 Which of the following are not polynomials and why? (iv) 3x2 + 3y

2.

2 (i) 3x + 5 (ii) x2 + 5x + 4 (iii) u 

(iv)

3y 2  11y  6 3

(vi) 4x2 + 5x–3 + 2

3.

4.

1 5 u

(vii) x2 + 2x + 1

1 10 6x3,  x , –3x5, 8x, 2x11, x7 5

5. 6. 7.

14.

1 2 4 9 x , x , 5 3 8x5, 7x6, 14x12, 2x16, x3 Find the zero of the polynomial P(x) = 3x. Find the value of the polynomial 5x – 4x2 + 3 at x = –1. Collect like terms and simplify the expression : 12m2 – 9m + 5m – 4m2 – 7m + 10

16.

17.

18.

19. 20. 21. 22. 23. 24.

x, 4x4, 9x7,

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1 1 3  5 find the value of x  3 . x x Find the following products without actual multiplication. (i) (3x – 4y + z) (3x – 4y – z) (ii) (5a + 3b + 2c) (5a – 3b – 2c) Find the division and verify your answer by using division algorithm. (2x3 – 7x2 + 8x + 5)  (2x – 3) Find the division and verify your answer by using division algorithm. (3x4 – 8x3 – 2 + 10x2 – 8x)  (3x2 – 2x + 5) Divide: x2 – 16x + 64 by x – 8 Divide : x4 –74x2 – 1131 by x2 + 13 Divide : 64x2 – 81z2 by 8x – 9z Divide : x9 – y9 by x3 – y3 Divide : 27a3 – 125b3 by 3a – 5b Factorize: (i) ax + bx + ay + by (ii) ax2 + by2 + bx2 + ay2 (iii) a2 + bc + ab + ac (iv) ax – ay + bx – by Factorize the following expressions: (i) a3x + a2(x – y) –a (y + z) –z (ii) (x2 + 3x) – 5 (x2 + 3x) –y (x2 + 3x) + 5y Factorize : (x + y) (1 – z) – (y + z) (1 – x) Factorize : (x + y) (a + bz) – (y + z) (a + bx)

15. If x 

(v) x–3 – 3x2 + 6

(viii) x3 + 2x2 + 5x +4 (ix) x 2  2 x  1 (x) y2 – 2y4 + y Find the degrees of each of the following : (i) 3x2 + 2x + 1 (ii) x + 1 (iii) 2y + 3 – 5y3 (iv) u7 – 3u5 + 2u (v) y4 + 2y3 – 3y2 + 8 Write the following monomials in the descending order of their degrees : a) 4x2, 5x9, –7x14,

b)

13.

From the sum of 2y2 + 3yz and – y2 – yz – 2z2, yz + 2z 2 subtract the sum of 3y 2 – z 2 and –y2 + yz + z2. Subtract 24ab – 10b – 18a from 30ab + 12b + 14a. (a) What should be added to 3x2 – 2x + 6 to get 2x2 + 6x – 5 ? (b) What must be subtracted from a2 + b2 + c2 – 3abc to get 2a2 – b2 – 3c2 + abc? Multiply (2x2 – 3x + 4) by (3x2 – 2x + 1) Find the products: (i) (3x2 – 5x + 4) × 2x (ii) (6x2 – 4x + 3) × 4x3 (iii) (3x2 – 5x + 6) × (4x – 3) (iv) (x2 + x + 1) (x2 – x + 1) Expand the following (i) (2a + b)3 (ii) (3a + 4b)3 (iii) (2a – b)3 (iv) (a2 – 4)3 (v) (101)3 (vi) (1002)3 3 (vii) 98 If a – b = 2, ab = 15, find the value of a3 – b3.

25.

26. 27.

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77

28. Factorize the following expressions : (i) 9x2 – 4y2 (ii) 36x2 – 12x + 1 – 25y2 (iii) a2 – 1 + 2x – x2 29. Factorize : x6 – 1. 30. Factorize : (i) 4x2 – 4xy + y2 – 9z2 (ii) 16 – x2 – 2xy – y2 (iii) x4 – (x – z)4 31. Factorize : 8x3 – 60x2 + 150x – 125 32. If a3 + b3 + c3 = 3abc, show that either b + c = 0 or a = b = c. 33. Factorize : 6x2 + 7xy + 11xz + 7yz + 3z2 .

4.

5.

a+

1 34. If the value of a  4  119 , find the value of a 1 . a3 Find the value of k if (x + 1) is a factor of x8 + kx3 – 2x + 1. Factorize : (5x + y)2 + (5x + y) (x + 2y) – 20 (x + 2y)2 . Find the value of k if the expressions p(x) = kx3 + 4x2 + 3x – 4 and q(x) = x3 – 4x + k leave the same remainder when divided by x – 3. If x15 + ax13 + 5 is divisible by x + 1, find the value of a. If (x2 + x + 1) divides x4 – 5x2 – bx – 5 exactly, find the other factor. Find the remainder when 6x3 – 5x2 + 2x – 4 is divided by 2x + 1 .

35. 36. 37.

38. 39. 40.

HOTS Worksheet 1.

2.

3.

If A = 2x + 3, B = 3x – 5, C = x2 + 3x – 1 find (i) A × (B × C) (ii) (A × B) × C State whether the multiplication of polynomials is associative. Products must be in the form of a3+b3+c3–3abc. One expression is given. Find the second expression and also the result. (i) (2a + b – c) (ii) (3a – 2b + 4) 2 2 (iii) (4x + y + 25 – 2xy – 5y – 10x) The following products should be either in the form a3 + b3 or a3 – b3. Find by how much the following expressions should be multiplied to get that form. Also write the products. (i) 2x + 1 (ii) 3x – 2y (iii) 4x2 – 6x + 9 (iv) 16x2 + 20x + 25

1) a  b 2) b  c 3) a  b  c 4) a = b= c If a, b, c are three consecutive numbers, find a2 + b2 + c2 – (ab + bc + ca + 3).

6.

If x  2  1 , y  1  3 and z  3  2 ,find x3 + y3 + z3 – xyz.

7.

Solve x3 (y – z) + y3 (z – x) + z3 (x – y) + (y – z) (z – x) (x – y) (x + y + z). 2

8.

1 1   Factorize :  5x    4 5x    4, x  0 x x  

9.

Factorize : ax – (ax – by)2 + a2x + aby + by.

4

a3 

If a2 + b2 + c2 – bc – ca – ab = 0, then which of the following is true?

10. Factorize 2 2p3  8q 3  27r 3  18 2pqr 11. Factorize 8p3 – 27q3 – 36p2q + 54pq2 12. Factorize 2 2a 3  16 2b 3  c  12abc 13. Factorize : 12(p2 + 7p)2 – 8(p2 + 7p) (2p – 1) – 15 (2p – 1)2 14. Write down the quotient without acutal division. (8a3 + 27b3 – 125 + 90ab)  (2a + 3b – 5) 15. Find the quotients without actual division. (64x3 – 125)  (16x2 + 20x + 25)

3 for all 2 values of x. If P(5) = 2002, then find P(8).

16. P(x) is a polynomial satisfying P  x   x 

17. Find the remainder, when x100 is divided by x2 – 3x + 2. 18. Let L 1 and L 2 be the remainders when the polynomials x3 + 2x2 – 5ax – 7 and x3 + ax2 – 12x + 6 are divided by (x + 1) and (x – 2) respectively. If 2L1 + L2 = 6, find the value of a. 19. Find the quotient and remainder respectively when x2002 – 2001 is divided by x91. 20. Find the value of ‘k’ for which x3 – 7x + 5 is a factor of x5 – 2x4 – 4x3 + 19x2 – 31x + 12 + k. 21. What must be added to 6x4 – x3 + 3x2 + 4x – 2, so that the result is exactly divisible by (3x2 – 2x + 4)? Also obtain the expression that is exactly divisible. 22. For what value of k is 9l3 – 6l2 + 3l – k exactly divisible by l2 – 2l + 3?

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8th Class Mathematics

78

IIT JEE Worksheet If 4x4 – (a – 1) x3 + ax2 – 6x – 1 is divisible by 2x – 1, find the value of ‘a’. 2. Find the value of ‘p’ for which (x + 1) is a factor of x4 + (p – 3) x3 – (3p – 5) x2 + (2p – 9) x + 6. 3. When p(x) = ax3 + 3x2 – 13 and q(x) = 2x3 – 5x+a divided by (x + 2) give the same remainder, find the value of ‘a’. 4. If p(x) = ax3 + bx2 + x – 6 has (x – 1) as a factor and leaves a remainder 6, when divided by (x+1), find values of a and b. 5. If p(x) = ax3 + bx2 + 4x – 2 has (x + 2) as a factor and leaves a remainder 4, when divided by (x– 2), find the values of a and b and write p(x). 6. If x4 + 4x3 + 6px2 + 4qx + r is exactly divisible by x3 + 3x2 + 9x + 3, find the value of r(p + q). 7. When x94 – 7x19 + 5x4 – x + k is divided by x –1, the remainder is –1. Then the remainder obtained when it is divided by x + 1 is a) 0 b) 15 c) –1 d) 10 e) none of these 8. If f(x) = x3 + ax + b is divisble by (x – 1)2, then the remainder obtained when f(x) is divided by x + 2 is a) 1 b) 2 c) 3 d) – 1 e) none of these 9. What is the remainder when x11 is divided by 1 + x + x2 + ....... + x10? 10. One of the factors of (x – b)5 + (b – a)5 is a) a – b b) x – b c) x – a d) x + a 11. If x3 – (a + 4) x2 + (4a + c) x + d is divisible by x – a, then a is equal to _________. 12. If x15 + ax11 + 3 is divisible by x + 1, then it follows that ‘a’ is equal to a) 2 b) – 2 c) 4 d) – 4 1.



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Linear equations and Inequations Chapter - 76

Learning Outcomes

By the end of this chapter, you will understand    

Simple Linear equation Types of equations Types of linear equations Simultaneous linear equations

 Inequations  Types of inequations  Linear Inequation

1. Introduction Leonhard Euler (1707 – 1783) of Switzerland was arguably the greatest mathematician of the eighteenth century (his closest competitor for the title was Lagrange) and one of the most prolific of all time.

Basic Terms i) Variable A symbol which takes various values is known as a variable or literal. Any letter can be used to denote a variable. Example: x, y, q, etc. ii) Constant A symbol having a fixed numerical value is called a constant. Example: 3, 25, 8, etc.

Leonhard Euler

Euler’s important contributions were so numerous that terms like “Euler’s formula” or “Euler’s theorem” can mean many different things depending on the context. In mechanics, Euler’s equations for motion of fluids, and the Euler-Lagrange equation is used in calculus of variations. In the earlier classes, we have learnt the meaning of an equation and its solution. Most of the equations that we have worked with are integer coefficients and integer solutions. In this chapter, we shall deal with equations involving rational numbers as the coefficients and their solutions can also be rational numbers.

iii) Coefficient In the product of a variable and a constant, each is called the coefficient of the other. Example: 3x. Here 3 is the coefficient of x and also x is called as the coefficient of 3. iv) Term Numbers and variables are multiplied to form a product which is known as the Term. Further, a constant or a variable is also known as a term. Example: 3x, 4, y, 5y, etc. v) Expression (or ) Algebraic Expression A combination of constants and variables by the four basic mathematical operations (+, –, ×, ) is called an Expression or an Algebraic Expression. 1

Example: 3x2 + 4x – 3, 4x 3  3 x  5 , 4x–2 + 3x +5 In the expression 3x2 + 4x – 3 , x is the variable, 3, 4 are coefficients, –3 is the constant and the total number of terms in the expression are 3.

8th Class Mathematics

80

Equation

ii) Same number can be subtracted from both sides of an equation. Let x – 7 = 7x – 5 be an equation. We can subtract ‘3’ from both sides. i.e., (x – 7) – 3 = (7x – 5) – 3. iii) Both sides of an equation can be multiplied by the same non-zero number. Let 2x – 5 = 7x – 2 be an equation, we can multiply by ‘5’ on both sides. i.e., 5(2x – 5) = 5(7x – 2). iv) Both sides of an equation can be divided by the same non-zero number. Let x – 7 = 3x – 9 be an equation, we can

Let us consider two expressions 2x + 7; 7x + 11 Now, let’s connect these by the equality sign. 2x + 7 = 7x + 11 we call such statements equations. A statement of equality of two algebraic expressions is called an equation. An equation must involve at least one variable. Examples: i) 10x + 100 = 1000 ii) x2 + 2x + 5 = 0 ii) 7p + 5 = 11p – 2.

2. Simple Linear Equation

divide by ‘3’ on both sides. i.e.,

An equation which can be expressed in the form ax + b = 0, where a  0, is called a simple linear equation in the variable x. Thus, ax + b = 0, where a  0, is called the standard form of a simple linear equation. Examples: i) 5x – 3 = 9 – 5x ii) 2(y – 5) = 7

v) Any term of an equation may be taken to the other side with the sign changed. This process is called “transposition”. Let 7x – 9 = 3x + 7 By transposing ‘– 9’ to RHS and ‘3x’ to LHS, we get. 7x – 3x = 7 + 9   4x = 16

y 3 9

iii)

2y  5 

iv)

p  1 7p  5 7   6 9 2



Any value or values of the variable or variables when substituted in an equation makes both its sides equal, is called a solution (or root) of the equation. Thus,  is a root of the equation ax + b = 0.  a + b = 0. Note: Solving the equation means finding all its solutions (or roots ).

This process is called cross multiplication.

Formative Worksheet 1. 2. 3.

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Find out whether the solution x = 2 and y = 4 satisfy the equation x + y = 5. Solve 6x + 3 = 8x + 10. Solve the following equation (find out the solution sets of the following equations). (i)

Rules for solving an Equation i) Same number can be added to both sides of an equation. Let 2x + 5 = 7x – 1 be an equation. We can add ‘2’ on both sides, i.e., (2x + 5) + 2 = (7x – 1) + 2.

x =4

ax  b p vi) If cx  d  q , then q (ax + b) = p(cx + d)

Note: A first degree equation is called a Linear equation.

Solution of a Linear equation

x  7 3x  9  3 3

2x 10  (ii) 9 – 7x = 5 – 3x (iii) 4x = 15 – 3x 5 3

Conceptive Worksheet 1. 2.

Find the solution set of numbers such that x + y = 8. Nine times a number is 108. What is the number?

Linear Equations & Inequations

81

3. Types of Equations

1 5 7 2 or . Let’s take x  , , 7 3 9 7

Based on the degree, equations are of the following types: S.No

Degree of

 1 1 7   2  0  1 – 2 = – 1 : 7  7 LHS  RHS. For x 

Name of equation

an equation 01 02

1 2

Linear equation Quadratic equation

03 04

3 4

Cubic equation Bi – quadratic equation

5 5 7   2  0  For x  : 3 3 LHS  RHS.

7 7 49 7   2  0  : 20 9 9 9   LHS  RHS. For x 

Note: We have separate names for equations up to fourth degree. If the degree of an equation is greater than 4, then the equation is called the respective degree equation Examples: x5 + x4 + x3 + x2 + 1 is 5th degree equation. 10x10 + 100x100 + 1000 = 0 is 100th degree equation.

2 , the value of the expressions on 7 either side of the equation become equal. Such a value of the variable (x) is called the solution of the equation. Thus, when x 

Basically linear equations are of the following types: i) Linear equation in one variable. ii) Linear equation in two variables. Let us observe the degree and number of variables in the following equation: 5p – 7 = 3 Clearly, the degree of the above equation is one; i.e., it is a Linear equation. And number of variables is one, i.e., ‘P’. Linear Equation in one Variable: An equation of the form, ax + b = 0 or ax = c is a linear equation in one variable x, where a  0 and a, b, c are real numbers. Examples:  i) 2x + 7 = 0 ii) 4y – 7 = 0 iii) 9p + 3 = 0

Formative Worksheet 4. 5.

6.

15 – 7x = 9. 4 Bansi has 3 times as many two-rupee coins as he has five-rupee coins. If he has in all a sum of Rs 77, how many coins of each denomination does he have? Deveshi has a total of Rs 590 as currency notes in the denominations of Rs. 50, Rs 20 and Rs 10. The ratio of the number of Rs 50 notes and Rs 20 notes is 3:5. If she has a total of 25 notes, how many notes of each denomination has she? Solve

7.

Solve 5x 

8.

Solve

7 3  x  14 2 2

x  a  b  c

Solution for Linear Equation in One Variable Let us consider a linear equation 7x – 2 = 0 Let us find for which values of x, both sides of the equation 7x – 2 = 0 are equal.

2 7   2  0  2 – 2 = 0  7

2 : 7 LHS = RHS For x 

4. Types of Linear Equations

Linear Equation in one Variable

35 20 3

bc

9.

Solve

10. Solve



x  b c  a  ca



x  c a  b  ab

3

x 2  9 5  for positive value of x. 9 5  x2

 x  2 2x  3  2x 2  6 x 5

 2.

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8th Class Mathematics

82 11. A father is 7 times as old as his son. Two years ago, x x x the father was 13 times as old as his son. What are 11. Solve :    7 2 3 4 their present ages?

3t  2 2t  3 2 xb xb   t 12. Solve :  4 3 3 ab ab 13. The numerator of a fraction is 4 less than the 1 1 denominator. If 1 is added to both the numerator 13. Solve:  0. 2x  1 3x  1 and denominator, the fraction becomes 1/2. Find the 14. A number is 56 greater than the average of its third, fraction. quarter and one-twelfth. Find the number. th  4 15. Arvind has a Piggy bank. It is full of one-rupee and 14. Divide 34 into two parts in such a way that   7 fifty-paise coins. It contains 3 times as many fifty paise coins as one rupee coins. The total amount of th  2 the money in the bank is Rs 35. How many coins of of one part is equal to   of the other.. 5 each kind are there in the bank? 15. There are 90 multiple choice questions in a test. Linear Equation in Two Variables Suppose you get two marks for every correct Let us observe the degrees and number of variables answer and for every question you leave in the following equation. unattempted or answer wrongly, one mark is 2x – 7y = 5 deducted from your total score of correct answers. If you got 60 marks in the test, how many questions Since the degree is one, it is a linear equation. did you answer correctly? But the number of variables (x and y) are two. What do we call this type of linear equation ? 12. Solve :

Conceptive Worksheet x 5 3   . 3 2 2

3.

Solve :

4.

Solve: 3 +

5.

Solve:

4x  5 2  2x  7  3  .  2 6 3

6.

Solve:

1 9 {4a (1 + x – (a – x)} 3 4

x2 = 10. 6

1 16 = {3a (1 – x) – (a + x)}. 4 3 7. A chemist has one solution containing 50% acid and a second one containing 25% acid. How much of each should be mixed to make 10 litres of a 40% acid solution? 8. A boy covers a distance of 25 km. in 4 hours partly on foot at the rate of 3.5 kmph and partly on cycle at 9 kmph. Find the distance covered on foot. 9. On dividing a number by 8, the quotient is 11 and the remainder is 6. Find the number. 10. Two numbers are in the ratio 3 : 8. If the sum of the numbers is 165, then find the numbers.

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Such an equation is called linear equation in two variables. Linear Equations in Two Variables An equation of the form ax + by + c = 0 or ax + by = c, where a, b, c are real numbers; a  0, b  0 and x, y are variables, is called a linear equation in two variables. Examples: i) x + 3y = 5 ii)

iii)

2p  3q  10

3p 7q   15 3 11

Note: If in the equation ax + by = c, we have a = 0, b  0 or b = 0, a  0, then the equation reduces to y = c or ax = c respectively. In either case, we get an equation in one variable. That is why it is generally assumed that both ‘a’ and ‘b’ are non-zero as given in the definition.

Linear Equations & Inequations

5. Simultaneous Linear Equations Let us try to solve the following linear equation in two variables 3x + 5y = 8  3x = 8 – 5y

8  5y Can we solve this? 3 No, we cannot. Why because, the number of unknown variables are two and there is only one equation. How many equations do we need to solve for linear equations in two variables ? We need a pair of linear equations in two variables to solve for ‘x’ and ‘y’. What do we call such a pair of linear equations? Such a pair of linear equations are called simultaneous linear equations. 

x

Example: Consider two linear equations: 2x – y = 4 ; x + 7y = 9 Such a pair of linear equations in two variables is said to form a system of simultaneous linear equations in two variables. The general form of system of linear equations in variables is: a1x + b1y + c1 = 0 a2x + b2y + c2 = 0 Where a1  0, b1  0, a2  0, b2  0 and a1, b1, c1, a2, b2, c2 are real numbers.

Solutions of Simultaneous Linear Equation Consider the following system of linear equations: 2x + 5y = 12 3x + 7y = 17 When x = 1 and y = 2, let us observe both the equations: 2(1) + 5(2) = 12  2 + 10 = 12  12 = 12  LHS = RHS 3(1) + 7(2) = 12  3 + 14 = 17  17 = 17  LHS = RHS Thus, the pair of values x = 1 and y = 2, which satisfy both the equations is said to be the solution of the system of linear equations.

83

Methods to solve Simultaneous Linear Equations Basically, there are three methods to solve simultaneous linear equations. They are i) Substitution Method ii) Elimination by equating the coefficients.

Substitution Method The following steps are to be followed in this method: Step 1: Pick either of the equations and write one variable in terms of the other. i.e. express x in terms of y. Step 2: Substitute the value of ‘x’ in second equation. We will get the value of ‘y’. Step 3: Substitute the value of ‘y’ in either of the equations; we get the value of ‘x’.

Substitution Method procedure

- General

The general procedure involved in solving the system of equations. a1x + b1y = C1 and a2x + b2y = C2 Step 1: Express x in terms of ‘y’ from the first equation. i.e., x 

C1  b1 y a1

Step 2: Substitute the value of ‘x’ from step 1, in the second equation, to get the value of ‘y’. Step 3: Substitute the value of ‘y’ from step 2, in the equation obtained in step 1, to find the value of ‘x’. Hence, the solution of the system of equations is the value of ‘x’ and ‘y’.

Example: Let the following be the simultaneous linear equations: 7x – 15y = 2 __________ (1) x + 2y = 3 __________ (2) i) Let us take equation (2); write one variable in terms of the other. i.e., x + 2y = 3 x = 3 – 2y __________ (3) ii) Substituting the x value (i.e (3)) in (1), we get; 7(3 – 2y) – 15y = 2  21 – 14y – 15y = 2  21 – 29y = 2  – 29y = – 19 Therefore, y 

19 29 www.betoppers.com

8th Class Mathematics

84 iii) Substituting this value of ‘y’ in equation (3),

 19  we get: x  3  2    29   29x = 87 – 38 

29x = 49

 x

49 29

Therefore, the solution is x 

49 19 , y 29 29

iii) Now, substitute the value of ‘y’ obtained from step (ii) in equation – (i) i.e., 2x + 3(3) = 13  2x + 9 = 13  2x = 4 x =2 Hence the obtained values x = 2 and y = 3 are said to be the solution of the system of equations.

Formative Worksheet

16. Solve given pair of equations using substitution method. x + y – 5 = 0, y – 2x = 2x The general procedure involved in solving the system 17. Solve given pair of equations by elimination method of equations: a1x + b1y = C1; a2x + b2y = C2 3x – y = 5, 5x – 2y = 4 Step 1: Consider coefficients of one variable (x), 18. Find the solution sets of the system of equations. say a1 and a2. x y x y   5;  4 Step 2: Multiply 1st equation by the coefficient of x 3 2 7 14 nd or y of 2 equation and vice versa. So, we get two equations having same coefficients 19. If x  y  5 ; x  y  1, then find (x + y). 4 3 12 2 for one variable. Step 3: Now, we eliminate the terms having same 4p  9q 5q p  . and p, q are positive, find coefficients, either by adding or subtracting the 20. If p pq q equations. Hence, we obtain the value of one variable. x 2 y2  = 18, x + y = 12, find (x, y). Step 4: Substituting the obtained value of the 21. If y x variable in 1st equation, we get the value of the 22. If 3x + 6y = 4xy, 9x + 8y = 7xy, find (x, y). second variable. 23. If 4x + 5y = 82, 3x + 2z = 54 and 5y + 4z = 110, find Hence the solution of the system in the values of 5x + 2y + z. ‘x’ and ‘y’ obtained. 24. The sum of the numerator and denominator of a Example: Let the following be the simultaneous fraction is 8. If 3 is added to both the numerator linear Equations. and denominator, the fraction becomes ¾. Find the 2x + 3y = 13 __________ (i) fraction. 3x + 2y = 12 __________ (ii) 25. Seven times a two-digit number is equal to four times 2, 3 respectively. the number obtained by reversing the digits. The i) Let us multiply (i) by 3; (ii) by 2. difference of the digits is three. Find the number. i.e. (i) × 3 = 3(2x + 3y) = 3(13) 26. The sum of a two-digit number and the number  6x + 9y = 39 __________ (iii) obtained by reversing the digits is 66. If the digits of 2 × (ii) = 2(3x + 2y) = 2(12) the number differ by 2, find the number. How many  6x + 4y = 24 __________ (iv) such numbers are there? ii) We made coefficients of x equal, 27. The middle digit of a number between 100 and 1000 Now, let us do (3) – (4); we get is zero, and the sum of the other digits is 11. If the 6x + 9y = 39 digits be reversed, the number so formed exceeds 6x + 4y = 24 the original number by 495; find it – – – 28. A plane left 30 minutes later than the scheduled __________ time and in order to reach the destination 1500 km 5y = 15 away in time, it has to increase the speed by 250 __________ km/hr from the usual speed. Find its usual speed.  y=3

Elimination by Equating the Coefficients

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Linear Equations & Inequations

Conceptive Worksheet 16. Find the solution sets of the systems of equations 3x – y = 5; 3x + 3y = 21 17. Find the solution sets of the system of equations 2p + q = 56 ; 2p – q = 11 18. Find the solution sets of the system of equations

x y x y   3;  4 7 15 2 5 19. If 2x + 3y = 34 and 1 

x 13  , find 5y + 7x. y 8

1 1 7 20. If x  y  12 , xy = 12, find (x, y). 4 6 21. If x + y = 4 and y + = 6, find (x, y). x 22. x + y = 9, x2 + y2 + xy = 61, find (x, y). 23. In a fraction, if numerator is multiplied by 3 and denominator is reduced by 3 we get (18/11), but if numerator is increased by 8 & denominator is doubled, fraction becomes (2/5). Find the fraction. 24. The sum of the two numbers is 50 and their difference is 10. Find the numbers. 25. Find two numbers which differ by 4, such that onehalf of the greater exceeds one-sixth of the lesser by 8.

6. Inequations Let us consider two expressions, 3x – 40, 10x + 100; Connect the above expression by inequality sign (< or  or > or  )

 3x – 40 < 10x + 100. Such mathematical equations are called inequations. A statement of inequality between two algebraic expressions involving at least one variable is called an inequation Examples: i) 2x + 3x  100 ii) x < – 32 + 3x

85

7. Types of Inequations Basically, there are two types of inequations: i) Absolute Inequations ii) Conditional Inequations.

Absolute Inequation Let us observe the following inequation: x2 > 0 Now, let’s substitute some values for ‘x’; x = – 1  (– 1)2 > 0  1 > 0 (True) 2 x = – 2  (– 2) > 0  4 > 0 (True) 2 x = – 5  (– 5) > 0  25 > 0 (True) The above inequation is true for all values of ‘x’. Such an inequation is called Absolute inequation.

An inequation which is true for all values of variable is known as absolute inequation. Examples: i) x4 > 0 ii) y6 > 0 iii) t2 > 0 i.e., any variable raised to an even power greater than zero.

Conditional Inequation Let’s observe the following inequation: x + 5 > 6 x = – 1  (– 1) + 5 > 6  4 > 6 (False) x=1  (1) + 5 > 6  6 > 6 (False) x=2  (2) + 5 > 6  7 > 6 (True) The above equation is not true for all values of x. It’s true for some limited values of x, i.e. x  2 Such an inequation is called Absolute inequation. An inequation which is not true for all values of variable is known as conditional inequation. Examples: i) x + 9 > 20, it is true for x > 11. ii) 2x – 10 < 90, it is true for x < 50.

8. Linear Inequations First degree inequation is called Linear Inequation. Examples: i) x – 7 < 1 – 13 ii) x + 27 > – 38

iii) 2x 

3 1  5 10

iv)

5x  9 2x  3  7 4

iii) 10x + 100 > 0. iv) 2008x – 2009  x + 2.

Solution of Linear Inequation

Each of the above statements is an inequation.

Let us consider the following linear inequation. 7x – 40 > 9  7x > 9 + 40  7x > 49  x>7 Thus, the given inequation is true for x > 7. www.betoppers.com

8th Class Mathematics

86 Hence the solution of the given inequation 7x – 40 > 9 is x > 7. i.e., x can take 8, 9, 10 - - - - -  . Now, the set of all the above values is the solution set. The set of values of variable which satisfy the given inequation is said to be the solution set of the inequation. Example: 2x – 5 > 11  2x > 11 + 5  2x > 16

 x  

Let us divide inequation (1) by 4 on both sides:

2x  8 12  4 4 

16 2

x>8 Solution set = {9, 10, 11 - - - - -}

2.

By adding or subtracting the same numbers to or from each side of an inequation does not change the inequality. Let us observe the following inequation x – 7 > 23 __________ (1)  x > 23 + 7  x > 30 Let us add 5 on both sides of inequation (1) x – 7 + 5 > 23 + 5 x – 2 > 28 x > 28 + 2 x > 30 Now, let us subtract 5 on both sides of the inequation __ (1) (x – 7) – 5 > 23 – 5  x – 12 > 18  x > 18 + 12  x > 30 In both cases, the solution is the same. Thus, by adding or subtracting the same numbers to or from each side of an inequation does not change the inequality. By multiplying or dividing by the same number on each side of an inequation, does not change the inequality. Let us consider the following inequation 2x – 8 > 12 __________ (1)  2x > 12 + 8  2x > 20 x > 10

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x4 3 2  x–4>6  x > 10 Now, let us multiply by 3 on both the sides: 2x – 8 > 12  3(2x – 8) > 3(12)  6x – 24 > 36  6x > 36 + 24  6x > 60  x > 10 In both the cases, the solution is the same. Thus, by multiplying or dividing by the same number on each side of an inequation, does not change the inequality. By multiplying or dividing by the same negative number on each side of an inequation, changes the inequality. Let us consider the following inequation: – 12 > – 8 _________ (1) Multiply by ‘– 4’ on both sides of the above inequation: (– 12) × (– 4) < (– 8) × (– 4) 48 < 32. But the above statement is false as 48 is not less than 32. To make the above statement true, we have to reverse the inequality. i.e. changing 48 < 32 to 48 > 32. Now, let us divide inequation (1) by ‘– 4’ on both sides: 

Properties of Inequations 1.

2  x  4  12  4 4

3.

12 8   4   4  3 2. Thus, by multiplying or dividing by the same negative number on each side of an inequation, changes the inequality.

Linear Equations & Inequations

Formative Worksheet 29. 30. 31. 32.

Solve the inequation 2 – x > – 5 Solve the inequation 2x  6, x ? N Solve 6 – 2x < – 8 over Z (integers). Find the solution set subject to the given conditions. (i) b – 1 > 9, b is a prime number less than 20. (ii) 2c – 3 < 11, c is a +ve even number. (iii) 2x > – 8, x is a –ve integer. (iv) State the smallest integer n for which 4n > 17. 33. If 5 < x < 7 and 2 < y < 6, find the least value of x – y. 34. Solve: – x + 2  4, x ? N. 35. Solve: – 3x + 2  – 4 over R.

Conceptive Worksheet 26. Find the solution set subject to the given conditions (i) 3a + 1 < 16, a is a positive integer (ii) 0 < 2x < 40, x is divisible by 5. (iii) Find the integer x such that 2x – 5 5. 31. Given P = {x : 5 < (2x – 1) < 11, x  R} Q = {x : –1  3 + 4x < 23, x  I}. Find the solution set for P & Q. 32. i) Solve 7x – 4 < 17 over N ii) Solve 2 (3 – x) < 5 (2 – x) over W. iii) Solve x – 1  2x + 3 < 9 over Z.

9.

10.

11.

12.

13.

14.

HOTS Worksheet xy = 14, x2 + y2 + xy = 84, find (x, y).

1.

If x + y +

2.

1 1 1 1    Solve x  7 x  9 x  11 x  9

15.

3.

Solve the equation,



abx acx  c b

cbx 4x  1 a a bc 0.585x  0.975 0.6

Three prizes are to be distributed in a quiz contest. The value of the second prize is five-sixths the value of the first prize and the value of the third prize is four-fifths that of the second prize. If the total value of the three prizes is Rs 150, find the value of each prize. 50 kg of an alloy of lead and tin contains 60% lead. How much lead must be melted into it to make an alloy containing 75% lead? The length of a rectangle exceeds its breadth by 4 cm. If length and breadth are each increased by 3 cm, the area of the new rectangle will be 81 cm2 more than that of the given rectangle. Find the length and breadth of the given rectangle. A father is 7 times as old as his son. Two years ago, the father was 13 times as old as his son. What are their present ages? Anil and Sunil can do a piece of work in 8 days, while Anil alone can do it in 12 days. In how many days can Sunil alone do the same work? A & B are friends, and their ages differ by 2 yrs. A’s father D is twice as old as A, and B is twice as old as his sister C. The ages of D and C differ by 40 yrs. Find the ages of A and B. If x – y = 6, x2 + y2 = 180, find (x, y).

16. If

4p  9q 5q  p p  q and p, q are both positive,

find

p q.

17. If (x + y) (y + z) = 35, (y + z) (z + x) = 42, (z + x) (x + y) = 30, find xyz. 18. If (x + y)2/3 + 2 (x – y)2/3 = 3(x2 – y2)1/3 and 3x – 2y = 39, find (x, y).

4.

Solve 0.65x +

5.

1.56 0.39x  0.78  0.2 0.9 Solve : 1.32y + 0.02y = 1.19 + y

6.

Solve :

7.

The altitude of a triangle is five-thirds its base. If 20. If x  2 3  2 3 , find the value of 2x3 + 6x. the altitude were increased by 4 cm and the base 21. If x = 89, y = 87, z = 84, find x3 + y3 + z3 – 3xyz be decreased by 2 cm, the area of the triangle would remain the same. Find the base and the altitude of 22. Two places A and B are 120km apart from each other on a highway. A car starts from A and another the triangle. from B; in the same direction, they meet in 6 hours The sum of two numbers is 2490. If 6.5% of one and if they move in opposite directions, they meet number is equal to 8.5% of the other, find the in 1 hour 12 min. Find the speeds of the cars. numbers.



8.

x  2  11  x 1  3x  4     6 4 12  3

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1 1 19. If x + y = 1 and y + = 1, find the value of z z+

1 . x 1

1

Linear Equations & Inequations 23. A rectangular field is 16m long and 10m wide. There 4. is a path of uniform width all around it, having an area of 120 sq.m. Find the width of the path. 24. A man can row 8km upstream and 24km downstream in 4 hours. He can row 12km upstream and 12km downstream in 4 hours. Find the speed of the man in still water and find the speed of the 5. current. 25. A person spent Rs. 9.10 in buying oranges at the rate of 3 for 10 paise and apples at 25 paise for a dozen; if he had bought five times as many oranges 6. and a quarter of the number of apples he would have spent Rs. 26.50. How many of each did he buy? 26. If 3x – 5 < 10 < 2x + 5, then range of x is

27. 28.

29.

30.

31. 32.

89 If two electrical conductors of resistance R1 and R2 are connected in parallel, the resultant resistance

1 1 1 ‘R’ is given by the formula R  R  R . If R1 = 1 2 24 and R2 = 12, find R. (The unit of resistance is ohm.) Find the value of ‘x’ if 1 2 3   0 x  1 x  2 x  2 x  3 x  3         x  1 .

A two-digit number ‘xy’ can be written as xy = (10 x + y). For example, 23 = (2 × 10 + 3). a) The digits of a given two digit number are interchanged. Then the sum of the given number 3 5 and the resulting number is always divisible by (A) < x < 3 (B) – 6}. Find P Q . 18. Let A = {2, 4, 6}, B = {1, 2, 3, 5} and  = {1, 2, 3, 4, 5, 6, 7, 8} Verify that: (i)  A  B  '   A ' B '  (ii)  A  B  '   A ' B '  19. If X = {1, 2, 3, 4}, Y = {2, 3, 5, 7}, Z = {3, 6, 8, 9} and W = {2, 4, 8, 10}, find  X Y    Z W  . 20. If universal set is Z, the set of integers, find the complement of the set of integers, find the complement of the set of positive integers. 21. Let A and B be two sets such that n(A) = 20,

101 22. A town has a total population of 25,000 out of which 13,000 read “The Hindustan Times” and 10,500 read “The Indian Express” and 2,500 read both papers. Find the percentage of population who read neither of these newspaper. 23. In a class of 60 boys, there are 45 boys who play cards and 30 boys play carroms. Find the number of boys who play both games. 24. In a town of 10,000 families it was found that 40% families buy newspaper A, 20% families buy newspaper B and 10% families buy newspaper C. 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers, the number of families which buy (i) A only, (ii) B only, (iii) none of A, B and C, then find the increasing order of (i), (ii), (iii). 25. In a class of 60 students, it was found that 30 students read “The Hindu” and 35 read “Indian Express” and 5 read neither. Using the Venn diagram, find how many students read both “The Hindu” and “Indian Express”. 26. For a natural number n, F(n) denotes the set of all divisors of n, except 1. Find the value of x, satisfying F(18)  F(12) = F(x). Find the least value of y,, satisfying F(20)  F(16)  F(y). 27. In a city, there are three major newspapers A, B, C of which at least two are read by 35 percent of the population. It is known that newspaper C is read by 45 percent of the population, that newspapers A and B are read by 15 percentage and that all the three are read by 10 percent. What percent of the population reads only newspaper C? 28. How many elements are there in the product set A × A, where A = {a, b, c, d}? Also write A × A 29. If n(A) = 50, n(B) = 30 and n  A  B   10 , then find n  A Δ B  . 30. If n(A – B) = 30 + x, n(B – A) = x and n  A  B   2 x and n(A) = 2 n(B), then find x. 31. Out of 80 students who appeared in a combined test in Science and Mathematics, 64 passed in atleast one subject. If 45 passed in Science and 52 in Mathematics, find: i) how many passed in both the subjects: ii) how many passed in Science only; iii)how many failed in Mathematics.

n  A  B   42 and n  A  B   4 . Find n(B) +

n(A – B) + n(B – A). www.betoppers.com

8th Class Mathematics

102 32. If  = {x : x  N and x  24 }, A = {x : x is a multiple of 3} B = {x : x is a factor of 24} and C = {x : x is a prime number} Draw a Venn diagram to show the relationship between the given set. Hence, find i) A  C ii) B  C iii) A  B  C 33. In a group of athletic teams in a school, 21 are in the basketball team, 26 in the hockey team and 29 in the football team. If 14 play hockey and basket ball; 12 play football and basketball; 15 play hockey and football and 8 play all the three games. Find (i) how many players are there in all (ii) how many play football only. 34. Match the following: Group - A Group - B 1) If A  B ; then [ A– B = 2) A – (B  C) = [

] A) (A – B)  (A–C)

3) A      A is known as 4) A – (B  C)

[

] C) (A – B)  C

[

] D) n(A) + n(B)

5) If A and B are

[

] E) 

Identification of Elements of Sets from Venn Diagram Let A, B be two sets and  is a universal set, the venn diagram of these sets is as:

The elements of set A = { a, b } The elements of set B = { a, b, s } The elements of set  = { a, b, s, m, n, o, p }

Venn Diagram of a Subset

representation

We know that, if every element of A is present in B, then we say that A is a subset of B.

] B) 

disjoint sets, then

F) law of identity

n A  B 

G) (A – B)  (A – C) H) (A – B)  C

Venn Diagram representation of Union The collection of elements of set A and set B is called the Union of two sets, A and B.

10. Venn Diagrams These are simple closed figures used for representations. These figures were first used in 1880 by John Venn, an English Mathematician. These figures were also used by Leonard Euler (1700 – 1783). So, these closed figures are called Venn-Euler diagrams or simply Venn diagrams. A set is represented by circles or a closed geometrical figure inside the universal. An element of a set A is represented by a point within the circle which represents A.

Shaded area denotes A  B.

Venn Diagram representation of Intersection The set of elements that belong to both set A and set B is called the intersection of two sets A and B.

Example:

Shaded area denotes A  B. Note: The universal set  is represented by a rectangular region. www.betoppers.com

Sets and Relations

103 Venn Diagram representation of A B C U

U

Venn Diagram representation of Difference of two Sets The difference of two sets A and B is obtained by the elements which are present in A, but not in B.

Shaded area is A – B Venn diagram representation of A (B U C) U

Shaded area is B – A

Venn Diagram representation of Complement Set The elements of all that are not in A constitute a new set called the complement of A and denoted by A or Ac and read as “A complement”

Venn Diagram representation of Disjoints Sets

Venn diagram (A – B) U (B – A)

representation

of

If the intersection of two sets A and B is the empty Set, then A and B are called the disjoint sets.

Venn Diagram representation of A U B U C

Relationship among the cardinal numbers Let A, B be any two sets and n(A), n(B) cardinal numbers of sets A & B respectively. Then: U

i)

n(A U B) = n(A) + n(B) – n(A

B)

ii)

n(A U B U C) = n(A) + n(B) + n(c) – n(A B) – n(B C) – n(C A)+ n(A B C) U

U U

U U

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8th Class Mathematics

104 n(AUB) = n(A) + n(B)., if A and B are disjoint non-empty sets.

iv)

n(A – B) = n(A) – n(A B)

v)

n(A  B) = n(A) + n(B) – 2n(A

U

B)

vi)

n(A  B) = n(A) + n(B) – 2n(A

U

B)

50. If n (A) = n(B), find n (A the following figure.

U

iii)

B) and n (A U B), see

U

A

B

4x – 8 2x x + 7

vii) n(A ) = n(  ) – n(A) |

Let n(A) = p and n(B) = q. Then a) Minimum of n(A U B) = max {p, q} b) Maximum of n(A U B) = p + q c)

Minimum of n(A

U

B) = 0

d)

Minimum of n(A

U

B) = min {p, q}

Formative Worksheet 44. Illustrate the following through Venn diagram:

U

a) (A – B) U (A – C) b) (A B)| c) A| U B| 45. Indicate the following sets in Venn diagram.

U

U

i) AI (B U C) 46. Name the sets shaded:

ii) AI (B – C)

51. Let T = {1, 3, 5, 7, 9, 11, 13, 15} S = {2, 4, 6, 8, 10, 12} Find (i) S – T (ii) T – S iii) T U S iv) T S. 52. Draw a Venn diagram for three non-empty sets A, B and C so that A, B and C will have the following property: i) A  B, C  B, A C =  ii) If A  C, A  C, B C =  iii) A  B, C  B, A C =  iv) A  (B C), B  C, C  B, A  C

U

B)

U U U

B ) = n(  ) – n(A

B)

U

ix)

U

viii) n(A

|

U

|

U

vii) n(A| U B|) = n(  ) – n(A

Conceptive Worksheet 35. Shade the region in the given diagram for the following sets. i) (A – B)  (A – C) ii) A – (B  C) A

B

C

36. Using Venn diagram, write down the elements of A 8 1 10

B 2 4 6 12

5

47. If A and B are two overlapping sets, show that A B| = A – B by using Venn diagrams. 48. Let  be the universal set and A U B U C = . Then indicate [(A – B) U (B – C) U (C – A)]c. 49. 65% of students in a class like cartoon movies, 70% like horror movies and 75% like war movies. What is the smallest percent of student liking all the three type of movies? U

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i) A  B

ii) A – B



11

7

iii)  A  B  ' iv) B – A

37. In a class of 100 students, the number of students who passed in English only is 46, in Maths only is 46, In commerce only is 58. The number who passed in English and Maths is 16, Maths and Commerce is 24 and English and Commerce is 26, and the number who passed in all the subjects is 7. Find the number of students who failed in all the subjects.

Sets and Relations

105

38. A charitable coaching institute for poor children is imparting coaching for SSC Public examinations on three afternoons. The table below shows the number of candidates attending the same.

DAYS

1ST DAY ONLY

1ST DAY

2ND DAY

3RD DAY

1ST AND 2ND DAYS

1ST AND 3RD DAYS

2ND AND 3RD DAYS

No. of candidates

60

100

70

46

30

28

23

i) Find the number of candidates who attended on all the three days. ii) What percent of the candidates attended on any one day only? 39. Verify the distributive laws with the help of Venn diagrams:

U

U

1.

A U (B C) = (A U B (A U C)

2.

AU (B U C) = (A

U

B) U (A

C)

U

40. A = {1, 2, 3, 4} ; B = {2, 4, 5, 7}. Illustrate A – B through Venn diagram. 41. A = {1, 3, 5} ; B = {2, 4, 6}. Represent the set A and B with the help of Venn diagrams and find A U B. 42. Match the following: Group - A Group - B A B 1) [ ] A) A 1 3

2

4 5

A– B = 2)

A

[

B 1

3 4

] B) A11

2

3 2 5

n A B  C   3) a

x

c

b

y

d

 A  B  4) A Δ B= 5) (A1)1 =

[

] C) 15

[

] D) {1, 2}

[

] E) 20

[

] F)  A  B    A  B 

RELATIONS 11. Introduction to Relations Let’s observe table of pens and cost of pens: Number of pens

Cost of pens

1

10

2

20

3

30

4

40

5

50

The above table is shows the number of pens and their corresponding cost can be expressed as follows: (1, 10), (2, 20), (3, 30), (4, 40), (5, 50) We observe that: The first element in the brackets 1, 2, 3, 4, 5 represent the number of pens and second elements in the brackets 10, 20, 30, 40, 50 represent their respective costs. It is clear that the elements in the brackets are written in a specific order. i.e., the above table gives the relation between the number of pens and their cost.. In day-to-day life, we often talk of relation between two persons, between two straight lines (e.g., perpendicular lines, parallel lines), etc. Let A be the set of all male students in Delhi whose fathers live in Delhi. Let B be the set of all the people living in Delhi.

G)  A  B    A  B 

Let ‘a’ be a male student living in Delhi i.e., a  A. Let ‘b’ be the father of a. The b B and a is related to b under son-father relation.

H) {x, y} www.betoppers.com

8th Class Mathematics

106 If we denoted the son-father relation by symbol R and ‘a related to b’ under relation R, we can also express this by writing aRb. Here, aRb means ‘a is R-related to b’ i.e., ‘a is related to b under relation.’ Here, R denotes the relation ‘is son of’. So aRb means ‘a is son of b’. We can also express this statement by saying that the pair of ‘a’ and ‘b’ is in relation R i.e., the ordered pair (a, b)  R.

Example-1 Let A = {2, 4} B = {p, q, r} be two sets. The ordered pairs obtained by the product of two sets are {(2, p), (2, q), (2, r), (4, p), (4, q), (4, r)} The set of three pairs is the Cartesian product of the two sets A and B and denoted by: A × B = {(2, p), (2, q), (2, z), (4, p), (4, q), (4, r)} Note: i) The product of set A × B is empty when either A or B is an empty set. ii) In the above (2, p)  A× B, but (p, 2)  A × B

Ordered pair

i)

ii)

A pair of elements, when the order of elements is definite is known as an ordered pair. It is written by enclosing the pair of elements in brackets as (x, y). Example: (2, 3); (1, 4); (7, 10) …….. etc.., Note: The ordered pair (a, b) contains two elements a, b and they are arranged in such a way that a is the first element and b is the second element. The first element in the ordered pair is called first coordinate and second element in the ordered pair is called second coordinate.

Properties of Ordered Pairs i)

ii) iii)

(a, b)  (b, a) If (a, b) = (5, 7), then a = 5 and b = 7 We, therefore, conclude that a relation is a set of ordered pairs. We denote the relation by the letter R. If (a, b)  R, we can write this as aRb, meaning a is related to b. The concept of relation is an association of two objects.

Cartesian Product of Two Sets

Example-2 Let’s find the Cartesian product of sets A = {1, 2, 3} and B = {p, q} A × B = {(1, p), (1, q), (2, p), (2, q), (3, p) (3, q)}

(i)

Some results on Cartesian Product of sets For any three sets A, B, C (a) A  (B  C) =  A  B    A  C  (b) A   B  C    A  B    A  C  (c) A   B  C    A  B    A  C 

(ii) If A and B are any two non - empty sets, then A B  B A  A  B. (iii) For any sets A, B, C, D

 A  B    C  D   A  C    B  D

(iv) Let A and B be two non - empty sets having ‘n‘ Let A, B be two sets, the set containing all the elements in common, then A  B and B  A have ordered pairs where the first element is taken from ‘n2 ‘ elements in common. A and second element is taken from B which is Cardinal Number of A × B called the Cartesian product of two sets A and B. Let A = {p, q, r} and B = {1, 2} be two sets. Denoted by A × B and read as A cross B.   (A) = 3  n(B) = 2 n(A × B) = 3 × 2 = 6  n(A × B) = 6 The cardinal number of A × B is the product of number of elements in the set A and set B.

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Sets and Relations

Relation Let A and B be two sets. Then a relation R from A to B is a subset of A  B. Ex : Let A = {1, 3, 5, 7} and B = {2, 4, 6, 8} be two sets and let R be a relation from A to B defined by “ (x, y)  R  x > y” is given by R=

107 Given A = {5, 2, 1} A × A = {(5, 5), (5, 2), (5, 1), (2, 2), (2, 5), (2, 1), (1,1), (1, 5), (1, 2)} a)

B is the relation is greater than.  B = {(5, 2), (5, 1), (2, 1)}

b)

3,2  ,  5, 2 ,  5,4 ,  7, 2  ,  7,4  ,  7,6 

C is the relation is less than  C = {(1, 2), (1, 5), (2, 5)}

Domain, Co-domain & Range

Set-builder Form

The set of first co-ordinates of all the ordered pairs of a relation R is called the domain of R. Example: If R = {(1, 5), (7, 5), (9, 11)} Then the domain of R = {1, 7, 5}

In this method, the property connecting the first correcting the first coordinate and the second co-ordinate of every ordered pair of the relation is stated. Let’s understand this method by an example. If A = {6, 3, 1} write the following relations of A × A in the set-builder form. a)

B is the relation “is greater than”

b)

C is the relation “is less than”

Given A = {6, 3, 1} A × A = {(6, 6), (6, 3), (6, 1), (3, 3), (3, 6), (3, 1), (1, 1), (1, 6), (1, 3)} The set of second coordinate of all the ordered pairs of R is called the co-domain of R.  Co-domain of R = {5, 11} The set of mapped elements in second coordinate of all the ordered pairs of R is called the range of R. If R = {(x, 7), (1, p), (2, s)}, then the range of R = {7, p, s}

a)

B

= {(6, 3), (3, 1), (6, 1) = {(x, y) / (x, y)  A × A, x > y}

b)

C

= {(1, 3), (3, 6), (1, 6)} = {(x, y)/(x, y)  A × A, y > x}

Arrow Diagram Let the relation from X to Y be described as {(1, 2) (2, 5); (3, 5)}; then it can be represented in the form of arrow diagram as follows:

12. Representation of a Relation A relation is usually represented in the following ways: i) Roster form or list form ii) Set-builder form iii) Tree diagram iv) Graphical representation

Roster Form In this method of describing a relation, all the ordered pairs which satisfying the formula given in the relation are listed. Let’s understand this method by an example. If A = {5, 2, 1} write the following relations of A × A in the Roster form. a) B is the relation “is greater than” b) C is the relation “is less than”

Tree Diagram If the relation between two sets is described as {(1, 4) (1, 5), (4, 3) (5, 7)} It can be represented in the form of tree diagram as shown in below. www.betoppers.com

8th Class Mathematics

Graphical representation of A × B Let’s see this method of describing a relation by an example. Let A = {1, 2, 3, 4} B = {2, 3, 4} be two sets. Now, let’s represent A × B in a graph. i) Draw two lines, one horizontal and other vertical, which are perpendicular to each other, denote 1, 2, 3…. on the horizontal and vertical lines. ii) Now, A × B = {(1, 2) (1, 3) (1, 4) (2, 2) (2, 3) (2,4) (3, 2) (3, 3) (3, 4) (4, 2) (4, 3) (4, 4)}

iii)

iv)

v)

The intersecting point of the vertical line from point 1 and horizontal line from point 2 is denoted as (1, 2). Similarly, the intersecting point of the vertical line from point 1 and the horizontal line from point 3 is denoted as (1, 3). The set of points so obtained is called the graph of A × B.

Formative Worksheet 53. Let the sets A = {1, 2, 3, 4, 5} and B = {3, 4, 5, 6} Write the following relations from A to B. R1 = {(x, y) / x = y} R2 = {(x, y) / y = x + 1} R3 = {(x, y) / y = 2x} R4 = {(x, y) / y = x + 2} R5 = {(x, y) / y < x}

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54. Find, if A × B is same as B × A, where A = {1, 2, 3}, B = {5, 6} 55. In a city, there are three major newspapers A, B, C of which atleast two are read by 35 percent of the population. It is known that newspaper C is read by 45 percent of the population, that newspapers A and B are read by 15 percent and that all the three are read by 10 percent. What percent of the population reads only newspaper C? 56. Is {(a, x), (a, y), (b, x), (c, y)} a product set or not?

U

108

57. (a) If n(A) = x and n(B) = y and A B =  , is true that n(A × B) = xy? (b) If x ? {1, 2, 3, 4, 5, 6, 7, 8, 9} and y = xx – x such that ‘y’ is divisible by 10, find R where (x, y)  R. 58. If A = {1, 2, 3, 4, 5} and R is a relation in A defined as follows, write R in the Roster form. i) (x, y) ?R and y is a multiple of x. ii) (x, y) ? R and y is a factor of x iii) x, y ? R and x + y = 6 iv) (x, y) ? R and x + y < 7. 59. If A = {100, 99, 98}, write the following relations of A × A in the list form. a) B is the relation ‘is greater than’. b) C is the relation ‘is less than’. c) D is the relation ‘is equal to’ 60. If A = {100, 99, 98}, write the following relations of A × A in the set-builder form. a) B is the relation ‘is greater than’. b) C is the relation ‘is less than’. c) D is the relation ‘is equal to’

Conceptive Worksheet 43. Is {(5, a), (6, b), (5, b), (7, b)} a Cartesian product? 44. Find which of the following are relations i) from A to B ii) from B to A, if A = {x, y, z}, B = {a, b, c, d} a) R1 = {(z, a), (z, b), (z, c)} b) R2 = {(z, b), (y, d), (x, a)} c) R3 = {(x, d), (y, c), (x, a)} d) R4 = {(a, y), (b, y), (c, z), (d, x)}

Sets and Relations 45. Write the domain and the range of the relations given:

109 6.

A ={x : x2 = 9 and 2x = 3} B = {x : x2 – 5x + 6 = 0, 2x = 6} C = {x : x2 – 4x + 3 = 0} D = {x : x2 = 25} E = {x : 2x = 6 or x = 1}

i) R1 = {(x, y) / 2x = y, x, y  N }   1   1   1  ii) R2  1,1 ,  2,  ,  3,  ,  4,    2   3   4  

iii) R3 = {(1, 1), (2, 4), (3, 9), (4, 16), (5, 25)}

Summative Worksheet 1.

2.

Express or describe the following sets in Roster form: i) The set of letters in the word “Geometry”. ii) The set of all multiples of 2 less than 11. iii) The set of all the factors of 36. iv) The set of even primes. Write the following in set builder form. 1) A1 is the set of numbers 1, 3, 5, 7. A1 = {1, 3, 5, 7} 2) A2 is the set of all vowels in the English Alphabet a, e, i, o, u A2 = {a, e, i, o, u} 3) A3 is the set of people living in Andhra Pradesh. A3 = {Residents of Andhra Pradesh} 4) A4 is the set of capitals of Karnataka, Andhra Pradesh, Tamil Nadu, Kerala, Goa. A4 = {Bangalore, Hyderabad, Madras, Thiruvananthapuram, Panaji}

3.

Write the following sets in Roster form. 1) 2) 3) 4) 5)

4.

A = {x / x is an even natural number} B = {x / x + 3 = 6} C = {x / x  N and 0 < x < 5} D = {x / x  N and x lies between 10 and 20} E = {x / x is a prime number and x < 12}

State which of the following sets are finite and which are infinite. 1) A2 = {x / x  N and x  5 } 2) B = {5, 7, 9, ......}

5.

Which of the following sets are singleton sets? i) The set of even primes ii) The set of factors of 1. iii) The set of months whose names begin with the letter F.

Which of the following sets are empty, singleton, equivalent and equal sets?

7.

Show that the following sets A and B are equal i) A = {x / x is a letter in the word ASSASSINATION} ii) B = {x/x is a letter in the word STATION}

8.

What is the cardinal number of a set which has 63 subsets excluding itself?

9.

A = Set of letters in the word “VERSE” B = Set of letters in the word “SERVE” State whether A = B.

10. A = Set of letters in the word “DENOTED” B = Set of letters in the word “NOTED” State whether A = B. 11. Match the following: Group - A

Group - B

1) A  B 

[

] A)  x / x  Aand x  B

2) A  B 

[

] B)  x / x  B

3) A – B =

[

] C)  x / x  and x  A

4) A Δ B=

[

] D) x/x  A and x  B

5) A1 =

[

] E) x/x  B and x  A F)  A  B    B  A  G)  x / x  A or x  B

12. If A = {2, 4, 6}, B = {1, 2, 3}, C = {1, 3, 5}, find i) A  B , ii) A   B  C  iii) A   B  C  13. Pick out equal, equivalent overlapping and disjoint sets from the following sets. i) A = {x / x is a letter of the word “education”} B = {x / x = vowels in the English alphabet} ii) C = The set of even one-digit numbers D = The set of even numbers less than 9. iii) E = {x / x < 6 and x  W } F = {x / x < 7 and x  N } www.betoppers.com

8th Class Mathematics

110 iv) G = The set of vowels from the word “ MATHEMATICS”. H = The set of constants from the word “PROBLEM” 14. Let A = {2, 4, 6, 8, 10, 12, 14}, B = {3, 6, 9, 12, 15} and C = {6, 10, 12, 15, 16}. Verify that: i)

16.

17.

18. 19. 20.

21.

22.

23.

24. If A = {a, b, x, y}, B = {4, 9, 12, 15}, find i) A × B ii) B × A iii) (A × B)  (B × A) iv) (A × B)  (B × A)

v) A  B

vi) A  B

25. Out of 800 boys in a school, 224 played cricket, 240 played hockey and 336 played basketball. Of the total, 64 played both basketball and hockey, 80 If A = {1, 2, 3} and B = {2, 3, 4, 5}, then find A Δ B. played basketball and cricket, 40 played cricket and If A = {2, 3, 4, 5}, B = {1, 3, 4, 5, 6}, hockey, 24 played all the three games. Find the C = {1, 3, 4, 5, 7} and   {1, 2, 3, 4, 5, 6, 7} then number of boys who did not play any game. find: 26. Out of 50 students taking examination in i) A   B  C  ii) A Δ B Mathematics, Physics and Chemistry, 37 passed in Mathematics, 24 in Physics and 43 in Chemistry. iii) (A – B)  (A – C) iv)  A  B  ' 19 passed in Mathematics and Physics, 29 in in Mathematics and Chemistry and utmost 20 Physics v) A ' B ' vi) What do you notice? and Chemistry. Find the maximum possible number In a class of 60 students, 23 play hockey, 15 play that could have passed all three examinations. basket ball and 20 play cricket. 7 play hockey and 27. If n(U) = 80, n(A) = 36, n(B) = 64, then find least basket ball, 5 play cricket and basket ball, 4 play hockey and cricket and 15 students do not play any value of n  A  B  . of these games. Then find the number of students 28. If (U) = 72, n(A) = 45 and n(B) = 54, find the play hockey, basket ball and cricket. greatest value of n  A  B  . In the above problem find the number of students who play hockey but not cricket. 29. In a group of 36 persons 20 like coffee, 27 like tea In the above problem (6) find the number of students and 3 persons dislike both, Find the number of who play hockey and cricket but not basket ball is persons who like both tea and coffee. 30. In a population study of 1500 Indian rivers the A survey shows that 63 percent of the Americans following data were reported : 520 were polluted like cheese whereas 76 percent like apples. If x % by sulphur compounds, 335 were polluted by of the Americans like both cheese and apples, find phosphates, 425 were polluted by crude oil, 100 were the value of x. polluted by both crude oil and sulphur compounds; A survey of 500 television viewers produced the 180 were polluted by both sulphur compounds and following information; 285 watch football, 195 phosphates, 150 were polluted by both phosphates watch hockey, 115 watch basketball, 45 watch and crude oil and 28 were polluted by sulphur football and basketball, 70 watch foot ball and compounds, phosphates and crude oil. hockey, 50 watch hockey and basket ball, 50 do not i) Find the number of rivers which were polluted watch any of the three games. Find the number of by atleast one of the three impurities. viewers watch all the three games. ii) Find the number of rivers which were polluted In a group of 75 people, 20 like coffee, 5 like tea by exactly one of the three impurities. and coffee. Answer the following: 31. In a group of 760 persons, 510 can speak Hindi and i) How many like tea, but not coffee? 360 can speak English. Find: ii) How many like coffee, but not tea? i) how many can speak both Hindi and English In a village atleast 50% of the people read a ii) how many can speak Hindi only; newspaper. Among those who read a newspaper. iii) how many can speak English only. Among those who read a newspaper, at the most ii)

15.

A B  B  A

following statements follows from the statements we have given. Which one is it?

 A  B  C  A  B  C  .

25% read more than one paper. Only one of the www.betoppers.com

Sets and Relations

111

32. In a group of 52 persons, 16 drink tea but not coffee and 33 drink tea. Find: (i) how many drink both tea and coffee. (ii) how many drink coffee, but not tea. 33. Match the following: Group - A Group - B 1) A  A|

[

] a) A  B

38. If B  A , what is A  B ? 39. Draw Venn diagram to represent A  B when A B. 40. Draw Venn diagram to represent A  B when B A. 41. If R = {(x, y)/x + 2y = 10} is a relation in A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, find R.

2)

 A  B

|

[

] b) 

42. If A = {1, 2} and B = {a, b, c} how many distinct relations can be formed from A to B?

3)

 A  B

|

[

] c) A|

43. If R = {(x, y) / x = 2y} is a relation in

4) A  A|

[

] d) B

5) A   A  B 

[

] e) A|  B| f) A

P = {8, 7, 6, 5, 4, 3, 2, 1}, what are all the elements of R? 44. If R is a relation is “is less than” from A = {2, 4, 6, 8} to B = {3, 7, 9} find R and R–1 as a set of ordered pairs.

g) A|  B| h)  34. Draw overlapping sets and shade the following: 1) A  B

2) A  C

3) B  C

4) A   B  C 

5)  A  B    A  C 

35. With the help of Venn diagram prove that A  B  C    A  B   A  C 

45. Suppose that the number of elements in the set S is 105 and that S is split into n subsets 11m + 2 elements each. If m is an integer, find m. (No two subsets have common elements).

HOTS Worksheet 1.

36. In an exam 60% of the candidates passed in Maths and 70% candidates passed in English and 10% 2. candidates failed in both subjects. If 300 candidates passed in both the subjects, find the total number of candidates who appeared in the exams, if they took the stest in only two subjects (Maths and English). 3. 37. Write the following statements in set language. i) The intersection of a set A and its complement is the null set. ii) The cardinal number of the intersection of sets A and B is equal to the sum of the cardinal numbers of the sets A and B less the cardinal number of the union of the sets A and B. iii) If A and B are disjoint sets, then the intersection of their union and their intersection is the empty 4. set. iv) If A is a subset of B and B is a subset of C, then 5. if x’ is an element of A , it is also an element of C. v) The number of elements in the union of two sets A and B is equal to the sum of the cardinal numbers of A and B, provided A and B are disjoint sets.

Given A = {1, 121, 12321, 1234321} B = {1, 112, 1112, 11112 } State whether A  B or B  A or A = B Given A = {23 + 1, 24 + 1, 25 + 1, 26 + 1} and B = {x / x is an odd number} State with reasons whether A  B or B  A or A = B. Show that there is one-to-one correspondence between the following sets. i) A = {5, 6}

B = {7, 8}

ii) X = {1, 2, 3}

Y = {p, q, r}

iii) X = {8, 9, 10}

Y = {7, 8, 9}

iv) A = {p, q, r}

B = {q, p, r}

v) A = {1, 2, 3}

B = {3, 2, 1}

If A and B have 3 and 6 elements, then find the minimum number of elements in A  B . Which of the following is a singleton set? (A) {x : x2 = 7, x  N } (B) {x : |x| = 4, x  N } (C) {x : |x| = 8, x  Z } (D) {x :x2 + 2x + 1 = 0, x  N } www.betoppers.com

8th Class Mathematics

112 6.

Which of the following is an empty set? 2 (A)  x : x  R and x  1  0 2 (B)  x : x  R and x  x  2 2 (C)  x : x  R and x  1  0

(D)  x : x  R and x  9  0 2

7.

8.

If A and B are two sets, then A  B ifff (A) B c  Ac

(B) Ac  B c

(C) B = A

(D) B c  Ac

If A and B are any two sets, then (A) A  A  B (B) A  B  A (C) Both (A) and (B) (D) none

9.

Given μ = {x : x is a natural number} B = {2x : x  } and C = {2x + 1 : x  }, then

 B  C   U  C   10. If A = {x, y}, B = {a, b, c}, C = {s, b, d} verify whether the following statements are true. i) A ×  B  C    A  B    A  C 

17. The maximum number of sets obtainable from A and B by applying union and difference operations is 18. A has 3 elements and B has 6 elements such that A  B . Find the number of elements in A  B . 19. If Na = {ax / x  N }, find N 6  N 8 . 20. If the sets A and B are defined as

1 A = {(x, y) / y  ,0  x  R } x B = {(x, y) / y = – x, x  R }, then (A) A  B = B (C) A  B = A

(B) A  B =  (D) A  B = 

21. If aN = {aX : x  N }, find the set 3 N  7 N . 22. If A = { x  R : 0 < x < 3} and B = { x  R :1  x  5 }, then find A  B . 23. Let A = {(x, y): y = ex, x  R }, B = {(x, y): y = e–x, x  R }. Then (A) A  B   (C) A  B = 

(B) A  B = R (D) None

24. For any three sets A, B, C prove that:

ii) A ×  B  C    A  B    A  C 

n A  B  C 

iii) A ×  B  C    A  B    A  C 

 n  A   n  B   n  C   n  A  B  C   –

iv) A ×  B  C    A  B    A  C  v) (B – C) × A = (B × A) – (C × A) 11. If R = {(x, y) / x + 2y = 10} is a relation in A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, find R. 2 4 12. If A =  x  C : x  1 and B =  x  C : x  1 , then

find A Δ B . 13. If A  B  A  C and A  B  A  C , then (A) B = C only when A  C . (B) B = C (C) B = C only where A  B. (D) None 14. If n(A) = n, find n {(x, y, z): x, y, z  A , x  y , y  z , z  x }. A = {x : x  I, 2  x  2 }, B = {x  I, 0  x  3 }, C = {x : x  N 1  x  2 } and D = {(x, y)  N × N : x + y = 8}, find n(D). 16. If A = {(4n – 3n – 1) / n  N } and B = {9n – 9: n  N } find A  B . 15. If

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 n  A  B   n  B  C   n  A  C   25. A shop has only red, green and blue carpets: 60% of the carpets have red colour, 30% have green colour and 50% have blue colour. If no carpet has all the three colours, what percentage of the carpets have only one colour? 26. In the following diagrams, the universal set is shown by a square and A, B and C are the sets of points in the square. Draw diagrams to show: i)

A B

ii)

A–B

iii)

B–C

iv)

B C

vi)

C–A

v)

|

B A

vii) A  (B  C)

viii) B  (A  C)

B A

Sets and Relations

113 7. C

8. 27. The set of points inside the circle. A is the set of 9. points in the upper semi-circle and B is the set of points in the lower semi-circle. C is the set of points inside the triangle. Identify and shade the region (B 10. – C)  (A – C). A

11.

C B

28. If A is a finite set. Let n (A) denote the number of elements in A. A and B are finite sets, A  B and 12. n(A) = n(B). Show that n(A  B) < n (A). 29. If X = {{1}, {1,2,}, 3,2,1}, find (i) P = {x/x  X, x  Y for some y  X} 13. (ii) Q = {{x}|x  X and x  Y for some y  X} (iii) P  Q 30. For subsets A, B of a set X defined A × B as A × B = (A  B)  (X – A)  (X – B), mark in Venn diagram (i) X – B (ii) X – (A × B)

IIT JEE Worksheet 1.

2. 3. 4. 5. 6.

Two sets A and B are such that n(A  B) = p and n (A  B) = p. How many elements does A contain? Find the number of all subsets of {a, b, c} with atmost two elements. Prove that a) n (A – B) = n(A) – n(A  B) b) n (A  B) = n(A) + n(B) – 2n (A  B) Let n (A – B) = 25 + x, n(B – A) = 2x and n (A  B) = 2x. If n (A) = 2(n (B)) then find x. If n (  ) = 2006, n (A) = 1002 , n(B) = 1576 then find i) Greatest value of n(A  B) ii) Least value of n(A  B) iii) Least value of n (A  B) iv) Greatest value of n (A  B) Verify the distributive laws with the help of Venn diagrams: 1. A  (B  C) = (A  B)  (A  C) 2. A  (B  C) = (A  B)  (A  C) In the following diagrams, the universal set is shown by a square and A, B and C are the sets of points in the square.

B A

Let A be any point in a plane. Let the sets be: X = {all points within the distance of 5 cm from A}and Y= {all points within the distance of 6 cm from A} Describe the sets X  Y and X  Y IF A  B =  , what can you say about A – B? | Simplify A  (A  B) Given that n (A) = 12, n(B) = 5 ; what is the greatest possible value of n(A  B)? If n (A) = 10, n (B) = 6, what is the least possible value of n(A  B)? Can two infinite sets be disjoint? Give as many examples as possible.

C C Draw diagrams to show: i) A  B iii) B – C | v) B  A viii) A  (B  C)

ii) A – B iv) B  C vi) C – A viii)B  (A  C)

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8th Class Mathematics

114 14.

In each Venn diagram shown in Fig, shade the region representing each of the following:

V

U

U

W

V

W (b)

(a) V

U W

.

15.

16.

(c) | i) V  W ii) W | iii) W – V iv) V  W | | | v) V – W vi) V  W Let A and B be the subsets of the universal set | ‘U’. If n (A ) = 15, n(B) = 5, n (A  B) = 3 and n (U) = 30. Find | | i) n (A) ii) n (A  B) iii) n(B – A ) If n (A) = n(B), find n (A  B) and n (A  B) use the fig.

A 4x – 8

B 2x

x+7



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By the end of this chapter, you will understand     

Even & Odd Numbers Prime and Composite Numbers Squares & Square roots Cube & Cube roots Divisibility Rules

    

HCF & LCM Concept of Factorial Finding unit digit of the Resultant Value Types of Numbers Congruences

1. Introduction Over 3,000 years ago the ancient Egyptians used special symbols known as pictographs to write down numbers. Later, the Romans developed a system of numerals that used letters from their alphabet instead of special symbols. Today, we use numbers based on the Hindu-Arabic system. We can write down any number using combinations of up to 10 different symbols (0, 1, 3, 4, 5, 6, 7, 8 and 9). The ancient Egyptians developed number systems to keep accounts of what was bought and sold.

What is a Number ?

Chapter - 98

Number Theory

Learning Outcomes

he provided the first proof that all equations formed by setting a fundamental theorem of algebra. In 1801 the completely restructured the number theory. Elementary number theory includes classic and elegant results such as Quadratic Reciprocity’s counting results, using the Mobius Inversion Formula (and other multiplicative number theoretic functions); and also the Prime Number Theorem, asserting the approximate density of primes among the integers, which has difficult but elementary proofs and combinatory and further more abstract algebra.

We use numbers every day and tend to take them 2. Even & Odd Numbers for granted. But how did the idea of numbers arise? All numbers divisible by 2 are called Even numbers. Did every culture develop the ideas of counting and Numbers which are not divisible by 2 are called numbers separately or have these ideas arisen in Odd numbers. Even numbers can be represented only a few cultures and then spread, to others? Is as 2n where n is an integer. Odd numbers can counting intuitive or did it arise to solve particular be represented as 2n + 1 (Where n is an integer). problems? Even and Odd numbers are defined for integers. Some of the oldest evidence of counting so far For example, – 2, – 4, 0, 2, 4 are even and – 3, –1, discovered comes from ancient artifacts belonging 1, 3 are odd. to groups of hunters and gatherers. For example, a Some facts about Even and Odd Numbers: wolf bone, dated about 30,000 BC, has been discovered with a series of notches carved in it, 1. The sum or product of any number of even numbers is even. which seem to represent a tally of some kind. 2. The difference of two even numbers is even. Number Theory 3. The sum of odd numbers depends on the number Number theory is one of the oldest and largest of numbers. branches of pure mathematics. Elementary number a) If the number of numbers is odd, the sum is theory involves divisibility among integers. The odd. division algorithm, the Euclidean algorithm, b) If the number of numbers is even, the sum elementary properties of primes, including Fermat’s is even. theorem, and Wilson’s theorem extending it. Gauss is ranked with Archimedes and Newton as 4. If the product of a certain number of numbers is even, then atleast one of them is even. the greatest of mathematicians. At the age of 19, he brought and a regular polygon with 17 sides, the 5. If the product of a certain number of numbers is first major new figure since Greek times. In 1799 odd, then none of the numbers is even. i.e., the product of any number of odd numbers is odd.

8th Class Mathematics

116

Formative Worksheet 1.

2. 3.

4.

5. 6.

7.

8.

4.

If a, b, c, d,...... are first four consecutive odd numbers, then 58th positive odd number is (A) 117 (B) 115 (C) 119 (D) 113 For n = 3, n3 – n represents an ______ number (A) Odd (B) Even (C) Prime(D) Perfect If ‘a’ and ‘b’ are the 4th and 8th multiple of 3 and 7 respectively, then the sum of their successive multiple is (A) Prime(B) Even (C) Odd (D) None If sum of 3 consecutive positive even integers is 144 then the 2nd number is divisible by ______ number (A) Odd (B) Even (C) Odd or even (D) None If 2a = 3b = 7c = 5d = 210 then a + b + c + d is (A) Odd (B) Even (C) Prime(D) Negative If a, b, c, d are first 4 positive odd integers then their sum is (A) 52 (B) 22 (C) 42 (D) 62 For what values of k and m is (19k – 3k) (18m – 11m) divisible by 112. (k, m  N) is (A) When k is odd, m is even (B) When k is even m is odd (C) For all values of k and m (D) Cannot be determined There are 120 pages in a novel. Then the sum of all odd and even numbers is (A) 7263 (B) 7260 (C) 7261 (D) 7264

5.

6.

 even 

7.

8.

Prime Number

2.

2009

odd

  2010 

2010

A

is odd and  even 

even

is ____________

(A) Odd (C) Even or odd www.betoppers.com

A number which is not a prime is called composite number. (or) A number which has factors besides other than 1 and itself is called a composite number.

is odd number

Some Important Points is odd

(A) Both Statements are true, Statement II is the correct explanation of Statement I. (B) Both Statements are true, Statement II is not correct explanation of Statement I. (C) Statement I is true, Statement II is false. (D) Statement I is false, Statement II is true. A C

Composite Number

(D) ab + 3

Statement-I:  2009  Statement-II:  odd 

3.

A number which does not have any factors other than 1 and itself is called a prime number. The set of prime numbers is {2, 3, 5, 7, 11, 13.........}

If a and b are odd numbers then which of the following is/are even? (A) a + b + 1 (B) a + b (C) 2  a  b 

(B) Even (D) Both a and b

d)  even  even  is (If the (D) greatest dividend is multiple of one digit number divisor e) neither prime nor composite If p, q, r and s are consecutive odd numbers, then (p2 + q2 + r2 + s(B) is always divisible by (A) 5 (B) 7 (C) 3 (D) 4 4n + 2 For n  N, 2 + 1 is (A) Always divisible by 5 (B) Divisible by 5 for odd values of n (C) Divisible by 5 for even values of n (D) Name of these

3. Prime and Composite Numbers

Conceptive Worksheet 1.

D  A is ______________ C (A) even (B) odd (C) even or odd (D) both a and b AB + CD is ____________ (A) even (B) odd (C) even or odd (D) both a and b Column - I Column - II a) odd  odd (If the dividend (A) odd is multiple of divisor) b) even  odd (B) 9 c) first odd composite number (C) even

1. 2. 3.

4. 5.

1 is neither prime nor composite. 2 is the only even prime number (all other even numbers are composite). Any prime number greater than 3 can be written in the form of either 6k + 1 or 6k – 1, where k is a natural number. a3 – b3 = (a – b) (a2 + ab + b2) If the difference of two prime numbers is ‘2’, then they are called Twin primes. Example: 3, 5 ; 5, 7 ; 11, 13 etc.,

Number theory

117

6.

There are only two primes which are consecutive integers. They are 2 and 3. 7. The primes 3, 5 and 7 are called prime triplet. 8. If a and b are prime numbers, then their product ab will have only l, a, b and ab as factors. 9. The set of prime numbers is infinite. 10. Two numbers are said to be co-primes or relatively prime numbers, if their H.C.F is 1. 11. The co-primes need not necessarily be prime themselves. 12. A pair of co-primes may consist of a) both primes (3, 5) b) one prime and one composite (7, 6) c) both composite (8, 15) d) The co-primes need not necessarily be prime themselves e) If two numbers are not co-primes then they must have a common factor other than 1.

10.

11.

12.

13.

Conceptive Worksheet

How many prime numbers are there between 80 and 105? (A) 3 (B) 4 (C) 5 (D) 8 10. Which of the following statements are true? (A) An even number greater than 3 can never be a prime number ormative orksheet (B) A square number can never be a prime If N, N + 2 and N + 4 are prime numbers, then the number number of possible solutions of N is (C) A number which is greater than 5 and whose (A) 1 (B) 2 (C) 3 (D) none units digit is 5 can never be a Prime number Let x and y be positive integers such that ‘x’ is (D) A number ending with 0 can be a Prime prime and ‘y’ is composite. Then which of the number following is true ? 2 3 4 11. Statement-I: 2x × 3y × 5z where x, y, z are positive (A) y – x cannot be even and 2x = 6y = 4z = 12 then the number of (B) xy cannot be even divisors is 37 × 9 × 82 xy Statement-II: If N is a composite number such that (C) cannot be even (D) none x a p  bq  c r  ....... where a, b, c ....... are prime factors of N and p,q,r ……..are positive integers, a, b, c, d are first four odd prime numbers then which then the number of factors N is given by the of the following is true? (A) a + b + c is not composite odd expression  p  1 q  1 r  1 ...... (B) b + c + d is prime (A) Both Statements are true, Statement II is the (C) abc is even correct explanation of Statement I. (D) a + b + c + d is not even composite (B) Both Statements are true, Statement II is not correct explanation of Statement I. If 5x + 11y is a prime number for natural number values of x and y, then what is the minimum value (C) Statement I is true, Statement II is false. of x + y? (D) Statement I is false, Statement II is true. (A) 2 (B) 3 (C) 4 (D) 5 12. Two prime numbers whose product is 33 are The sum of three prime numbers is 100. If one of (A) 4, 9 (B) 15, 17 (C) 11, 3 (D) 61, 7 them exceeds another by 36, then one of the number 13. Smallest even and odd composite numbers is respectively are (A) 7 (B) 29 (C) 41 (D) 67 (A) 4, 9 (B) 2, 3 (C) 4, 5 (D) 2, 5 14. If sum of two prime is 68, then the number are (A) 60, 8 (B) 55, 13 (C) 61, 7 (D) 27, 41

F 9.

14. If x = 7, y = 5, z = 3 then which of the following is true (A) xyz + 1 is composite (B) xyz – 4 is prime (C) x + y + z + 10 is a perfect square (D) All of these 15. If 2n – 1 is a prime, then n is (A) negative integer (B) prime (C) composite (D) none 16. If p, 8p – 1 are primes then 8p + 1 is (A) prime (B) composite (C) negative integer (D) fraction

9.

W

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8th Class Mathematics

118 15. Column - I a) Number of primes between 200 to 230 are b) Number of primes between 100 to 110 are c) Number of primes between 1 to 25 are d) Number of primes between 50 to 70 are

Column - II (A) 3

iii)

(B) 5 (C) 4 (D) 9

(E) 7 16. If a, b, c, d are first 4 two-digit composite numbers. If a + b + c + d + e is a perfect square, then the value of e is (A) 3 (B) 5 (C) 2 (D) 7

4. Squares & Square Roots Perfect Square

iv)

v)

A natural number is said to be perfect square, If it is the square of some other natural number. Examples: 81 = 92, 4 = 22, 36 = 62, 10000 = 1002.

Identifying a Perfect Square A given number is perfect square, if it can be expressed as the product of pair of equal factors. Note: The factors can be found by prime factorisation method. Example: Is 225 a perfect square ? Resolving 225 into prime factors, 5

225

5 3 3

45 9 3 1

vi)

we get, 225 = 5 × 5 × 3 × 3 as it is expressed as the product of equal factors. Hence, it is a perfect square.

Properties of Squares of Numbers i)

ii)

The square of an even number is always an even number. Examples: 6 is an even number and 62 = 36 which is even. 8 is an even number and 82 = 64 which is even. The square of an odd number is always an odd number. Examples: 7 is an odd number and 72 = 49 which is odd. 11 is an odd number and 112 = 121 which is odd.

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The square of a proper fraction is less than the proper fraction. Examples: 2

3 is a proper fraction and 4

9 3 3      4  16 4

5 is a proper fraction and 6

25 5 5     36 6 6

2

The square of a decimal fraction less than 1 is smaller than the decimal. Examples: 0.2 < 1 and (0.2)2 = 0.04 < 0.2 0.4 < 1 and (0.4)2 = 0.16 < 0.4 A natural number ending with 2, 3, 7 or 8 is never a perfect square. Examples: 52, 77, 93, 18 are not perfect squares. (or) All perfect squares end with 0, 1, 4, 5, 6, (or) 9 only. Example: 16, 225, 1024, 900 are perfect squares. Note: Though all perfect squares ends with 0, 1, 4, 5, 6, (or) 9 all numbers ending with 0, 1, 4, 5, 6, (or) 9 need not be perfect squares. Examples: 56, 10, 39, 84, 71, 95 are not perfect squares. Any natural number ending with odd number of zeros is never a perfect square. Examples: 1000, 500000...... Note: Perfect squares always end with an even number of zeros but all numbers ending with even number of zeros are not perfect squares. Examples: 900 is ending with 2 zeros, it is a perfect square. 800 even though ending with 2 zeros, is not a perfect square.

Interesting patterns of Square Numbers: i) ii)

Sum of first n odd natural numbers is n2. The square of a natural number having all digits as 1 (number of ones are £ 9) results in a palindrome.

iii)

The numbers 49, 4489, 444889, . . . obtained by inserting 48 into the middle of the perceiving number are squares of integers of 7, 67, 667, 6667, . . . respectively.

Number theory

iv)

vi)

119

The fun is in being able to find out why this happens. May be it would be interesting for you to explore and think about such questions even if the answers comes some years later. The difference of odd squares of successive numbers is equal to their sum. Example: 172 – 162 = 17 + 16 The squares of natural numbers like 11, 111, ….., have a nice pattern as shown below 121 × (1 + 2 + 1) = 484 = 222 112 × (sum of the digits in 112) = (2 ×11)2 12321 × (1 + 2 + 3 + 2 + 1) = 110889 = 3332 1112 × (sum of the digits in 1112) = (3× 111)2 1234321 × (1 + 2 + 3 + 4 + 3 + 2 + 1) = 19749136 = (4444)2 11112 × (sum of the digits in 11112) = (4 × 1111)2  Square of a number   having all ones

  sum of digits  ×     in the square 

 number of digits  = × the number   in the number 

Square Root The square root of a number n is that number which when multiplied by itself gives n as the product.

Finding Square Root 1. 2.

Properties of Square Root 1. 2. 3.

ii)

2m is even then the Pythagorean triplet is (2m, m2 – 1, m2 + 1).

pq  p  q

p p  q q

ii)

5. Cube & Cube Roots Perfect cube In the above table, 1, 8, 27, ..., 729 are called perfect cubes or perfect third powers of 1, 2, 3 , ..., 9 respectively. A natural number is said to be a perfect cube if it is the cube of some natural number. If small numbers are given, we can identify whether it is a perfect cube or not. But if a larger number is given then it is difficult to do so. Hence, we need a method to check whether the number is a perfect cube or not.

Pypthagorean Triplet

 m2  1 m2  1  ,  m,  2 2 . 

If a number ends in an odd number of zeros, then it does not have a square root. The square root of an even square number is even and square root of an odd square number is odd. If p and q are perfect squares (q  0), then i)

2

Let us observe the sum of squares of 3 and 4 i.e., 32 + 42 = 9 + 16 = 25  32 + 4 2 = 5 2 Here, the sum of squares of 3 and 4 is square of another number 5.  The sum of squares of two numbers is again square of another number.  The three numbers (3, 4, 5) are said to be Pythagorean triplets. In general, a triplet (m, n and p ) of natural numbers m, n and p is said to be a Pythagorean triplet if m2 + n2 = p2. Note: if m > 1 and i) If m is odd, then the Pythagorean triplet is

One way to find the square root of perfect squares is by resolving the given number into prime factors. The square root of a number may also be found by division method.

Points to be noted 1. 2. 3. 4. 5. 6. 7. 8.

A number ‘n’ is a perfect cube if there is an integer ‘m’ such that n = m × m × m. If a prime p divides a perfect cube then p3 also divides this perfect cube. In the prime factorization of a perfect cube every prime occurs three times or a multiple of 3 times. Perfect cubes up to 1000 are 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000. If ‘n’ is even, then n3 is also even. If ‘n’ is odd, then n3 is also odd If ‘m’ is even and ‘n’ is odd, then m3 × n3 is even. m is called cube root of n if n = m3 and written as m 3n .

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8th Class Mathematics

120 22. Column – I a) 112 b) 1112 c) 11112 d) 152

Column – II (A) 12321 (B) 1234321 (C) 121 (D) 325 (E) 225 23. When a number is divided by 25, its cube root is x. When it is multiplied with 5, then its cube root is y. If x + y = 36, then the number is _____________

Properties of Cubes 3

x×y = 3 x × 3 y .

3

x 3y

3.

3

x  3 y  3 xy .

4.

3

x 3 y  3 xy .

5.

3

1   1 .

6.

3

1. 2.

3

x y.

24. If 3a = 4b = 6c and a + b + c = 27 29 , then

1  1.

a 2 + b 2 + c 2 is

7. 8.

The cube root of a negative perfect cube is negative. The units digit of the cube root of a perfect cube can be determined with the help of the units digit of the perfect cube. 9. Cubes of 1, 4, 5, 6 and 9 in unit place are the numbers ending in the same digit. 10. Cubes of 2 and 8 in unit place ends in 8 and 2 respectively. 11. Cubes of 3 and 7 in unit place ends in 7 and 3 respectively.

Formative Worksheet 80 + 6 5 is (A) 13.41 (B) 20.46 (C) 21.66 (D) 22.35 3 28 x  1426  of 2872 , then the value of x is 4 (A) 576 (B) 676 (C) 1296 (D) 1444

19. The value of (A)

21

2797 is x then the value of x is 3364

271 232 59 (B) (C) 4 (D) none 58 58 38





2 20. The square of 5  y  25 is

(A) y 2  5 y 2  25

(B)  y2  10 y 2  25

(C) y 2  10 y 2  25

(D) y 2 

10y 

2

  50 

(D) None of these

25. If x =

2 2 2 2.... and y = 3 3 3 3.... , then the true statements in the following are I) x + y  5 II) x2 + y2 = 6xy III) x2y2 = 6xy (A) I only (B) III only (C) I and III (D) None of these

17. Statement – I: Sum of the squares of first three natural numbers is 14. Statement – II: Sum of the square of 3 consecutive numbers is 3n2 + 2. (A) Both Statements are true, Statement II is the correct explanation of Statement I. (B) Both Statements are true, Statement II is not correct explanation of Statement I. (C) Statement I is true, Statement II is false. (D) Statement I is false, Statement II is true. 18. What should be added to 1752 to make it a perfect square number? (A) 6 (B) 10 (C) 12 (D) 18 19. By what least number 4320 be multiplied to obtain a number which is a perfect cube? (A) 55 (B) 52 (C) 50 (D) 150 20. The cube root of 0.000216 is

2

21. What should be subtracted from 1300 to make it a perfect square number? (A) 7 (B) 4 (C) 8 (D) 10

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(B) 81

Conceptive Worksheet

17. If 3 5 + 125 = 17.88 , then the value of

18.

(A) 3 29 (C) 87

(A) .6

(B) 0.06 (C) 0.006 (D) None of these

32.4  2 , then x is x (A) 4 (B) 8.1 (C) 4.1

21. If

(D) 8

Number theory

121

22. The square root of (2722 – 128(B) is (A) 144 (B) 200 (C) 240 (D) 256 23. If 5x is a perfect cube, then the least number of ‘5’s that must be contained in the prime factorization of x is (A) 1 (B) 2 (C) 3 (D) none of these 24. The number that must be multiplied to 128 to make a perfect cube is (A) 2 (B) 4 (C) 8 (D) 12

7.

6. Divisibility Rules

9.

1.

In a × b = c ; a and b are factors and c is the product

2.

In c  b  a ; c, b and a are called dividend, divisor and quotient respectively.

8.

Test of Divisibility 1.

2.

3.

4.

5.

6.

Divisibility By 2: A number is divisible by 2, if its unit’s digit is any of 0, 2, 4, 6, 8. For example, 84932 is divisible by 2, while 65935 is not. Divisibility By 3: A number is divisible by 3, if the sum of its digits is divisible by 3. For example, 592482 is divisible by 3, since sum of its digits = (5 + 9 + 2 + 4 + 8 + 2) = 30, which is divisible by 3. But, 864329 is not divisible by 3, since sum of its digits = (8 + 6 + 4 + 3 + 2 + 9) = 32, which is not divisible by 3. Divisibility By 4: A number is divisible by 4, if the number formed by the last two digits is divisible by 4. For example, 892648 is divisible by 4, since the number formed by the last two digits is 48, which is divisible by 4. But, 749282 is not divisible by 4, since the number formed by the last two digits is 82, which is not divisible by 4. Divisibility By 5: A number is divisible by 5, if its unit’s digit is either 0 or 5. Thus, 20820 and 50345 are divisible by 5, while 30934 and 40946 are not. Divisibility By 6: A number is divisible by 6, if it is divisible by both 2 and 3. For example, The number 35256 is clearly divisible by 2. Sum of its digits = (3 + 5 + 2 + 5 + 6) = 21, which is divisible by 3.Thus, 35256 is divisible by 2 as well as 3. Hence, 35256 is divisible by 6. Divisibility By 8: A number is divisible by 8, if the number formed by the last three digits of the given number is divisible by 8. For example, 953360 is divisible by 8, since the number formed by last three digits is 360, which is divisible by 8. But, 529418 is not divisible by 8, since the number formed by last three digits is 418, which is not divisible by 8.

10.

11. 12. 13.

14. 15. 16.

Divisibility By 9: A number is divisible by 9, if the sum of its digits is divisible by 9. For example, 60732 is divisible by 9, since sum of digits = (6 + 0 + 7 + 3 + 2) = 18, which is divisible by 9. B u t , 68956 is not divisible by 9, since sum of digits = (6 + 8 + 9 + 5 + 6) = 34, which is not divisible by 9. Divisibility By 10: A number is divisible by 10, if it ends with 0. For example, 96410, 10480 are divisible by 10, while 96375 is not. Divisibility By 11: A number is divisible by 11, if the difference of the sum of its digits at odd places and the sum of its digits at even places, is either 0 or a number divisible by 11. For example, the number 4832718 is divisible by 11, since : (sum of digits at odd places) – (sum of digits at even places) = (8 + 7 + 3 + 4) – (1 + 2 + 8) = 11, which is divisible by 11. Divisibility By 12: A number is divisible by 12, if it is divisible by both 4 and 3. For example, Consider the number 34632. i. The number formed by last two digits 32, which is divisible by 4. ii. Sum of digits = (3 + 4 + 6 + 3 + 2) = 18, which is divisible by 3. Thus, 34632 is divisible by 4 as well as 3. Hence, 34632 is divisible by by 12. Divisibility By 14: A number is divisible by 14, if it is divisible by 2 as well as 7. Divisibility By 15: A number is divisible by 15, if it is divisible by both 3 and 5. Divisibility By 16: A number is divisible by 16, if the number formed by the last 4 digits is divisible by 16. For example, 7957536 is divisible by 16, since the number formed by the last four digits is 7536, which is divisible by 16. Divisibility By 24: A given number is divisible by 24, if it is divisible by both 3 and 8. Divisibility By 40: A given number is divisible by 40, if its is divisible by both 5 and 8. Divisibility By 80: A given number is divisible by 80, if it is divisible by both 5 and 16. Note: If a number is divisible by p as well as q, where p and q are co-primes, then the given number is divisible by pq. If p and q are not co-primes, then the given number need not be divisible by pq, even when it is divisible by both p and q.

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8th Class Mathematics

122

17.

18.

19.

20

21. 22. 23. 24. 25. 26.

27. 28. 29.

30.

For example, 36 is divisible by both 4 and 6, but it is not divisible by (4 × 6) = 24, since 4 and 6 are not co-primes. If a number is divisible by p as well as q where p and q are coprimes then the given number is divisible by pq. xn + yn is divisible by x + y when n is odd. For example, (20n + 17n) is divisible by (20 + 17) if n = 1, 3, 5, 7...... xn – yn is divisible by x + y when x is even. For example, 48n – 36n is divisible by (48 + 36) when x = 2, 4, 6...... xn – yn is divisible by x – y for all values of ‘n’. For example, 68n – 36n is divisible by (68 – 36) when n = 1, 2, 3, 4,...... For any integer n, n3 – n is divisible by 3. For example,33 – 3 is divisible by 3. For any integer n, n5 – n is divisible by 5. For example, 25 – 2 is divisible by 2. For any integer n, n11 – n is divisible by 11. For example, 211 – 2 is divisible by 2. For any integer n, n13 – n is divisible by 13. For example, 213 – 2 is divisible by 13. If a number is divisible by another number, then it is divisible by each of the factors of that number. If a, b, c are three natural numbers such that ‘a’ is divisible by ‘b’ and and b is divisible by c then ‘a’ is divisible by c also. If two numbers b and c are divisible by a then b + c is also divisible by a. If a number is a factor of each of the given numbers then it is a factor of their difference. Starting from right hand side make pairs of two digits each. Add all the pairs and divide the sum by 99. If the sum is divisible by 99, then the number is divisible by 99. Example: 297  2 97  2 + 97 = 99 is divisible by 99 8217  82 17  82 + 17 = 99 is divisible by 99 73854  73854  7 + 38 + 54 = 99 is divisible by 99. Starting from right hand side make pairs of 3 digits each. Add all the pairs and divide the sum by 999. If the sum is divisible by 999, then the number is divisible by 999. Example-1: 4995  4 995  4 + 995 = 999 is divisible by 999

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Example-2: 17982  17 982  17 + 982 = 999 is divisible by 999 31. When a positive number A is divided by an other positive number B and if B > A then the remainder will be A itself . Example: Remainder of 5 ÷ 12 = 5.

Formative Worksheet 26. The largest number of the digits exactly divisible by 88 is (A) 9944 (B) 8888 (C) 9988 (D) 9999 27. What value must be given to x to make 8597x 65 is exactly divisible by 11? (A) 0 (B) 2 (C) 4 (D) 7 28. A number 6x + 8y is divisible by 8 where x and y are positive integers. If x and y take least values then the value of x + y is (A) 4 (B) 5 (C) 6 (D) 7 29. Which of the following numbers is exactly divisible by 45: (A) 4581 (B) 4185 (C) 4158 (D) 5985 30. Which of the following is divisible by 15? (A) 462 (B) 365 (C) 465 (D) 460 31. Column – I Column – II a) 253 (1) divisible by 2 b) 714 (2) divisible by 3 c) 3256 (3) divisible by 7 d) 321000 (4) divisible by 8 (5) divisible by 11 32. The sum of cubes of three successive natural numbers is divisible by (A) 7 (B) 4 (C) 2 (D) 9 3n 33. If n = 1, 2, 3,.......n then 2 – 1 is divisible by (A) 6 (B) 3 (C) 9 (D) 7 34. The number of numbers lying between 1 and 200 which are divisible by either of 2, 3, or 5 is (A) 146 (B) 145 (C) 158 (D) none of these

Conceptive Worksheet 25. Which of the following numbers is exactly divisible by 45: (A) 4581 (B) 4185 (C) 4158 (D) 5985 26. Which of the following number is exactly divisible by 14? (A) 1000 (B) 1004 (C) 1006 (D) 1008 27. Which of the following number is divisible by 18? (A) 1122 (B) 1234 (C) 1456 (D) 1296

Number theory

123

2

28. n(n – (A) is always divisible by _____ (A) 5 (B) 7 (C) 6 (D) 8 29. The product of any four consecutive natural numbers is always divisible by (A) 26 (B) 28 (C) 20 (D) 24 30. What is the least number that should be added to 37301 so that it is exactly divisible by 27? (A) 21 (B) 6 (C) 14 (D) 23

Finding H.C.F by Prime Factorization Method We first find the prime factorization of each of the given numbers. Then the product of all common prime factors, using the least power of each common prime factor, is the HCF of the given numbers. Example: Lets find the HCF of 200 and 320

75

31. The remainder when 7575 is divided by 37. (A) 0 (B) 1 (C) 3 (D) can’t be determined 32. A number when divided by 14 leaves a remainder of 8, but when the same number is divided by 7, it will leave the remainder: (A) 3 (B) 2 (C) 1 (D) can’t be determined

2 200

2 320

2 100

2 160

2

50

2

80

5

25

2

40

5

2

20

2

10 5

7. HCF & LCM

200 = 2×2×2×5×5 200 = 23 × 52

Highest Common Factor (HCF) The number that divides all the given numbers is called as a Common Factor. The greatest among the common factors of the given numbers is called Highest Common Factor (HCF). Example: The common factors for 24 and 12 are 2, 4, 6, 12. Among the factors, 12 is the highest and is the HCF of 24 and 12 . HCF is also called as: GCD – Greatest Common Divisor (or) HCD – Highest Common Divisor (or) GCF – Greatest Common Factor (or) HCF – Highest Common Factor Note: 1. 2. 3.

4.

320 = 2×2×2×2×2×2×5 320 = 26 × 5

 HCF = product of lowest power prime factors

= 23 × 5 = 40

Finding HCF by Long Division Method:

1. 2. 3.

4.

The following steps should be used to find the HCF by long division method. If two numbers are given, divide the greater number by the smaller one. Divide the divisor by the remainder. Repeat the process of dividing the preceding divisor by the remainder last obtained, till the remainder zero is obtained. The last divisor is the required HCF of the given numbers. Example: Let us find the HCF of 136, 170 and 255, using the long division method.

If a divides b, then a is called factor of b and b is called multiple of a. 136) 170 (1 34) 255 (7 A factor of a given number is called a prime factor, 136 238 if it is a prime number. 34) 136 (4 17) 34 (2 A number which divides each one of the given 136 34 numbers exactly, is called a common factor of 0 0 the given numbers. H.C.F or G.C.D of two or more numbers is the  HCF of 136, 170 and 255 = 17 greatest number that divides each one of them Note: exactly. 1. Two numbers are said to be relatively prime or coprimes, if their H.C.F is 1. 2. The largest number which divides the numbers p, q and r to give remainders of s, t and u respectively www.betoppers.com

8th Class Mathematics

124

3.

will be the H.C.F of any two of the three numbers (p – s), (q – t) and (r – u). The largest number which divides the numbers p, q and r and gives the same remainder in each case will be the HCF of the differences of any two of the three numbers (p – q), (q – r) and ( p – r).

Least Common Multiple Least number that is divisible by all the given numbers is called Least Common Multiple. For example, the common multiples of 7 and 2 are 14, 28, 42, etc, Among these, 14 is least and is the LCM of 7 and 2.

Note: 1. Dividend = Divisor × Quotient + remainder. 2. Two numbers are called relatively prime or coprimes if their H.C.F is 1. 3. Any number which when divided by p, q or r leaving the same remainder ‘s’ in each case will be of the form N = K (L.C.M of p, q and r) + s, where K = 0, 1, 2, 3.....

HCF and LCM of Fractions i) HCF of fractions =

HCF of numerators LCM of denominators

LCM of numerators ii) LCM of fractions = HCFof denominators

Finding of LCM by Prime Factorization Relation between LCM and HCF Method In order to find the LCM of two or more given numbers we write the prime factorization of each of the given numbers. Then, the required LCM of these numbers is the product of all different prime factors of the numbers, now multiplying common prime factors having greater powers, will get L.C.M.

Finding of LCM by Long Division Method

1. 2.

3. 4.

In order to find the LCM of any amount of numbers, we use the long this method. We arrange the given numbers in a line, in any order. We divide by a number which divides exactly at least two of the given numbers and carry forward the numbers which are not divisible. This process is repeated till no two of the given numbers are divisible by the same number. The product of the divisors and the undivided numbers is the required LCM of the given numbers. Example: Let’s find the LCM of 16, 8 and 4. 2 16, 8, 4 2 8, 4, 2 2 4, 2, 1 2, 1, 1  LCM of 16, 8 and 4 = 2  2  2  2 = 16

Note: We must take prime numbers as divisors like 2, 3, 5, 7............

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Consider any two numbers a and b. Let their LCM and HCF be x and y respectively. Then ab = xy Product of any two numbers

=

Product of their LCM and HCF

Note: Product of ‘n’ numbers

=

(HCF for each pair)n–1 × LCM of ‘n’ numbers.

Formative Worksheet 35. The number of number-pairs lying between 40 and 100 with their H.C.F as 15 is (A) 3 (B) 4 (C) 5 (D) 6 36. Three bells toll at the interval of 6, 15 and 20 minutes respectively. If all the three bells toll together at 9 a.m. After how much interval, they will toll together again? (A) 65 min (B) 70 min (C) 1 hr (D) 60 min 37. The least number which when divided by 6,7,8, 9 & 12 leaves the remainder 1 is (A) 502 (B) 505 (C) 504 (D) 506 38. Column – I Column – II a) L.C.M of

1 5 2 4 , , , is 3 6 9 27

1) 220

20 3 35 3) 100

b) H.C.F of 1.75, 5.6 and 7 is 2) c) Ratio of two no is 3:4 & HCF = 4, then L.C.M d) Product of two no’s is 1320 & their H.C.F is 6, then L.C.M is

4) 48 5) 0.35

Number theory

125

39. The L.C.M of two numbers is 48. The numbers are in the ratio 2 : 3. Then the sum of the numbers is (A) 28 (B) 32 (C) 40 (D) 64 40. L.C.M. of the fractions

12 9 4 , , is 25 35 15

1 36 36 36 (B) (C) (D) 525 5 25 35 41. The greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case is (A) 4 (B) 7 (C) 9 (D) 13 42. The L.C.M of two numbers is 48. The numbers are in the ratio 2 : 3. Then the sum of the numbers is (A) 28 (B) 32 (C) 40 (D) 64 (A)

Conceptive Worksheet 33. The product of two numbers is 2028 and their H.C.F is 13. The number of such pairs is (A) 1 (B) 2 (C) 3 (D) 4 34. The least number which when divided by 20,25,35 & 40 leaves remainders 14, 19, 29 & 34 respectively is (A) 1394 (B) 1349 (C) 1943 (D) 1249 35. The smallest number which when diminished by 7, is divisible by 12, 16, 18, 21 and 28 is : (A) 1008 (B) 1015 (C) 1022 (D) 1032 36. The L.C.M. of two numbers is 72 times their H.C.F. The sum of the LCM and the HCF is 584. If one of the number is 72, then the other number is ____________ 37. Three numbers are in the ratio 1 : 2 : 3 and their H.C.F is 12. Then their sum is (A) 24 (B) 30 (C) 144 (D) 72 38. If L.C.M. and H.C.F. of two numbers are p and q respectively, then L.C.M. of p and q is

41. Two bridges have to be build of length 1024 m and 960 m respectively. The largest length of concrete beams that is required to built the both bridges as the combination of beams exactly fit for the required lengths, is (A) 12 m (B) 16 m (C) 32 m (D) 64 m

8. Concept of Factorial Factorial of the given positive integer N: Factorial of N is the product of the first N natural numbers. That is N (factorial) = 1 2  3  4  5  6  ........  N . N (factorial) is denoted by N! All the factorial values of N will end in zeros when N is atleast 5. It becomes interesting to find the number of zeros at the end of the factorial of the given number N.

Finding the number of zeros at the end of N! Let us understand the definition of ‘Greatest Integer Function’ The greatest integer function (or floor function) will round any number down to the nearest integer. The notation for the greatest integer function is shown here. That is ‘[N]’. Example: [0.7]  0

and

[1.8]  2

Let N! is given to us . To find the number of zeros at the end of it. The formula to find is N  N  N  N  5    52    53   ......   5n  where 5n < N.

Product of k consecutive integers is

p divisible by k! (A) p + q (B) (C) pq (D) p q Examples: 39. If L.C.M. of two numbers is 5 times to one of the 1. i) 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 40320 numbers and the other number is 25, then H.C.F. 16! 16  15  14! of the two numbers is   240 ii) 14! 14! (A) 5 (B) 25 (C) 125 (D) 175 8! 8  7  6  5! 40. There are two ropes each measures 496 m and   168 iii) 624 m respectively. The largest scale that should 5! 2! 5! 2  1 be used to measure both the ropes exactly is of iv) 4  5  6  7  8  (A) 4 m (B) 8 m (C) 16 m (D) 24 m 1  2  3  4  5  6  7  8 8!  1 2  3 3! www.betoppers.com

126 v)

LCM of 3!, 4! 3 × 2 × 1, 4 × 3 × 2 × 1 24 LCM of 3! and 4! is 24

vi)

2. 3. 4. 5.

6. 7.

8. 9.

10.

51! 51  50!   51 50! 50! 75! 85! 75  74! 85  84!     75  85  160 74! 84! 74! 84! 21! 15! 21 20! 15  4!     21  15  6 20! 14! 20! 14! If (n + 2)! = 60(n – 1)! Put n = 3 (3 + 2)! = 5! = 5 × 4 × 3 × 2 × 1 = 60 × 2 × 1 = 60 × (3 – 1)!

 n  2 !   n  2  n  1!  n  2  n  1!  n  1!

Formative Worksheet

(n + 4)(n + 3) × (n + 2)(n + 1)(n)(n – 1)........ = (n + 4)

 n  5!  6  n  4 ! ;

(n + 5) ×

 n  4 !  6  n  4 !

n + 5 = 6; n=1 5! : 4! 5×4×3×2×1:4×3×2×1 120 : 24 = 5 : 1

 n  1!  3  9 ;  n  2 !

 n  1

 n  2 !  3  9  n  2 !

3n – 3 = 9;

3n = 12;

11.

1 1  ; 8! 7!

1 1 1 8 9    8  7! 7! 8  7! 8  7!

12.

1 1 x   ; 9! 10! 11!

11 x x  ; 11 = ; 10 11 10 11 13.

 n-5! n!

÷

 n  5 !  n!

 n  4 ! n!



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1 11 x 1    9! 10 11 10 9! x = 121.

1 then the value of n is 6

n! 6  n  4 ;

n – 5 = 6 n = 11.

n=4

1 1 x   9! 10! 9! 11 10  9!

1 1 x 1 1    ;  9!  10  11  10 9!

n  5

8th Class Mathematics 14. Number of zeroes in the product 25 × 40 = 1000 – 3 125 × 8 = 1000 – 3 250 × 40 = 10000 – 4 200 × 5 = 1000 – 3 100 × 10 = 1000 – 3 20 × 5 = 100 – 2 75 × 4 = 100 – 2 500 × 20 = 1000 – 4 1250 × 8 = 10000 – 4 5 × 2 = 10 – 1 2500 × 4 = 10000 – 4 15. The product (5 × 2)n will decide number of zeroes in product Example: 2000 × 5 = 10000 – number of zeroes 4.

 n  4 !  6  n  4 !

50! is 48! (B) 2460

43. The value of (A) 2450

(C) 2470

(D) 2440

10! is 7!3! (A) 100 (B) 120 (C) 140 (D) 160 45. Which of the following has zero at the units place? (A) 3! (B) 4! (C) 5! (D) 9! 44. The value of

46. If

 n  4 !  20 , then the value of n is  n  2 !

(A) 8 (B) 7 (C) 6 (D) 5 47. LCM of 4!, 6! is (A) 480 (B) 690 (C) 720 (D) 960 48. If (n + (C)! = 56(n + (A)! then n = (A) 6 (B) 5 (C) 3 (D) 2 49.

 n  3 !   n  5 !   n  8 !   n  2 !  n  4 !  n  7 ! (A) n = 3

50.

(C) n = 5

(D) n = 1

(C) n = 5

(D) n = 8

 n  12 !  17 , then  n  11! (A) n = 4

51.

(B) n = 4

25, then

(B) n = 6

 n  8 !  n  7 ! n!



n!



1 , the value of n is 4

31 33 37 35 (B) (C) (D) 4 4 4 4 52. The number of zeroes at the end of the product 12 × 18 × 15 × 40 × 25 × 16 × 55 × 105 is (A) 7 (B) 8 (C) 5 (D) 6 (A)

Number theory

127 Observation: If the number whose last digit is 5, is multiplied by even number (including zero), the unit digit of the product will always be zero. For example: 82 15 = 1230, 156 45 = 7020, 62 13 65 = 52390, etc.

Conceptive Worksheet 42. 5 × 6 × 7 × 8 × 9 × 10 × 11 × 12 = (A)

12! 12! 12! 12! (B) (C) (D) 4! 2! 3! 5!

43. If n! 

 n  4 !  n  1!

(A) 5 (B) 18 44. The HCF & LCM (A) 12! & 32! (C) 26 & 403 45. If 46.

Now lets make few more observations in which the unit digit of the resultant value depends upon the unit digits of all the participating numbers . Observe the following.

then the value of n is (C) 6 (D) 9 of 13! & 31! are respectively (B) 13! & 31! (D) 46 & 36

12 + 17 + 13 + 47 = 89 2 + 7 + 3 + 7 = 19 Thus it is clear that the unit digit of the resultant value 89 depends upon the unit digits 2, 7, 3, 7

1 1 x + = , then the value of x = _________ 6! 7! 8!

48! 53!   is 47! 52! (A)102 (B) 104

Similarly, 63 × 87 × 59 = 317898 3 × 7 × 8 = 168

(C) 103

(D) 101

19! 11!   is 47. 18! 10! (A) 7 (B) 8 (C) 9 (D) 4 48. If (n + (A)! = 90(n – (A)! then n = (A) 11 (B) 10 (C) 8 (D) 9

9. Finding Unit digit of the Resultant Value Observe the following : 1×5=5 3 × 5 = 15 5 × 5 = 25 7 × 5 = 35 9 × 5 = 45 11 × 5 = 55 ........................ ........................ Observation: If the number whose last digit is 5, is multiplied by any odd number, the unit digit of the product will always be 5. For example: 13 15 = 195, 19 35 = 665, etc. Now, observe the following: 2 × 5 = 10 4 × 5 = 20 6 × 5 = 30 8 × 5 = 40 10 × 5 = 50 12 × 5 = 60 .................... ......................

Observation: So we can find out the unit digit of the resultant value only by solving the unit digits of the given expression.

Cyclicity Observe the following: 31 = 3 32 = 9 33 = 27 34 = 81 35 = 243 36 = 729 37 = 2187 38 = 6561 39 = 19683 etc. Similarly, 41 = 4 42 = 16 43 = 64 44 = 256 45 = 1024 46 = 4096 Observation: The unit digit follows a periodic pattern that is after a particular period it repeats in a cyclic form.

Note: 1. 2.

The unit digit of 21, 25, 29, 213....... is the same which is 2. The unit digit 22, 26, 210, 214,...... etc. is 4, www.betoppers.com

8th Class Mathematics

128 3. 4. 5. 6. 7. 8. 9. 10.

The unit digit of 31 is 3. The unit digit of 32 is 9. The unit digit of 33 is 7 The unit digit of 34 is 1. The unit digit of 35 is 3. The unit digit of 36 is 9. The unit digit of 37 is 7. The unit digit of 38 is 1. Observation: The last or unit digit must follow a pattern. It can be seen that the unit digit of 2 repeats after every four steps , the unit digit of 4 repeats after every 2 steps and the unit digit of 3 repeats after every four steps.

Formative Worksheet 53. The number in the units place of (62(C)36, (62(C)38, (62(C)39 respectively are (A) 1, 3, 7 (B) 1, 9, 7 (C) 1, 3, 9 (D) 3, 7, 9 54. The number in the unit place of (12(B)20, (12(B)22, (12(B)23 respectively are (A) 2, 4, 6 (B) 2, 4, 8 (C) 6, 4, 8 (D) 6, 8, 6 55. The last digit in the product

3  3 1

Examples: 1. Find the unit digit of 123 + 345 + 780 + 65 + 44. Sol. We can find the unit digit just by adding the unit digits 3, 5, 0, 5, 4 as 3 + 5 + 0 + 5 + 4 = 17. So the unit digit (or the last digit) of the resultant value of the expression 123 + 345 + 780 + 65 + 44 will be 7. (you can verify it by doing the whole sum) 2. Find the unit digit of 676 × 543 × 19. Sol. We can find the unit digit of the product of the given expression just by multiplying the unit digits (6, 3, 9) instead of doing the whole sum. Thus 6 × 3 × 9 = 162. Hence, the unit digit of the product of the given expression will be 2. (you can verify it by doing the complete sum). 3. Find the unit digit of 135 × 361 × 970. Sol. The unit digit can be obtained by multiplying the unit digits 5, 1, 0. Then 5 × 1 × 0 = 0 thus the unit digit will be zero. 4. Find the unit digit of the product of all the odd prime numbers. Sol. The prime numbers are 3, 5, 7, 11, 13, 17, 19.... etc. Now we know that if 5 is multiplied by any odd number it always gives the last digit 5. So the required unit digit will be 5. www.betoppers.com

 33  .......  310  is _______

(A) 1 (B) 2 (C) 3 56. Column - I a) The last digit of 38 10

Some Important Observations 1. The last digit (or unit digits) of 0, 1, 5, 6 is always the same irrespective of their powers raised on them. 2. The last digit of 4 and 9 follows the patter n of oddeven i.e., their cyclic period is 2. 3. The last digit of 2, 3, 7, 8 r epeats after every 4 steps i.e., their cyclic period is 4.

2

b) The last digit of  21

(D) 7 Column - II 1) 5 2) 6

c) The last digit of 216 d) The last digit of 520 57.

58. 59.

60.

61.

3) 5 4) 20 5) 1 The number in the units place of (98)40, (98)42 and (98)43 respectively are (A) 6, 4, 8(B) 2, 4, 8 (C) 6, 2, 4 (D) 6, 4, 2 The units digit in 217 × 818 × 193 is (A) 0 (B) 1 (C) 7 (D) 8 The respective digits in the unit’s place in the expansions of 77 and 177 are (A) 2, 6 (B) 3, 3 (C) 1, 4 (D) 9, 9 When 4157 is multiplied with 7113 , the digit in the least significant place is (A) 7 (B) 4 (C) 8 (D) 2 The last digit of the number obtained by multiplying 1750and 1976 is (A) 1 (B) 2 (C) 3 (D) 9

Conceptive Worksheet 49. The number in the unit place of 72958 is (A) 1 (B) 7 (C) 9 (D) 5 9 50. The unit digit of 3 is _______ (A) 3 (B) Least odd prime number (C) 7 (D) 9 51. The last digit of 31002 is _______ (A) 3 (B) 9 (C) 7 (D) 1 1002 52. The last digit of  3  10  is _______

(A) 3 (B) 9 (C) 7 (D) 0 53. The last digit of the multiplication 3153 × 7162 is _____

Number theory

129

54. The units digits in the square of 2 digit numbers ending in 2, 4, 6, 8 are (A) 4, 6 (B) 2, 4 (C) 4, 8 (D) 8, 6 55. The digit in the units place of the square root of 9216 is (A) 4 (B) 2 (C) 9 (D) 6 56. The digit in the units place of the cube root of 729 is (A) 3 (B) 4 (C) 9 (D) 1

10. Types of Numbers 1. 2. 3. 4. 5. 6. 7. 8.

Set of odd numbers : 1,3, 5......... Set of even numbers : 2, 4, 6....... Set of prime numbers : 2, 3, 5, 7, 11, 13....... Set of co prime numbers : 8, 25; 2,3 ; 3, 14 Twin prime : 3,5 ; 5, 7 ; 11, 13..... Perfect numbers : 6, 28, 496 Armstrong number : 153 [13 + 53 + 33 = 153] Factorial numbers = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720 9. Nice number = 6 [6 has exactly 4 factors 1, 2, 3, 6] 10. Palindrome number : 1347007431, [Reverse number is also the same number] 11. Harshad number

Number 24 4 = sumof digits 6

Numerator digits total 2 + 4 = 6 Denominator digits total is also 6 12. Powerful number 33 + 44 + 33 + 55 = 3435 (Here each power is equal to its base) 13. Friendly numbers or pair of amicable numbers Example: 220, 284

Numbers according arrangement structure: 14. 15. 16. 17. 18. 19. 20. 21. 22.

to

their

Triangular numbers, 1, 3, 6, 10, 15.......... Square numbers 4,9........ Rectangular numbers 6, 12....... Cube numbers 1, 8........ Tetrahedral number is 4. Pyramidal number is 5. Three squares which are in A.P. are - 1, 25, 49......... If 42 + 52 + 62 = x2 + y2 + z2 then x = 8, y = 3, z = 2 If 1x + 4x + 5x + 6x + 9x = 2x + 3x + 3x + 7x + 7x + 8x for several powers n = 1, 2, 3....... such a relation is called a multigrade.

23. Squares numbers containing all the 9 digits un repeated. Example: 118262 = 139854256. 24. Square numbers containing all the 10 digits un repeated. Example: 456242 = 2081549376. 25. Difference of two squares containing the ‘9’ digits 111132 – 2002 – 123458769 26. Numbers which are in A.P the sum of whose squares is a square Example: 22 + 52 + 82 + 112 + 142 + 172 + 202 + 232 + 262 = 482 27. The number 48 has a peculiar property. That is unity is added to it the sum is a square (48 + 1 = 49 = 72) and unity is added to its half (24)the result is 25 which also a square 28. Relation between cubes 13 + 6 3 + 8 3 = 9 3 93 + 123 + 153 = 183 13 + 3 3 + 4 3 + 5 3 + 8 3 = 9 3 29. Fascinating facts about 4 th powers of certain numbers 304 + 1204 + 2724 + 3154 = 3534 44 + 64 + 84 + 94 + 144 = 154 30. Teasing relationship 5th powers of certain numbers 45 + 55 + 65 + 75 + 95 + 115 = 125 31. Crazy 6th powers of certain numbers 16 + 26 + 46 + 56 + 66 + 96 + 126 + 136 +156 + 166 + 186 + 206 + 216 + 226 + 236 = 286 32. 1302 has a unique property 1302 = 12 + 142 + 242 + 272 + 522 + 572 + 632 + 742 = 52 + 102 + 202 + 312 + 482 + 612 + 672 + 702 33. 4 is a confusing number as: 22 = 2 × 2 = 2 + 2 = 4 34. Certain numbers can also be expressed as the difference between 2 squares 3 = 22 – 12 , 5 = 3 2 – 22 35. If a number is a square number it has to be the sum of consecutive odd numbers starting from 1 1 + 3 = 4 = 22, 1 + 3 + 5 = 9 = 32, 1 + 3 + 5 + 7 = 16 = 42 36. The square of any odd natural number other than 1 can be expressed as the sum of two consecutive natural numbers. Example: 32 = 9 = 4 + 5, 52 = 25 = 12 + 13 www.betoppers.com

8th Class Mathematics

130 37. Relations between squares 12 + 2 2 + 2 2 = 3 2 22 + 3 2 + 6 2 = 7 2 32 + 42 + 122 = 132 n  n +1 . 2 39. Sum of squares of ‘n’ natural numbers is

38. Sum of ‘n’ natural numbers is n  n +1 2n +1 . 6

40. Sum of cubes of ‘n’ natural numbers is

n 2  n +1 4

3 3 3 3 64. The sum of n terms 1 , 2 ,3 , 4 ........... is

(A)

n  n  1 2

(B)

n  n  1 n  2  6 2

 n  n  1   (C) (D)  2  4  65. The sum of the squares of the first ‘n’ natural numbers is n 2  n  1

Sn   n 2 

2

1 3  n  n 2  1 3

(B)

1 n  n  1  2n 2  2n  1 6

(C)

1  2n 2  2n  1 3

(D)

1 n  n  1  2n 2  2n  1 3

.

62. Observe the following statements (A) 13 + 123 = 93 + 103, B) 33 + 43 + 53 = 63, (C) 93 + 123 + 153 = 173 , D) 93 + 123 + 153 = 183 of these the true statements are (A) B, D, C (B) A, C, D (C) A, B, D (D) A, B, C 63. The sum of first 10 natural numbers is (A) 55 (B) 58 (C) 56 (D) 54

2

(A)

2

Formative Worksheet

2

68. If 12345679 × 9 = 111111111, 12345679 × 18 = 222222222, then 12345679 × 63 = (A) 666666666 (B) 555555555 (C) 888888888 (D) 777777777 69. The sum to n terms of the series 12 + (12 + 3(B) + (12 + 32 + 5(B) + .... is

n(n  1)(2n  1) , then 6

112 +122 + ...... + 202 = _________ (A) 2850 (B) 2860 (C) 2485 (D) 2495 66. If a = 2n + 1, b = 2n2 + 2n, c = 2n2 + 2n + 1 and n = 10, then the sides of right angled triangle are (A) 22, 220, 221 (B) 23, 220, 221 (C) 21, 220, 221 (D) 21, 224, 224 67. If 10989 × 9 = 98901, 109989 × 9 = 989901, then 1099989 × 9 = (A) 9899901 (B) 9999901 (C) 9989991 (D) 9889991

Conceptive Worksheet 57. The sum of squares of first 11 natural numbers is (A) 504 (B) 502 (C) 500 (D) 506 58. The sum of the squares of the first ‘n’ natural numbers is

Sn   n 2 

n(n  1)(2n  1) , then, 6

112 +122 + ...... + 302 = _________ (A) 9425 (B) 9435 (C) 9445 (D) 9070 59. The sum of the squares of the first ‘n’ natural numbers is

Sn   n 2 

n(n  1)(2n  1) , 6

then,

12 + 22 + 32 + ....... +1002 = _________ (A) 338350 (B) 339350 (C) 337350 (D) 337050 60. Column-I Column-II a) 1+ 2 + 3 + 4 +.... +10 = _____ 1) 55 b) 12 + 22 + 32 + 42 + .... +102 = _____ 2)

2

c) 13 + 23 + 33 + ....... +103 = _________ 3) 385 3

3

3

3

d) 1 + 2 + 3 + ....... + n = _________ 4) 3025 5)

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 n 

n 2 (n  1) 2 4

Number theory 61. The sum of 51  52  53  ......  100 is ____________ 62. Find the sum to n terms of the series 3 + 6 + 10 + 16 + ..... n  n  1  1 (B) n(n + (A) + 2n – 1 2 (C) n(n + (B) + 1 (D) 3(2n + (A) – 2n 63. The consecutive cubes whose sum is a cube is (A) 23, 33, 43 (B) 43, 53, 63 (C) 33, 43, 53 (D) 53, 63, 73 64. The sum of 100 terms of the series 1 – 2 + 3 – 4 + 5 – 6 + .... ? (A) 100 (B) 50 (C) –550 (D) –50

(A)

11. Congruences If a and b are two integers and m is a positive integer, then a is said to be congruent to b modulo m if m divides a – b denoted by m|(a – b). In notation form we express it as a  b mod m or a  b  0 mod m

131

Formative Worksheet 70. Which of the following is true? A) 481  1 (mod 12) B) 111  1 (mod 11) C) 111  12 (mod 11) D) 100  10 (mod 20) (A) A, C (B) A, D (C) A, B (D) C, D 71. The remainder when 625 is divided by 13 is (A) 1 (B) 7 (C) 3 (D) 9 72. (1! + 2! + 3! + 4!)  p (mod 10) then p is (A) 3 (B) 4 (C) 2 (D) 1 + 98 100 73. The remainder when 2 ÷ 7 is (A) 1 (B) 2 (C) 3 (D) 4 74. The remainder when 3100 ÷ 7 is (A) 1 (B) 2 (C) 3 (D) 4 32

75. The last digit of 3232 is (A) 5 (B) 6 (C) 7

(D) 8

Conceptive Worksheet

65. If 125  x (mod 20) then the value of x is (A) 4 (B) 6 (C) 5 (D) 10 Note:  66. If 28 – p (mod 16) ; 23  q (mod 5) then p and q is 1. a  b mod m, then m|(a – b) or (a – b) is a multiple of m. (A) p = – 12, q = 3 (B) p = 12, q = 3 (C) p = – 12, q = –3 (D) p = 24, q = 3 2. If m|(a – b) [m does not divide (a – b), then a is said to be incongruent to b mod m and this fact is 67. Which of the following statement is true? expressed as a is not congruent to b mod m. (A) 45  –5 (mod 10) 3. If m|a, then a  0 mod m (B) 157  7 (mod 15) (C) 17 is not congruent to 3 and 5. Example: (D) 72  8 (mod 9) 1. 13  1 mod 4 ( 4|(13 – 1) = 12) 68. If 100  x (mod 7), then least positive value of x is 2. 4  1 mod 5 ( 5|4 – (–1)) = 5) (A) 1 (B) 2 (C) 3 (D) 4 3. 12  0 mod 4 ( 4|12) 69. p  q (mod l) if 4. 17 is not congruent to 3 mod 5 ( 5|(17 – 3)) (A) (p – q) |l (B) l| (p – q) Properties: (C) l|p (D) l|m 1. If a  b mod m, then 70. If 58  4 (mod n), then n can be (A) 12 (B) 18 (C) 27 (D) 16 i) a + c  b + c mod m 71. Observe the following columns. ii) ac  bc mod m, where c is any integer Column-I Column-II 2. If a  b mod m and c  d mod m, then a) 3x  5 (mod 6] then x 1) 3 b) The remainder when 2) 11 i) a  c  b  d mod m 231 is divisible by 5 ii) a  c  b  d mod m c) 4832718 is divisible by 3) 2 iii) ac  bd mod m d) 7x 5 mod 8 then x 4) No solution k k 5) 3 3. If a  b mod m, then a  b mod m for every 72. The unit digit of 111! (factorial 111) is positive integer k. (A) 5 (B) 6 (C) 7 (D) 8 www.betoppers.com

132

8th Class Mathematics

11. Observe the following columns. Column-I Column-II 1. The smallest divisor (other than 1) of a composite a) General form of 1) 2n , where n  N number is a/an even number (A) odd number (B) even number b) General form of 2) 2n + 1 , where n  W (C) prime number (D) composite number odd number 2. If a2 – b2 is prime, then the correct option is c) Even prime is 3) 2n – 1, where n  N (A) a2 – b2 = a – b (B) a2 – b2 = a + b d) Sum of two even 4) Even (C) a2 – b2 = ab (D) a2 – b2 = 2 numbers 3. If p is a prime and n is a positive integer, then GCD of p and n is 5) 2 (A) 1 (B) p (C) n (D) 1 or p n 12. The least value of n, as 22  1 is a prime number,, 3 3 4. If x – 1 = 1,330 and y – 8 = 1,720. Then which of is ____________ the following is false? 13. lf n3 – 1 is a prime number, where n  N, then (A) y is multiple of x or x is a multiple of y (A) (n + (A)2 is prime (B) n(n + (A) is prime (B) Both x and y are even numbers (C) n – 1 is prime (D) n2 + n + 1 is prime (C) Both x and y are odd numbers 14. a, b, c are three prime numbers their sum and (D)(x + y)2 + 96 is a perfect square. product are 15 and 105 respectively, then the prime 5. Statement-I: If a and b are twin primes and numbers are a2 – b2 = 120, then their average is 30. (A) 2, 3, 5 (B) 3, 5, 7 Statement-II: If a2 – b2 is positive and a, b are twin (C) 5, 7, 11 (D) 7, 11, 13 primes, then a – b = 2. 15. If p = 5 and a = 2 then ap – a is divisible by (A) Both Statements are true, Statement II is the (A) p + 4 (B) p + 2 (C) p + 3 (D) p correct explanation of Statement I. (B) Both Statements are true, Statement II is not 16. Cube root of 2744 is (A) 14 (B) 13 (C) 17 (D) 16 correct explanation of Statement I. 17. The square of a prime number is a (C) Statement I is true, Statement II is false. (A) prime number (B) composite number (D) Statement I is false, Statement II is true. (C) perfect number (D) can’t say 6. Average of first 15 natural numbers is 18. If a, b are twin primes and b, c are twin primes, (A) 2 (B) 4 (C) 8 (D) 6 then the value of a + c is 7. The sum of digits of a 2 digit odd integer is 9. If (A) 10 (B) 15 (C) 5 (D) Can’t say their difference is 3 then the odd number is 4 2 2 2 3 19. 2 × 3 × 7 and 3 × 9 × 54 are two numbers. If (A) 63 (B) 54 (C) 27 (D) 72 the two numbers are multiplied, then resultant is 8. If x, y and z are odd positive integers, then (A) Odd (B) Even i) x2 + y2 + z2 + 12 is even (C) Negative integer (D) None ii) x2y2 + y2z2 + (z – x)2 – 45 is odd 5 2 2 3 20. 3 × 7 and 9 × 5 are two numbers. If the two iii) x2[y2 + z2 – (x – y)2 – (y – z)2 + 125] is even numbers are added, then the sum obtained is iv) x3[(y3 + z3 – (y – x)3 – (z – x)3] + 73 is odd (A) Odd (B) Even (A) (i) and (ii) are true (C) Negative integer (D) None (B) (i) and (iii) are true 21. If sum of first 9 odd numbers is added to sum of (C) (ii) and (iii) are true first 11 odd numbers, then the resultant answer is (D) (ii) and (iv) are true. (A) 203 (B) 204 (C) 202 (D) 201 9. Average of first 15 even natural numbers is 22. Find the least number which when multiplied by 882 (A) 16 (B) 17 (C) 15 (D) 19 gives a perfect square number. 10. Average of first 15 odd natural numbers is (A) 1 (B) 2 (C) 3 (D) 4 (A) 19 (B) 17 (C) 15 (D) 16 23. Which of the following numbers is exactly divisible by all the prime numbers between 10 and 20? (A) 75294 (B) 46189 (C) 35273 (D) 48391

Summative Worksheet

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Number theory 24. Statement – I: 5 should be added to 583171 to make it exactly divisible by 11. Statement – II: A number is divisible by 11 only when the sum of its digits is divisible by 11. (A) Both Statements are true, Statement II is the correct explanation of Statement I. (B) Both Statements are true, Statement II is not correct explanation of Statement I. (C) Statement I is true, Statement II is false. (D) Statement I is false, Statement II is true. 25. H.C.F of 4 × 27 × 3125, 8 × 9 × 25 × 7 and 16 × 81 × 5 × 11 × 49 is (A) 1260 (B) 540 (C) 360 (D) 180 26. If a = 142 – 5, b = 142 – 3 are two numbers, then they are (A) even numbers (B) odd numbers (C) twin primes (D) Both B & C 27. Statement – I : L.C.M of two number is 432 and their H.C.F is 72. If one no. is 144, then the other number is 216. Statement – II : Product of the two numbers = product of their LCM & GCD. (A) Both Statements are true, Statement II is the correct explanation of Statement I. (B) Both Statements are true, Statement II is not correct explanation of Statement I. (C) Statement I is true, Statement II is false. (D) Statement I is false, Statement II is true. 28. If n2 + 9n + 20 is the product of two prime numbers, for n being some integer, then n2 + 1 = (A) 0 (B) 2 (C) 5 (D) can’t say 29. Which of the following are correct i) xn + yn is divisible by x + y, only if n is an odd number ii) xn – yn is divisible by x – y, only if n is an even number iii) xn – yn is divisible by x + y, only if n is an even number iv) xn + yn is not divisible by either x + y or x – y, if n is an even number (A) (i) and (ii) (B) (ii), (iii) (C) (i), (iii) and (iv) (D) All are correct 30. Find the number of zeroes in the product of Column – I Column – II a) 10! 1) 14 b) 10 + 100 + 1000 + 10000 2) 10 c) 10×100×1000×10000 3) 1 3 5 4 7 7 2 d) 2 × 5 × 2 × 5 × 2 ×5 4) 2 5) 100/2

133 31. The number of prime numbers that can be written as a sum and difference of two primes is (A) 1 (B) 2 (C) 5 (D) 0

a2 b4 c6    12 , then number of 2 4 6 factors of ‘abc’ are (A) 20 (B) 40 (C) 45 (D) 50 x y x+2 y+1 33. If 864 = 2 × 3 then 2 × 3 × 12 represents a composite number. Then the composite number is (A) 2x + 4 × 3y + 6 (B)2x + 4 × 3y + 2 (C) 2x + 4 × 3y + 4 (D) 2x + 2 × 3y + 4 34. If A = 1112, B = 101112, C = 111112 then the number of factors of ABC is (A) 8 (B) 4 (C) 6 (D) 9 32. If

35. By how much does (A)

2 4 3

(C) 2( 3  2)

12  18 exceed (B)

3 2?

32 2

(D) 3( 3  2)

36. If 729x 9 is divisible by 3, then the value of x (A) 0 (B) 3 (C) 6 (D) 9 37. Statement – I: The least number which is a perfect square and which is also divisible by 16, 18, 15 is 3600. Statement – II: Perfect square is a number which is a square of some another natural number. (A) Both Statements are true, Statement II is the correct explanation of Statement I. (B) Both Statements are true, Statement II is not correct explanation of Statement I. (C) Statement I is true, Statement II is false. (D) Statement I is false, Statement II is true. 38. Given A = 36, B = 48, C = 80, D = 1025. If A 2  B2 = x, then the value of x is

(A) 62

(B) 60

(C) 59

(D) 61

39. In the above problem, if x 2  c 2 = y, then the value of y is (A) 110 (B) 90 (C) 100 (D) 120 40. Statement – I: If

 n  2 !  10!  6 , then n = 14.  n  1! 9!

Statement – II: n! = n × (n – (A). (A) Both Statements are true, Statement II is the correct explanation of Statement I. (B) Both Statements are true, Statement II is not correct explanation of Statement I. www.betoppers.com

8th Class Mathematics

134 (C) Statement I is true, Statement II is false. (D) Statement I is false, Statement II is true. 41. In the above problem, if y 2  1025 = z, then the value of z is (A) 105 (B) 75 (C) 95 (D) 115 42. Column – I Column – II Factorial Value

16! 15! 16! b) 14! 16! c) 0! 16! d) 13! a)

1) 240 2) 16! 3) 3360 4) 16

5) 123 43. The units digit in the square of a 2 digit number ending in 1, 3, 5, 7, 9 is (A) 1, 5, 9 (B) 2, 5, 9 (C) 3, 5, 9 (D) 7, 5, 9 44. Statement–I : The last digit of the sum  210  310  510  is “8”. Statement–II : The last 1110  2110  3110  is “3”.

47. If x  2 (mod (C) where x < 9 then the value of x is (A) x = 8 (B) x = 7 (C) x = 6 (D) x = 2 48. Statement-I: The least positive integer to which 79 X 125 congruent modulo 11 is 8. Statement-II: If a  b mod m then ac bc mod m. Where c is any integer. (A) Both Statements are true, Statement II is the correct explanation of Statement I. (B) Both Statements are true, Statement II is not correct explanation of Statement I. (C) Statement I is true, Statement II is false. (D) Statement I is false, Statement II is true. 49. If 17  –4 (mod (C) and 28  7 (mod (C) if 17 + 28  p (mod (C) the value of p is (A) 2 (B) 3 (C) 4 (D) 1 50. Column-I Column-II a) 1000  x (mod 21) 1) 13

digit of the sum

(A) Both Statements are true, Statement II is the correct explanation of Statement I. (B) Both Statements are true, Statement II is not correct explanation of Statement I. (C) Statement I is true, Statement II is false. (D) Statement I is false, Statement II is true. 45. The sum of n terms of the series 1 + (1 + (C) + (1 + 3 + 5) + .... is :

 n  n  1  (A)    2 

(C) Statement I is true, Statement II is false. (D) Statement I is false, Statement II is true.

n  n  1 2n  1 n  n  1 (D) 6 2 46. Statement–I : The sum of the cubes of first 50 natural numbers is 1625625. Statement–II : The sum of the cubes of first ‘ n’

1.

(mod 1235)

3) –1

d) 17  x

(mod 5)

4) 3

Show that odd perfect squares are of the form

2.

Find the value of

3.

3 5 7 2005  2 2  2 2  .......  2 1 .2 2 .3 3 .4 10022.10032 If 1  2  3  4  .......  n  n! (read as ‘n’ factorial) then show that ‘N’ where N= 1!  2!  3!  4!  5!  6!  7! 8!  9!  10! is not a perfect square. 2

4.

5.

Given that ‘a’ and ‘b’ are consecutive and c  ab then show that D is always odd provided that

D  a 2  b2  c2 . Given that K   a  b  c  b  c  a  c  a  b  a  b  c  and it is known that ‘K’ is even. Show that the

number  a  b  c  is even , if a, b, c are positive integers.

natural numbers is

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2) 10

 8k  1 .

(C)

n 2 (n  1) 2 . 4 (A) Both Statements are true, Statement II is the correct explanation of Statement I. (B) Both Statements are true, Statement II is not correct explanation of Statement I.

(mod 100)

HOTS Worksheet

2

(B) n2

b) 2010  x c) 1234  x

6.

2 2 2 If a  b  c then show that

an integer.

 b  c  a 2

is always

Number theory 7.

Prove that for every positive integer n ( 1n  8n  3n  6n ) is divisible by 10.

8.

Show that 44n  53n is divisible by 131, n being any positive integer. Show that the square of an integer cannot be in the form 4n + 3 or 4n + 2 where n  N Show that no square number can end with 4 ones or 4 nines. Prove that the product of four consecutive positive integers increased by 1 is a perfect square. Can one form a “magic square” out of the first 36 prime numbers?. ‘A magic square’ here means a 6 x 6 array of boxes, with a number in each box and such that the sum of the numbers along any row, column, or diagonal is constant. Prove that the number (n3+2n) is divisible by ‘3’ for any natural number ‘n’. Prove that (n3-n) is divisible by 24 for any odd ‘n’. Find the last digit of the number 19891989. Find the last digit of the number 250 × 350?. What is the last digit of 777777?. Find the remainder of 2100 when divided by 3. Find the remainder when the number 31989 is divided by 7. If it is known that (a+1) is divisible by 3 then prove that (4+7a) is also divisible by 3. Can sum of two odd perfect squares be another perfect square?. Can sum of three squares of odd natural number be a perfect square?. Suppose m and n are positive integers with mn = 40000. Suppose further that neither ‘m’ nor ‘n’ is divisible by 10.What is the value of (m+n)?. Find the natural numbers n for which the fraction

9. 10. 11. 12.

13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23.

24.

15n 2  8n  6 is a natural number . n 25. Prove that a) the sum (ab+ba) is a multiple of 11. (ab is a two digit number) b) a three digit number written by one and the same digit is entirely divisible by 37.

26. Prove that the difference 1025  7 , is divisible by 3. 27. Prove that for any prime p>3 the number ( p 2  1 ) is divisible by 24.

135 29. Find the two digit number equal to the triple sum of its digits. 30. Compare the two fractions

161 9999 and . 160 9998

IIT JEE Worksheet 1.

2. 3.

4. 5.

6. 7. 8. 9.

Show that

12007  22007  32007  42007  .......  20062007 i s divisible by 2007. Find the units and tens digits of S, where S  1! 2! 3! 4! ....  2006! . All two digit numbers are written from 10 to 99 consecutively such that N =1011121314....979899. Show that 9 divides N. Can the number A consisting of 600 sixes and some zeros be a square.? The symbol n! is used to represent the product n(n1)(n-2)----(3)(2)(1). For example 4!= 4  3  2 1 Determine ‘n’ such that n! = 215  36  53  72  11 1  13. Prove that there are infinitely many primes. Find the last digit of the number 12+22+...+992. Show that x n  y n is divisible by x  y only when ‘n’ is odd natural number. Show that x n  y n is divisible by x  y for all the natural values of ‘n’.

10. Prove that 11n  2  122n 1 is divisible by 133 for any natural value of ‘n’. 11. Does there exist any natural number ‘n’ such that

n 2  n  1 is divisible by 1995. 12. Find a three digit number all of whose integral powers end with the same three digits as does the original number. 1 13. Find all positive integers ‘n’ so that is the n repeating decimal 1  .abcabcabcabcabcabc.........  .abc with a,b and n c distinct digits between 0 and 9. 25 14. If 100  25 is written in decimal representation, find the sum of the digits.

15. Prove that 1110  1 is divisible by 100. 16. Show that the equation x 2  y 2  174 has no solution in integers.

28. Prove that the square of any prime number p  5 , when divided by 12, gives ‘1’ as remainder . www.betoppers.com

8th Class Mathematics

136 17. A point (x,y) is called integral if both ‘x’ and ‘y’ are integers. How many points on the graph

18. 19. 20. 21.

1 1 1 + = are integral points? For example (x,y)= x y 4 (2,-4) is one such integral point. Compute the units digit of 171983 + 111983 – 71983. The sides of a right triangle are all integers. Two of these integers are primes that differ by 50. Compute the smallest possible value for the third side. Find all ordered triples (x, y, z) such that x, y and z are (positive) primes and xy + 1 = z. If 5 4  4 5  4 5  4 5 6 5  6 5  6 5  6 5  6 5  65   2n , 5 5 5 5 5 3 3 3 2 2

compute the positive integer ‘n’. 22. Complete this “cross-number puzzle” by putting the proper digit into each box. All answers to clues are 3-digit numbers (thus no answer begins with a zero). 1

2

3

4 5

Across Down 1. A prim 1. An integral power of 5 4. A composite 2. An integral power of 3 5. A square 3. An integral power of 2 23. Compute the integer N, where (3!) (5!) (7!) = N! 24. Since a two-digit palindrome has identical digits, we need only consider 112 and 222 [332 > 1000]. Both 1 and 484 are palindromes, so the answer is 2. 25. Compute the smallest positive integer K such that 1988K is a perfect square. (Pass back the number K)



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Mensuration

Learning Outcomes Triangles Area of Triangles Perimeter of Triangles Quadrilaterals Square Rhombus Rectangle

     

Parallelogram Quadrilateral Trapezium Polygons Area of Polygons Circles

1. Introduction

2. Triangles

Observe the following plane figures:

Triangle

Pentagon

Square

Circle

Rectangle

Semi-circle

Figures like triangle, square and rectangle are made of straight lines. Figures like semi-circle, circle are made of curves. Based on this, these figures are classified into two categories.

A triangle is a closed figure bounded by three straight lines. Mathematically, it is denoted by a symbol ‘D’. The point of intersection of sides is called Vertex denoted by upper case letters A, B, C……. And the sides are denoted by lower case letters. a, b, c ……. corresponding to its opposite vertex.

Altitude or Height of a Triangle: It is the length of perpendicular drawn from any vertex of triangle to its opposite side. Examples: A

Height

Categorization of Figures c

Figures

Rectilinear figures

Curvilinear figures

D

B

b

C

a

A

The figures that are formed with line segments are called Rectilinear figures.

The figures that are formed with both line segments and curves or only curves are called Curvilinear figures.

Height

      

Chapter - 10 9

By the end of this chapter, you will understand

b c

D

Examples: i) Triangle ii) Square iii) Rectangle iv) Polygons

Examples: i) Circle ii) Semi-circle iii) Arc iv) Sector

B

a

C

AB = c = length of side of AB BC = a = length of side of BC AC = b = length of side of AC Base of the Triangle: The side to which the perpendicular is drawn is called the base of the triangle.

8th Class Mathematics

138

Isosceles Triangle

Example:

A triangle having two sides of equal, is called an isosceles triangle. Example:

Height

A

B

A C

D Base

Classification of Triangles Triangles are broadly classified into two types. They are:

B

6cm

C

AB = AC  BC Isosceles triangle

Triangles

Equilateral Triangle Type - I

Type - II

Based on lengths of sides of Triangles

Type - III

A triangle having all the three sides of equal length, is called an equilateral triangle. Example: A

Based on angles and sides in Triangle.

Based on angles in Triangle.

B

6cm

C

AB = BC = CA

I.

Types of Triangles based on Lengths of Sides Based on lengths of sides of a triangle, they are classified into three types. They are: i) Scalene triangle ii) Isosceles triangle iii) Equilateral triangle

Scalene Triangle If all the three sides of a triangle are of the different lengths then the triangle is called a scalene triangle. Example: A

Equilateral Triangle

II. Types of Triangles based on Angles in Triangle Based on angles in a triangle, they are classified into three types. They are: i) Acute – angled triangle ii) Right – angled triangle iii) Obtuse – angled triangle

Acute-angled Triangle A triangle with all the three angles acute (i.e., < 90°) is called an acute-angled triangle. Example: In  ABC, we have, A = 60°, B = 45° and C = 75°

B

7cm AB  AC  BC

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C

  ABC is an acute angled triangle.

Right-angled Triangle A triangle is called right angled if the measure of one of its angles is 90°.

Mensuration

139

Example: In the adjacent figure,  DEF is right angled at E. (  E = 90°)

III. Types of Triangles based on Angles, Sides in Triangle

D

AB = BC  AC

4cm

A

90 E

F

4cm

B

Note: i) The side opposite to the right angle is called the hypotenuse of the triangle. ii)

Example: In  DEF, DF is the hypotenuse. In a right-angled triangle, one angle measures 90°, and each one of the remaining two angles is acute(< 90°).

Right-angled Isosceles triangle

Pythagoras Theorem: It states that in a right-angled triangle the square of the hypotenuse is equal to the sum of the squares of the base and height. Example: A

3cm

Obtuse-angled Triangle A triangle having one obtuse angle(> 90°) is called an obtuse-angled triangle. In the  PQR

 Q = 130° (> 90°)   PQR is an obtuse angled triangle.

C

4cm

90 B

4cm

C

By Pythagoras theorem, we have (Hypotenuse)2 = (height)2 + (base)2 (AC)2 = (AB)2 + (BC)2 52 = 32 + 42= 25 = 9 + 16 = 25 = 25

Identification of Triangles Let’s see how we can identify the triangles. Case – I: If all the sides are unequal. Let A, B, C be the three vertices and AB, BC and AC the sides of a  ABC. Let’s see how we can identify the triangle, if all the sides are unequal. All sides are unequal i.e., AB  BC  AC Check if the longest side (say AC) is equal to the sum of the other two sides i.e., AC = AB + BC

Yes

Then, A, B, C are collinear A

B

Yes C

Then, ABC is a right triangle, right angled at B A

B

No

Check if the square of the longest side is No equal to the sum of the squares of the other two sides i.e., AC2 = AB2 + BC2

Then, it is a scalene triangle A

C B

C

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8th Class Mathematics

140 If the two sides are equal.

Case – II:

The two sides are equal i.e., AB = BC  AC Yes

Check if the longest side AC is equal to the sum of the other two sides i.e., AC = AB + BC

Then, A, B, C are collinear A B C

Yes

No

Check if the square of the longest side is equal to the sum of the squares of the other two sides i.e., AC2 = AB2 + BC2

Then, ABC is a right-angled isosceles triangle

ABC is an isosceles triangle A

A

B

No

C

B

C

If all the sides are equal.

Case – III:

If all the three sides are equal, then the triangle is equilateral.

Summary of Types of Triangles Name of the figure

About sides

Figure

Special relation

A Scalene

AB  BC  CA

A  B  C B

C A

Isosceles

Angles opposite to equal sides are equal.

AB = AC  BC B

C A

Equilateral

All the three angles are the equal (i.e., 60 )

AB = BC = AC B

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C

Mensuration

141 Name of the figure Right-angled isosceles triangle

About sides

Figure A

B = 90 AB = BC

90 B

Acute-angled

C A

(A , B and C) < 90 B

90 E

F

PQR > 90 P and P + R < 90

Obtuse-angled

R

ii. Area of a Right Angled Triangle:

3. Area of Triangles Area: The area of a simple closed figure is the measure of the surface enclosed by it. Units: cm2 (read as sq.cm) m2 (read as sq. m)

i.

D

DEF = 90 and D + F = 90

Right-angled

C

Area of Scalene Triangle Let  ABC be a scalene triangle,

A d

h b

B

1 Area   base  height  2

1 2

Area of Triangle    base    height 

Here, AB = base, CD = height (or) S  S  a S  b  S  c  (Heron’s formula)

1 1 1    BC    AB    b  h  bh 2 2 2

iii. Area of an Isosceles Triangle:

abc where, s  2

AB = AC = s A

(or. S = half perimeter of triangle)  Area of scalene triangle



1  base  height  or 2 C

A

h

D

s

S  S  a S  b  S  c 

a

C

1 Area of triangle   base  height 2

a

c

s h

B

Height

b

C

B

1 1   a  h  ah sq units 2 2 www.betoppers.com

8th Class Mathematics

142 Let  ABC be a right-angled isosceles triangle,  ABC = 90° and AB = BC = a

ii. Perimeter of Right-angled Triangle: A

1   base  height 2

Area of triangle

h 90

1   BC  AB 2

B b C AB = h cm= height ; BC = b cm= base; AC = AC cm = hypotenuse

1 1   a  a  a 2 sq. units 2 2 A

Perimeter = sum of all the sides = (b + h + AC) cm

iii. Perimeter of Isosceles Triangle: A

a 90 B

C

a

s

 Area of Right-angled isosceles triangle 

1 2 a sq. units 2

s

B C a AB = AC = s cm s cm, s cm, a cm are the lengths of sides Perimeter = sum of all the sides = s + s + a = (2s + a) cm

iv. Area of Equilateral Triangle Let  ABC be an equilateral triangle, AB = BC = AC = a AD = h = height A

iv. Perimeter of Right-angled Isosceles Triangle: A

a

a h

a B Area of triangle is:

D

d

C

i)

3 2 a (if only side is given) 4

ii)

h2 (if only height is given) 3

3 2 a  Area of equilateral triangle  4

C B a AB = BC = a cm a cm, a cm, d cm are sides of the triangle Perimeter = sum of the all sides = a + a + d = (2a + d) cm

v. Perimeter of Equilateral triangle: h2  3

AB = BC = CA = a cm

A

4. Perimeter of Triangles i.

Perimeter of a Scalene Triangle:

a

a

Let a cm, b cm, c cm are sides of a scalene triangle.

c

b

B C a Perimeter = sum of the sides = (a + b + c) cm www.betoppers.com

C B a Where a, a, a are the sides of the triangle Perimeter = sum of all the sides = a + a + a = 3a cm

Mensuration

143

Formative Worksheet 1.

13. The area of an Isosceles right angled triangle is 32 square cms. Find the lengths of its sides?

Find the area of the following triangles

14. The area of an equilateral triangle is 64 3 sq.cm . Find its perimeter and it’s height.

L

Conceptive Worksheet

2cm

P

2cm

1. Q

2. 3.

4.

5. 6. 7.

S

4 cm

R

O

M

3cm

N

In  PQR, PR = 8 cm, QR = 4 cm and PL = 5 cm. Find: (i) the area of the  PQR (ii) QM The cost of painting the surface of a triangular board at 80 paise per square metre is Rs. 176.40. If the height of the board measures 24.5m, find its base. A ladder 40m long is placed so as to reach a window 24m high on one side of a street, and on turning the ladder over the other side of the street, it reaches a window 32m high. Find the breadth of the street. Find the sides of a isosceles triangle which are in the ratio 2 : 2 : 3, if the perimeter is 56cm. Find the perimeter of a right angled triangle if its base is 3cm and height is 4cm. Find the area of the triangle whose sides are 5, 6 and

2.

3.

4. 5. 6. 7.

61

8. The area of an equilateral triangle is 64 3 sq. cm., Find its perimeter and its height. 9. The base of an isosceles triangles is 12cm and its 9. perimeter is 32cm. Find its area. 10. The adjoining figure shows an equilateral triangle of side 10cm each and a right angled triangle BDC whose side CD = 8 cm and  D = 90°. Find the 8.

area of the shaded region.



3  1.732

The base of an isosceles triangle is 12cm and its perimeter is 32cm. Find its area. The base of a triangular field is 540m and its height is 235m. Find the cost of levelling the field at Rs. 25 per sq.m. A ladder 25 m long stands upright against a wall. Find how far the bottom of the ladder must be pulled out from the wall so as to lower the top by one metre. Find the area of a triangle whose sides measure 13 cm, 14 cm and 15 cm. Find the area of a right-angled triangle whose base is 12cm and hypotenuse is 13 cm. The perimeter of a right-angled triangle is 60 cm. Its hypotenuse is 26 cm. Find the area of the triangle. If the perimeter of an isosceles right triangle is

 6  3 2  m , find the area of the triangle. One side of a right-angled triangle is twice the other, and the hypotenuse is 10 cm. Find the area of the triangle. The height of an equilateral triangle is 10 cm. Find its area.



A

D

B

10cm

C

11. In two triangle, the ratio of the areas is 4 : 3 and the ratio of their heights is 3 : 4. Find the ratio of their bases. 12. Find the length of the altitude of an equilateral triangle of side 3 3 cm www.betoppers.com

8th Class Mathematics

144

5. Quadrilaterals Till now, we have seen three-sided figures. Let’s observe some four-sided figures, called Quadrilaterals. C

D

D

D

C

C

B

A

A

D

C

A

D

C

A

B

A

B

B

B D

C

A

B

In the above figures, some have all equal sides, some have unequal sides and some sides are parallel and some are not parallel to each other. Examples D

C

A

B

A square is a quadrilateral having all sides and diagonals equal and each angle measuring 90 .

Square D

C

A rhombus is a quadrilateral having all sides equal, but diagonals are not equal and perpendicular to each other.

Rhombus A

B

D

C

A

B

A rectangle is a quadrilateral having pairs of opposite sides parallel and equal, each angle measuring 90 . Note: The two diagonals are equal in length.

Rectangle D

A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel, is called parallelogram.

C

A

B

Note: The two diagonals are not equal in length and bisecting each other

Parallelogram C

D

Trapezium

A

A trapezium is a quadrilateral having one pair of opposite sides such parallel.

B D C

Quadrilateral

A quadrilateral is a figure whose sides are not equal and not parallel.

B

A

Type-II: Opposite sides are equal. i.e., AB = CD and BC = DA Example:

Types of Quadrilaterals Type-I: The four sides are not equal. AB  BC  CD  DA. Example: D

D

D

C

D

A

B

A

C

C

C

A

B

A

B

Note: The difference between quadrilateral and trapezium is that trapezium has two parallel sides www.betoppers.com

B

Note: The difference between rectangle and parallelogram is that diagonals in rectangle (AC and BD) are equal. Where as in a parallelogram, they are not equal.

Mensuration

145

6. Square

Type-III: If all the sides are equal. AB = BC = CD = DA Example: D

D

C

Let ABCD be a square. a

D

C

C

a

B

A

A

B

Note: Difference between square and rhombus is that diagonals of square are equal whereas in a Rhombus, they are not equal

Identification of Quadrilaterals Case – I:If all the sides are unequal.

Check if one pair of opposite sides are parallel

No

It is a quadrilateral D

Yes

Length of diagonal  2 a cm Length of side  Area cm

Trapezium D

C

7. Rhombus A

B

B

Note: If the non-parallel sides are equal, it is an isosceles trapezium.

Case – II:

1  diagonal 2 sq. cm 2

Perimeter of square = sum of all sides = a + a + a + a = 4a cm

C A

A a B AB = BC = CD = DA = a cm BD = diagonal Area of square = side × side = a × a = a2 sq.cm Area 

Check if all the four sides are unequal i.e., AB  BC  CD  DA

Let ABCD be a rhombus. a A

The opposite sides are equal. a

Check if opposite sides are equal i.e., AB = CD and AD = BC Yes

D

No

Check if the diagonals are equal i.e., AC = BD

It is a rectangle

A

Case – III:

B

C

It is a parallelogram C D

B

A

Check if the diagonals are equal i.e., AC = BD

It is a square D C

B

d2 a

a

C

AB = BC = CD = DA = a cm AC = d1 cm = diagonal BD = d2 cm= diagonal

1 Area of Rhombus    product of diagonals  2

If all sides are equal.

Yes

d1

D

B

1 1 1   AC  BD   d1  d 2  d1 d 2 sq.cm 2 2 2

Check if all the four sides are equal i.e., AB = BC = CD = AD.

A

a

Perimeter of Rhombus = sum of all the sides = a + a + a + a = 4a cm

No

It is a rhombus D C

A

B

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8th Class Mathematics

146

8. Rectangle

Formative Worksheet 15. Find the area of a square, one of whose diagonals is 3.8m long. 16. The diagonals of two square are in the ratio of 2 : 5. Find the ratio of their areas. 17. What is the perimeter of a square, if the length of

A

its diagonals is 12 2cm? The perimeters of two squares are 40 cm and 32 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares. A housing society has been allotted a square piece of land measuring 2550.25 sq.m. What is the side of the plot? Find the number of marble slabs of size 20 cm × 30 cm required to pave the floor of a square room of side 3 metres. One of the diagonals of a rhombus is double the other diagonal. Its area is 25 sq.cm. Find the sum of the diagonals. If the diagonals of a rhombus are 24 cm and 10 cm, the area and the perimeter of the rhombus are respectively. The perimeter of a rhombus is 56 m and its height is 5 m. Its area is: Find the perimeter, diagonal and area of a square

B

18.

19.

20.

21.

22.

23. 24.





a

b

b a

10. The area of square ABCD is 16 cm2. Find the area of the square joining the mid-points of the sides. 11. Find the area of a rhombus having each side equal to 13 cm and one of whose diagonals is 24cm. 12. If the area of a rhombus is 24 cm2 and one of its diagonals is 4 cm, find the perimeter of the rhombus. 13. The area of a rhombus is 150cm2. The length of one of its diagonals is 10cm. Find the length of the other diagonal. 14. A square lawn is surrounded by a path 2 m wide around. If the area of the path is 120 m2, find the area of the lawn. 15. The perimeter of a square grassy plot is 122m. Find the cost of cutting the grass at Rs. 4 per sq. metre. 16. The length of the diagonal of a square is 32 cm. Find the area and perimeter of the square.

 Take



2  1.41

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C

Let ABCD be a rectangle. AB = DC = b cm ; BC = AD = a cm Area of rectangle = length × breadth = BC × DC = a × b = ab sq. cm Perimeter of rectangle = sum of all the sides =a+b+a+b = 2(a + b) cm Length of diagonal  a 2  b 2 cm

9. Parallelogram a

A

b

B

D

h

having each side 8m long. Take 2  1.41

Conceptive Worksheet

D

E

b

a

C

Let ABCD be a parallelogram. AB = CD = b cm ; AD = BC = a cm AE = h cm Area of parallelogram = base × height = BC × AE = a × h = ah sq. cm ( height = distance between to parallel lines) Perimeter = sum of all the sides= a + b + a + b = 2(a + b) cm

Formative Worksheet 25. Length of a rectangle is 12 m and its diagonal is 15 m. Find the breadth and area of the rectangle. 26. The perimeter of a rectangle is 36cm and its breadth is 8.4cm. Find the length and area of the rectangle. 27. A rectangular grassy lawn measuring 38 m ´ 25 m has been surrounded externally by a 2.5 m wide path. Calculate the cost of gravelling the path at the rate of Rs. 4.50 per sq. metre.

Mensuration

147

28. The area of a rectangular park is 2160 m2. The 24. One diagonal of a parallelogram is 70 cm and the length and breadth of the park are in the ratio 5 : 3. perpendicular distance of this diagonal from either Find the cost of fencing the park at Rs. 5 per metre. of the outlying vertices is 27 cm. Find the area of 29. A field in the form of a parallelogram has one of its the parallelogram. diagonals 42m long and the perpendicular distance 25. One side of a parallelogram is 18 cm and its distance of this diagonal form either of the outlying vertices from the opposite side is 8 cm. Find the area of the is 10.8m. Find the area of the field. parallelogram. 30. A parallelogram has sides of 15 cm and 12 cm. If the distance between its shorter sides is 7.5 cm, 10. Quadrilateral find the distance between its longer sides. Let ABCD be an quadrilateral. 31. Find the area of a parallelogram whose two adjacent d D sides are of length 5cm and 7cm, and (i) if diagonal C L is 12cm (ii) if diagonal is 8cm. h2 32. The length of a room is twice its breadth and its a c breadth is equal to its height. The cost of papering h1 M the walls at 50 paise per square metre is Rs. 48. Find the height of the room. b A B 33. If the perimeter of a parallelogram to 32cm. Find AB BC CD DA    the length and the breadth which are in the ratio 3: BL = h = Altitude (perpendicular) 1. 1 DM = h = Altitude (perpendicular) 34. The length and the breadth of a rectangular field 2 are in the ratio 4 : 3. If the area of the field is 8112 Area of Quadrilateral m2, then find the cost of fencing the field at Rs. 1 4.00 per metre.    length of diagonal  2 35. A hall in the form of a rectangular region is 14 m × 12 m. How many marble slabs 8 cm × 6 cm are   sum of lengths of perpendiculars  needed to cover the floor of the hall? 1 1   AC   h1  h 2   AC  h1  h 2  sq.cm onceptive orksheet 2 2 17. The cost of carpeting a room 18 metres long with a 11. Trapezium carpet 75 cm wide at Rs. 4.50 per metre is Let ABCD be a trapezium. Rs. 810. Find the breadth of the room. D C 18. Find the perimeter and area of a rectangle with length 7.5cm and breadth 3.6cm. h 19. Find the area of a parallelogram whose base is 3.5 m and height 80 cm. 20. One side of a rectangular field is 15 m and one of its diagonals is 17 m. Find the area of the field. A L B 21. If the diagonal of a rectangle is 17 cm long and its Base: Each of the two parallel sides of trapezium perimeter is 46 cm, find the area of the rectangle. is called a base of the trapezium 22. The perimeter of two squares are 40 cm and Altitude (or) Height: The distance between the 32 cm. Find the perimeter of a third square whose two parallel sides (bases) is called Altitude or height. area is equal to the difference of the areas of the In the figure, diagonal AC divides the trapezium two squares. ABCD into D ABC and ? ADC 23. The ratio between the length and the breadth of a  Area of trapezium = (Area of  ABC) rectangular park is 3 : 2. If a man cycling along the + (Area of  ADC) ________ (1) boundary of the park at the speed of 12 km/hr 1 completes one round in 8 minutes, find the area of Area of  ABC   AB  height the park (in sq.m). 2 www.betoppers.com

C

W

8th Class Mathematics

148

1 Area of  ACD   DC  height 2  Area of trapezium ABCD

Note:

1  1     AB  height     DC  height  2  2 

ii)

i)

An important property of a regular polygon is that it fits exactly into a circle with the same centre and such a polygon is called a circumscribed polygon. The line joining the centre (polygon or circle) to any vertex is called circum-radius OE = ‘R’. A

1    AB  DC   height 2

4 B

1    sum of parallel sides  2

A simple closed plane figure formed by three or more line segments is called a polygon. Regular Polygon : All sides of polygon are of equal length such polygon is called regular polygon. The line segments are called sides of the figure. Each end point of a side is called a vertex of the polygon. Note: The minimum number of line segments required to form a closed figure is three Examples: P o ly g o n

F ig u re

4

4

C

iv)

12. Polygons

O

4

iii)

E

R

  distance of parallel sides 

Perimeter of trapezium = sum of all the sides = (AB + BC + CD + DA) cm.

4

D

An important property of a regular polygon is that it fits a circle into the polygon with the same centre and such a polygon is called a inscribed polygon. The perpendicular drawn from side AE to the centre ‘O’ of an inscribed circle is called in- radius (r) A 4

4 rR

B

E

O

4

4

4

C

D

13. Area of Polygons 1. Area of a Regular Polygon Case-1: Area of regular polygon in terms of its side and radius of the inscribed circle. D

E T ri a n g le (3 si d e s )

F

C

O R G

Q u a d ri l a te ra l (4 s i d e s )

r R H

P e n t ag o n (5 s i d e s )

H e x ag o n ( 6 s i d es )

H e pt a g o n ( 7 s i d es )

O c ta g o n (8 si d e s )

A

Let ‘G’ and ‘H’ be any two vertices of a regular polygon Let ‘O’ be the centre of the inscribed circle. OG and OH be the bisectors and OP be the perpendicular drawn from O to GH, Then OP = r (in-radius) We will also get the polygon divided into n equal triangles like  OGH.  Area of polygon = Area of  OGH × Number of sides of the polygon

1 1   OP  GH  n   r  a  n 2 2 ( each side = a units) 

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B

P

nar sq.units 2

Mensuration  Area of regular polygon

149

1   perimeter × in-radius 2 ( perimeter of a polygon of n sides, each of length a units = na) 1  na  r.units 2 Case-2:Area of regular polygon in terms of the radius of the circumscribed circle. 

D

E F

 Area of the Hexagon = 6 × area of ?OAB 3 2  side  4 ( DOAB is an equilateral triangle)  6

3 3 3 3 2 2 a sq.units  side   2 2 Case-2: Let ABCDEF be the hexagon with OA = r as radius and length of each side = a. 

E

C

O

D

R G

20

r R

B A

H

R 

 r

2

C a

h P

A

B

Let OP = b be the perpendicular drawn from centre O to AB. From equilateral triangle AOB,

2

1 1 Area  AOB   AB  OP   a  h 2 2

2

n  Area of hexagon  6  a = 3ah sq. units.

a    [ GH = a units)  2

a  r  R   2

r

a

From the figure, we have OP = r and circum radius = OA = R, let GH = a units. From the right-angled triangle POH, we have (OH)2 = (OP)2 + (PH)2 [ by Pythagoras theorem) 2

O

F

P

1 2

2

3. Area of an Octagon

 Area of the polygon  1 1 a   n  a   r    na   R 2    2 2  2 

2

   

Let ABCDEFGH be the given Octagon, O the in- centre, and OP = r and length of each side = a. F

E

2



na a R 2    sq. units 2 2

G

D O

2. Area of a Regular Hexagon Case-1: Let ABCDEF be a regular Hexagon with OP as its in-radius (r) and the length of each side = a. E

r

H

C

a

a A

P

D

B

a

 The area of a regular Octagon

   2 1  2  a sq.units 2

 2 1  2  side  sq.units

O

F

C

2

a

a A

P

B

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8th Class Mathematics

150

Formative Worksheet

Conceptive Worksheet

36. Find the area of quadrilateral PQRS.

26. Find the area of following quadrilateral.

R

1.5 cm C

. A

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h2

8cm

4 cm

h1

M 13cm

B

27. Find the area of a trapezium whose parallel sides are of lengths 10 cm and 12 cm and the distance between them is 4 cm. 28. The area of a trapezium is 440 cm2. The lengths of the parallel sides are respectively 30 cm and 14 cm. Find the distance between them. 29. The area of a trapezium-shaped field is 480 m2, the height is 15 m and one of the parallel sides is 20 m. Find the other side. 30. Find the area of the following figure. 24 cm F 12 cm

E

15cm D

C 10 cm

37. The diagonal of a quadrilateral is 20 m in length and the perpendiculars to it from the opposite vertices are 8.5 m and 11 m. Find the area of the quadrilateral. 38. In quadrilateral ABCD shown in AB || DC and AD || AB. Also, AB = 8m, DC = BC = 5m. Find the area of the quadrilateral. 39. The parallel sides of a trapezium are 18 cm and 15 cm and the distance between them is 5.6 cm. Find the area of the trapezium. 40. The area of a trapezium is 126 sq.cm. The height of the trapezium is 7cm. If one of the bases is longer than the other by 6cm, find the lengths of the bases. 41. The lengths of the parallel sides of a trapezium are in the ratio 5 : 3 and the distance between them is 12.5cm. If the area of the trapezium is 450 cm2, find the lengths of its parallel sides. 42. The area of a field in the shape of a trapezium measures 1440 m2 . The perpendicular distance between its parallel sides is 24 m. If the ratio of the parallel sides is 5 : 3, find the length of the longer parallel side. 43. The two parallel sides of a trapezium are 1.5 m and 2.5 m respectively. If the perpendicular distance between them is 6.5 metres, find the area of the trapezium. 44. Find the altitude of a trapezium, the sum of the lengths of whose bases is 6.5 cm and whose area is 26 cm2. 45. In the adjoining figure AB || DC and DA is perpendicular to AB. Further DC = 7cm, CB = 10 cm and AB = 13 cm. Find the area of the quadrilateral ABCD. 7 cm

C

5 cm

P

D

4 cm

S

Q

m 2c

2.5 cm

5 cm

A

B

21 cm

31. The area of a trapezium is 180 cm2 and its height is 12 cm. If one of the parallel sides is double that of the other, find the two parallel sides.

14. Circles A circle is a plane figure contained by a line traced out by a point which moves, so that its distance from a certain fixed point is always the same.

r O

 The fixed point is called the centre (O) and the bounding line is called the circumference.  The distance between centre and any point on the circumference is called radius denoted by ‘r’. Any circle with the centre ‘O’ and the radius ‘r’ is denoted as c(O, r)

Mensuration

151

Note: Two circles are congruent if and only if, they have equal radii.

Chord:

Semi Circles A semi circle is a figure which is formed by cutting the centre of a circle along its diameter.

A line segment whose end-points lie on the circle is called a chord of the circle. A O

x

r

y

Area of semi-circle  A

B

AB is a line segment whose ends lie on the circle. A chord passing through the centre of a circle is called the diameter. xy is the chord passing through centre ‘O’. Note: i) The diameter of any circle is its longest chord. ii) The diameter of any circle is its longest chord. It is very clear that the diameter of a circle is equal to twice its radius. i.e., d = 2r , where d is the diameter and r is the radius.



1 r 2   2

r

O



B

1  Area of circle  2

r 2 sq. units  r is radius  2

Circumference of semi-circle



1  circumference of circle   diameter 2



1  2r   d  r  2r units ( r is radius) 2

Concentric Circles Observe the following figure.

Area of a Circle It is a circle with centre at O and radius(r) = OA = r cm.

r

A

O

The area of circle =  r2 sq. cm

It is very clear that centre of all the circles is same. What do we call such circles? They are called concentric circles. The region between two concentric circles is called Annulus or ring. B

22  3.14 (r is the radius) Where  is a constant  7

R O

Perimeter of a Circle Let’s take a rope and wind it around the circle. Now, measure length of the rope that covered the circle. This length is perimeter of the circle. Note: The perimeter of a circle is called its circumference. Circumference of circle = 2  r cm ( r = radius of a circle) =  d cm ( d = 2r) (Where d = diameter)

 = constant 

22  3.14 7

r

A

Let R, r be the radii of bigger and smaller circles respectively. The width of the ring = difference of the radii. i.e., width (w) = R – r Area of ring = difference between areas of the two circles.

A

=  R2 –  r2 =  (R2 – r2) =  (R + r) (R – r) sq. units

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152

Formative Worksheet 46. Find the circumference and area of a circle of radius 10.5 cm. 47. The circumference of a circle is 22cm. Find its area. 48. The area of a circle is 55.44 cm 2 . Find its circumference. 49. A wire is bent into the form of a square of side 27.5 cm. The wire is straightened and bent into the form of a circle. What will be the radius of the

 22  circle so formed? Take      7  50. The radius of a wheel is 35 cm and it takes 5 minutes to make 250 rotations. Find the speed of the wheel 22   in km per hour.  Take    7   51. In figure, find the area of the shaded region. (take p = 3.14) A

B 8cm 6cm

D

C

52. The sides of a triangle are a = 13cm, b = 14cm, c = 15cm, the sides a and b are the tangents to a circle, whose centre lies on the third side. Find the circumference of the circle. Note: i) Tangent is a line touching the circle at one point ii) Radius is perpendicular to the tangent. 53. If the radius of a circle is a rational number, its area is given by a number is__. 54. Two concentric circles form a ring. The inner and outer circumference of the ring are

2 3 50 m and 75 m respectively. Find the width of 7 7 the ring. 55. What will be the area of a semi-circle whose perimeter is 36cm? 56. Find the number of revolutions a wheel of diameter 40cm makes in travelling a distance of 176 m. 57. If a wire is bent into the shape of a square, then the area of the square is 81 sq.cm. When the wire is bent into a semi-circular shape, then find the area of the semi-circle. www.betoppers.com

8th Class Mathematics 58. There are 4 semi-circular gardens on each side of a square-shaped pond with each side 21m. Find the cost of fencing the entire plot at the rate of Rs. 12.50 per metre.

Conceptive Worksheet 32. Find the circumference and the area of a circle whose radius is 12.5 cm. Also find the perimeter of the semicircle.    3.14  . 33. Find the diameter of a circle whose circumference is 28.6 cm. 34. The circumference of a circle is 110 cm. Find its area. 35. If a circle of area is ‘a’ sq.units, Then find its circumference. 36. The ratio of the radii of two circles is 1 : 4. Find the ratio of their areas. 37. The area of a circular field is 13.86 hectares. Find the cost of fencing it at the rate of Rs. 4.40 per metre. 38. The inner circumference of a circular race track, 14m wide, is 440 m. Find the radius of the outer circle. 39. The ratio of the outer and the inner perimeters of a circle path is 23 : 22. If the path is 5 metres wide, the diameter of the inner circle is: 40. In the given figure, diameter of the bigger semicircle is 108 cm and diameter of the smallest circle is 36cm. Calculate the area of the shaded portion.

41. Find the area of the largest circle, that can be drawn inside a rectangle with sides 18 cm × 14 cm. 42. The circumference of a circle is 44 cms. Find its area. 43. The radius of a circular field is 77m. A path of width 7m is laid outside around it. Find area of the circular path.

Mensuration

153

Arc

Area of a Sector

Let OA and OB be line segments of length r(radius).

Let ‘r’ be the radius, and ‘  ’ is the angle subtended at the centre ‘O’ by the arc AB.

O  Let ‘q’ be the angle of the related sector at centre ‘O’. Area of segment

 1  r 2  r 2 sin  sq. units 360 2 Perimeter of segment 



A Area of sector 

B  r  r 2 sq. units  360 2

Where ‘  ’ = length of arc and ‘r’ is radius perimeter of sector 

   2r  2rsin units 360 2

  2r  2r units 360

=  + 2r units   2r Length of the arc  360 ormative orksheet Note: The region enclosed by the arc and the chord 59. Find the area of a sector of a circle of radius 5 cm, of the circle is called the segment of a circle. formed by an arc of length 3.5 cm Sector 60. In a circle of radius 7cm, an arc subtends an angle of 108° at the centre. Find the area of the sector. Let’s have a look at the circle whose centre is at 61. Find the perimeter of segment of a circle if the O. angle subtended at the centre is 90° and the radius of the circle is 12 cm. O 62. The length of the arc of a circle having measure 36° is 22cm. Find the circumference and the area r r of the circle. A B 63. Find the area and the length of the arc of the OA = OB = r segment of a circle of radius 21cm if the arc of the What is the region enclosed by OAB called? segment has a measure of 90°. The region enclosed by an arc of the circle (AB) 64. An arc of a circle subtends an angle of 90° at the and two radii to its end-points is called the sector centre. If the radius of the circle is 14cm, find the of a circle. In the above figure, region OAB is length of the arc. Also find the perimeter and the sector. area of the circle. Note: 65. If the radius of a sector is doubled and its arc is reduced to half, then its area is ________. i) The angle subtended by the arc of a sector at the centre is called the sector angle or the central angle 66. A shed of length 20m and br eadth 10m is of the arc. constructed in a grass field. A cow is tied with a rope of length 10m, outside the shed in a corner. ii) The degree increase of an arc is the degree Find the area of field that the cow can graze. measure of the central angle of the arc.

F

iii) iv)

W

The degree measure of an arc is denoted by m(AB). Two arcs of a circle are said to be congruent if, and only if, they have the same degree measure.

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Conceptive Worksheet 44. Find the area of a sector whose arc length is 26.5 cm and radius 63 cm. 45. The angle of a sector is 72° and its radius is 14cm, Then find the length of its arc.. 46. The radius of a sector is 7 cm, and its angle is 240°. Then the area of the sector. 47. The angle of a sector is 75° and its radius is 21 cm. Find its perimeter and area. 48. Find the angle subtended at the centre, if the perimeter of a sector of a circle is 72 cm and its radius 14 cm. 49. Find the area of the sector of a circle of radius 16cm, cut off by an arc of length 18.5 cm. 50. The length of the minute hand of a wall clock is 21 cm long. Find the area swept by it in 10 minutes.

10. The perimeter of a rhombus is 260cm and one of its diagonals is 66cm. Find the area of the rhombus and its other diagonal. 11. The dimensions of a room are 16 × 14 × 10 metres. There are 4 windows of 1.3m × 1.4m and 2 doors of 2m × 1m. What will be the cost of whitewashing the walls and painting the doors and windows, if the rate of whitewashing is Rs. 5 per m2 and the rate of painting is Rs. 8 per m2. 12. The shape of a garden is rectangular in the middle and semicircular at the ends as shown in the diagram. Find the area and perimeter of the garden. 7m

20 m

Summative Worksheet 1.

2.

3.

4.





The area of an equilateral triangle is 16 3 cm 2 . Find the length of each side of the triangle. The altitude drawn to the base of an isosceles triangle is 8 cm and the perimeter is 32 cm. Find the area of the triangle. Find the area of a triangle whose sides are 28 cm, 21 cm and 35 cm. Also find the length of the altitude corresponding to the longest side of the triangle. Find the area and height of an equilateral triangle





of side 8cm. Take : 3  1.73 . 5.

6.

7. 8.

9.

The base of a triangular field is three times its height. If the cost of cultivating the field at Rs. 367.20 per hectare is Rs. 4957.20, find its base and height. In the figure a square and a rectangular field have the same perimeter. Which one has a larger area? Square

Rectangle

60m

80m

The area of a square field is one hectare. Find the cost of fencing it at Rs. 6.50 per metre. The perimeter of a rhombus is 146 cm and one of its diagonals is 55 cm. Find the other diagonal and the area of the rhombus. Find the perimeter of a rhombus, if the length of its diagonals are 18cm and 12cm.

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13. The length and breadth of a rectangular field are in the ratio 3 : 2. If the area of the field is 3456 m2 , find the cost of fencing the field at Rs. 3.50 per metre. 14. The cost of turfing a rectangular field at 85 paise per square metre is Rs. 624.75. Find the perimeter of the field if its sides are in the ratio 5 : 3. 15. The total number of tiles used in a room floor are 50, whose length and breadth are 20cm and 15cm. Find the perimeter of each tile and also the perimeter of the total number of tiles. 16. The length of a wooden plank is twice its breadth. If the perimeter of the wooden plank is 90 cm, find the length and breadth of the plank. 17. The length of a rectangle is halved, while its breadth is tripled. What is the percentage change in area? 18. The ratio between the length and breadth of a rectangular field is 3 : 2. If only the length is increased by 5 metres, the new area of the filed will be 2600 sq. metres. What is the breadth of the rectangular field? 19. A parallelogram has sides 30m and 14m and one of its diagonals is 40 m, Find its area. 20. Find the sum of the lengths of the parallel sides of a trapezium whose altitude is 11 cm and whose area is 0.55 m2. 21. The area of the trapezium is 105 cm2 and its height is 7 cm. If one of the parallel sides is longer than the other by 6 cm, find the two parallel sides. 22. The parallel sides of a trapezium are 20 cm and 10 cm. Its non-parallel sides are both equal, each being 13 cm. Find the area of the trapezium. 23. If the perimeter of a trapezium be 52cm, its nonparallel sides are equal to 10 cm each and its altitude is 8 cm, find the area of the trapezium. 24. The area of a trapezium is 180 cm2 and its height is 9 cm. If one of the parallel sides is longer than the other by 6 cm, find the two parallel sides.

Mensuration

155

25. In the given figure AB || DC and DA is 33. The circumference of a circular path is 264m. Find perpendicular to AB. Further DC = 7cm, its area. CB = 10 cm and AB = 13 cm. Find the area of the 34. The circumference of two circles are in the ratio 2 quadrilateral ABCD. : 3. Find the ratio of their areas. D 7cm C 35. Four equal-sized maximum circular plates are cut off from a square sheet of paper sheet of area 784 cm2. Find the circumference of each plate. 10 cm 36. What is the area of a sector whose radius is 28 cm and the length of the arc l is 132 cm.? M 37. What is the length of the arc of a sector whose angle is 75° and radius is 42 cm? A B 13cm 26. An ant is moving around a few food particles of 38. A chord AB of a circle of radius 15cm makes an angle of 60° at the centre of the circle. Find the different shapes scattered on the floor. For which area of the major and the minor segments. food particles would the ant have to take a longer route? Take   3.14, 3  1.73



2.8 cm C

C

B

A

B

1.5 cm m 2c

2c m

E

2.8 cm

A

2.8 cm

39. Find the angle subtended at the centre of a circle, if the perimeter of a sector is 43cm and the radius of the circle 21cm. 40. Find the area of a sector whose arc length is 26.5 cm and radius 63 cm.

D O

27. Find the area of the shaded region in figure if ABCD is a square of side 14 cm and APD and BPC are 1. semicircles. B  7 cm  O  7 cm  c

2. P

3. A  7 cm  O  7 cm  D 4. 28. The area of a rectangular park is 2160 m2. The length and breadth of the park are in the ratio 5 : 3. Find the cost of fencing the park at Rs. 5 per metre. 29. A wire of 5024 m length is in the form of a square. It is cut and made into a circle. Find the ratio of the area of the square to the circle. 30. Find the ratio of the areas of the in-circle and circumcircle of a square. 31. The shape of a garden is rectangular in the middle 5. and semicircular at the ends as shown in the diagram. Find the area and perimeter of the garden.

7m 20 m



6.

HOTS Worksheet If the angles of a triangle are 30° and 45° and the included side





3  1 cm . Find the area of the

triangle. One side of a right angled triangle is 3925cm. The difference between the hypotenuse and the other side is 625cm. Find the hypotenuse and the other side. Can two isosceles triangles of sides 6.5, 6.5, 5 units and 6.5, 6.5, 12 units have the same area? Find the area of the shaded region, if ADE is also an equilateral triangle. A

8 D

B

8 4

8

E

C

If one leg of an isosceles right-angled triangle is increased by 6cm and that of the other leg decreased by 4cm, then the area of the triangle decreases by 24 sq.cm. Find the length of the leg of the original triangle. The area of a square and a rectangle are equal. If the side of the square is 40 cm and the breadth of the rectangle is 25 cm, find the length of the rectangle. Also, find the perimeter of the rectangle.

32. An athletic track 14 m wide consists of two straight sections 120 m long joining semicircular ends whose inner radius is 35 m. Calculate the area of the shaded region. www.betoppers.com

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ABCD is a square of side ‘x’ cm and E, F, G, H are 14. In measuring the sides of a rectangle, one side is the mid-points of its sides. J, K, L, M are the midtaken 5% in excess, and the other side 4% in deficit. points of the sides of the square EFGH, as shown Find the error percent in the area calculated from in the diagram. these measurements. D 15. The length of a blackboard is 8 cm more than its breadth. If the length is increased by 7 cm and M breadth is decreased by 4 cm, the area remains the H G same. Find the length and breadth of the blackboard J L 16. If a parallelogram with area P, a rectangle with area A C R and a triangle with area T are all constructed on the same base and all have the same altitude, then F E K which of the following statements is false? a) P = R b) P + T = 2R B c) P = 2T d) T = (1/2) R i) Show that the side of the second inner square 17. Find the area of a trapezium ABCD whose parallel x sides are AB = 19 cm, DC = 9 cm and the noncm EFGH  2 parallel sides are BC = 8 cm and DA = 6 cm. 18. The cross-section of a canal is in shape of a ii) Show that the side of the third inner square trapezium. The canal is 12 m wide at the top and 8 m wide at the bottom. If the area of the cross-section x JKLM  cm . is 840 sq.m., find the depth of the canal. 2 19. In the adjoining figure, ABCD is a trapezium is iii) Check whether the side of the nth inner which AD || BC, AD = 10 cm, BC = 18 cm and CD = 17 cm. Find the area of the trapezium. square  x  2 n1 / 2 C iv) If x = 8 cm, find the area of the 7th inner square. cm 8. Find the area of a rhombus having each side equal 17 to 13 cm and one of its diagonals equal to 24cm. 9. The perimeter of a rhombus is 146 cm and one of D E its diagonal is 55 cm. Find the other diagonal and the area of the rhombus. 10. A rectangular field is 50 m by 40.5 m. It has two crossroads through its centre, running parallel to its sides. The width of the longer and the shorter roads A B are 1.8 m and 2.5 m respectively. Find the total expenditure involved in cementing the road at Rs. 20. The parallel sides DC and AB of a trapezium are per 24 m2 laying grass in the remaining part at Rs. 12 cm and 36 cm respectively. Its non-parallel sides 1.60 per m2. are each equal to 15 cm. Find the area of the S P trapezium. 25 m 21. Diagram shown in figure of the adjacent of the picture frame has outer dimensions 24 cm × E H C D 28 cm and inner dimensions 16 cm × 20 cm. Find 1.8 m the area of each section of the frame, if the width B A of each section is same. F G 40.5 m

10 cm

18 cm

7.

A R Q 50 m

11. A rectangular grassy plot 110 m long and 65 m wide has a gravel path 3 m wide all round it on the inside. Find the cost of levelling the path at 80 paise per sq.metre. 12. The dimensions of a rectangle are increased by 20 percent each. Find the percentage increase in (i) the area and (ii) the perimeter. 13. Show that the diagonals of a parallelogram divide it into four triangles of equal area. www.betoppers.com

B A1

B1

20 cm

28 cm

16 cm D1 D

C1 24 cm

C

Mensuration

157

22. Length of the fence of a trapezium shaped field 33. If the area of a sector is ‘A’, and the length of the ABCD is 120 m. If BC = 48 m, CD = 17 m and AD arc is ‘l’, find its radius. = 40 m, find area of the field. Side AB is 34. If the perimeter of a sector is ‘p’ units and its radius perpendicular to the parallel sides AD and BC. is ‘r’ units, then find its area. 23. In the figure, find the area of the shaded region 35. If the radius of a sector is doubled and its arc is reduced to half, find its area. Use   3.14 . 36. A square shed, whose side is 14 m, is in a grass field. A buffalo is tied at one corner of the shed with a rope of length 21 m. What is the area that B A the buffalo can graze? 8 cm 37. A rectangular swimming pool of length 28 m and breadth 21 m is in a field. A goat is tied in a corner outside the pool with a rope of length 24.5m. Find 6 cm the area that the goat can graze. 38. A rectangular poultry farm of length 14 m and breadth 7 m is in a ground. A horse is tied with rope of length 10.5 m in a corner outside the farm. What D C is the area that the horse can graze? 39. Four carrom board strikers of radius 3.5cm are so arranged that each striker touches at least two other 24. If the circumference of a circle is increased by 50%, strikers. Find the area of the empty space between then its area increase by _____ %. the strikers. 25. What is the area of the largest triangle that can be 40. From a circular disc of radius 28 cm, a sector of inscribed in a semicircle whose radius is ‘r’? angle 45° has been cut out. Find the area of the 26. A circular piece of metal of maximum size is cut sector and the remaining part of the disc. out of a square piece and then a square piece of maximum size is cut out of the circular piece. The total amount of metal wasted is: orksheet 27. If the chord of the larger of the two concentric 1. The base of a triangle is 15 cm and height is 12 cm. circles is tangent to the smaller circle, then find the Find the height of another triangle of double the area of the annulus for chord being ‘a’ units. area having the base 20 cm. 28. If the radius of a circle is decreased by 50%, find 2. The area of a right-angled triangle is 40 times its the percentage decrease in its area. base. What is its height? 29. A cow is tethered in the middle of a field with a 14 feet long rope. If the cow grazes 100 sq.ft. per day, 1 1 1 then approximately what time will be taken by the : : . 3. The sides of a triangle are in the ratio of cow to graze the whole field? 2 3 4 30. ABC is a right-angled triangle, right-angled at B. If If the perimeter is 52 cm, find the length of the the semicircle on AB with AB as diameter encloses smallest side. an area of 81 sq.cm and the semicircle on BC with BC as diameter encloses an area of 36 sq.cm, then 4. The perimeter of a triangle is 30 cm and its area is the area of the semicircle on AC with AC as 30 cm2. If the largest side measures 13 cm, what diameter will be: is the length of the smallest side of the triangle? 31. In the given figure, what is the area of the shaded portion if the diameter AB of the larger circle 5. If x is the length of a median of an equilateral triangle, measures 28 cm? then find its area. 6. The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, find its area. 7. The length of a rectangular plot is 60% more than A B its breadth. If the difference between the length and the breadth of the rectangle is 24 cm, what is the area of the rectangle? 32. The area of a circular grass field is 2464 sq.cm. Find the cost for levelling the circular path width 1m. laid outside it, at a cost of Rs. 1.85 per sq.m.

IIT JEE W

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158 8.

A rectangle of certain dimensions is chopped off from one corner of a larger rectangle as shown. AB = 8 cm and BC = 4 cm. Find the perimeter of the figure ABCPQRA (in cm). P

A

Q

4 cm

R

C

8 cm

B

9.

The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle? 10. The diagonal of the floor of a rectangular closet is

1 1 feet. The shorter side of the closet is 4 feet. 2 2 What is the area of the closet in square feet? The length of a rectangular hall is 5 m more than its breadth. The area of the hall is 750 m2. Find the length of the hall. The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ Rs. 26.50 per metre is Rs.5300, what is the length of the plot in metres? Find the percentage increase in the area of a rectangle if each of its sides is increased by 20%. What will be the cost of gardening 1 metre broad boundary around a rectangular plot having perimeter of 340 metres at the rate of Rs. 10 per square metre? 7

11.

12.

13. 14.

15. A room is 15 feet long and 12 feet broad. A mat has

1 2 feet space from the walls. What will be the cost of the mat at the rate of Rs. 3.50 per square feet ? 16. A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is 100 m, find the altitude of the triangle. 17. The difference between two parallel sides of a trapezium is 4 cm. The perpendicular distance between them is 19cm. If the area of the trapezium is 475 cm2, find the length of the parallel sides. 18. Find the length of a rope by which a cow must be tethered in order that it may be able to graze an area of 9856 sq.metres. to be placed on the floor of this room leaving 1

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19. A man runs round a circular field of radius 50 m at the speed of 12 km/hr. What is the time taken by the man to take twenty rounds of the field? 20. A circle and a rectangle have the same perimeter. The sides of the rectangle are 18 cm and 26 cm. What is the area of the circle? 21. Find the circumference of a circle, whose area is 24.64m2. 22. If the circumference and the area of a circle are numerically equal, then the diameter is equal to: 23. A circular ground whose diameter is 35 metres, has a 1.4m broad garden around it. What is the area of the garden in square metres? 24. If the ratio of areas of two circles is 4 : 9, then the ratio of their circumferences will be: 25. Three circles of radius 3.5cm are placed in such a way that each circle touches the other two. Find the area of the portion enclosed by the circles.



3  1.732



26. If the radius of a circle is doubled, its area is increased by ___________ %. 27. If the radius of a circle is increased by 6%, then the area is increased by ___________ %. 28. If the radius of a circle is increased by 75%, then its circumference will increase by ________%.



By the end of this chapter, you will understand       

Basic Concepts of Geometry Properties of straight lines Kinds of Lines Angle Types of Angles Pair of angles Basic Theorems on Angles

      

Basic Theorems on Parallel lines Triangle and its angles Congruent Triangles Theorems on Congruent triangles Particular Type of Triangles Concurrent Lines in a Triangle Circles

1. Introduction Geometry means ‘Earth measurement’. Geometry provides knowledge, develops logical thinking, improves the power of reasoning. Nature consists of a number of animate and inanimate objects. These are usually trees, flowers, fruits, mountains, rivers, seas, sun, moon, stars. These are all basic objects of Nature, which we see, and they do not require any definitions. With the help of undefined objects, terms are created. For example, with the help of undefined object like sun, we define the term East. Such terms which are defined with the help of basic terms are said to be Defined Terms. With the help of undefined and defined terms, certain factual statement have been formed, which are true and do not require any proof or explanation. Such statements are called axioms. For example, The sun rises in the East.

Chapter - 11 10

Plane Geometry

Learning Outcomes

2. Basic Concepts Point A point is represented by a dot. It has no dimensions like length, breadth or thickness. It has only position. Points are denoted by capital letters A, B, C, D, P, Q, R etc. A point ‘A’ is represented as A.F o r example, a dot made by a sharp pencil is a point.

Line A geometrical line is a set of points that extends endlessly in both the directions. i.e., a line has no end points. It has only length. The arrowheads show that the line goes on endlessly on either side.

A line is also called a straight line. The line AB is  represented as AB . Another kind of line is a curved line.

History The Egyptians and the Babylonians used geometry for practical purposes and did very little to develop it as a systematic science. Euclid, the father of geometry, introduced the method of providing a geometrical result by deductive reasoning based on previously proved results and some self–evident specific assumptions called “axioms”.

For example, the edge of a ruler extending endlessly in both the directions. Lines are denoted by small letters l, m, n, …. etc.

Line–segment A line-segment is a line which has end points. In the above figure, the part of the line between the points A and B, including A and B, is a line-segment.

 The line – segment AB is represented as AB or segment AB. Geometry is the basis and inspiration for Indian mathematics.

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Plane

Midpoint

A plane is a set of points. It is a flat surface with length and breadth.

Given a line–segment AB, a point M is said to be the midpoint of AB, if M is an interior point of AB, such that AM = MB.

Perpendicular bisector of a Line– Segment A line ‘l’ passing through the midpoint ‘M’ of a line–segment AB and perpendicular to AB is called A geometrical plane extends endlessly in all the directions. Small letters are used to denote a plane. For example, the surface of a sheet of paper, surface of a wall and Surface of a matchbox.

the perpendicular bisector of the line–segment AB.

Coplanar Points The points that belong to the same plane are called coplanar points.

Collinear Points If three or more points lie on a straight line, then those points are called collinear points.

In the above figure, the points A, B, C, D, E and F are coplanar points.

A, B, C, D and E are collinear points.

Coplanar Lines

Non-collinear Points:

The lines that lie in the same plane are called coplanar lines.

In the above figure, the lines AB, CD, EF, GH and IJ are coplanar lines.

Space The universal set of points, lines and planes is called a space. It has no end.

The points which do not lie on the straight line are called non-collinear points.

In the above figure, the points ‘C’, ‘G’, ‘H’ do not lie on the straight ‘l’. Hence they are non-collinear points. A point ‘C’ is said to lie between ‘A’ and ‘B’, if a) A, B, C are collinear and b) AB = AC + CB.

Points and Lines Incidence Properties The relation between a point and a line is called an incidence property. Let us consider a line ‘l’ and a point ‘p’. If p  l, then we say that ‘p’ is incident on ‘l’. The line ‘l’ passes through the point ‘p’.

Distance between two Points The distance between two points A and B is the length of the line–segment joining them. The distance between A and B is denoted by d(A, B) or AB.

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Ray Let ‘’ be a line and A, B be two distinct points on ‘’. A ray is a part of the line ‘’ which has one end point at ‘A’ and contains the point ‘B”. A ray is  denoted by the symbol AB

Plane Geometry

Opposite Rays

161

4. Kinds of Lines Parallel Lines

  Consider two rays AB and AC . The two rays   AB and AC are said to be opposite rays, if they   are collinear and , AB  AC  A i.e., the point A

is the only common point of the two rays.

3. Properties of Straight Lines 1. 2.

On a single line l, there exists infinite number of points. An infinite number of lines pass through a single point P. (concurrent lines)

The lines which do not intersect each other when produced? and do not have any point in common are called parallel lines.

The lines ‘l’ and ‘m’ are parallel lines. Parallel lines are denoted as ‘||’.

Concurrent Lines When three or more lines pass through a single point, they are called concurrent lines.

In the above figure the lines k, l, m and n are concurrent lines. Two rays or two line–segments or a ray and a line–segment are said to be parallel if the lines containing them are parallel. 3.

4.

Only a single line passes through two distinct points A and B.

Given two different lines ‘l’ and ‘m’, there exists only one point ‘p’, which lies on both the lines ‘l’ and ‘m’. i.e., the two lines intersect at P. Note: Two distinct lines cannot have more than one common point

Transversal A line which intersects two or more lines at distinct points is called a transversal of the given lines.

In the figure above, AB is the transversal to the lines CD and EF.

Degree If the line–segment AO rotates about ‘O’ through a complete rotation, it comes back to its original position. This complete rotation is divided into 360 equal parts. Each part is called a degree and is denoted as ‘°’.

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Measure In the given figure, the line-segments AO and BO meet at O; O is the vertex of the AOB and AO and BO are called the measure of AOB. It is denoted as m AOB. For example, in the given figure, if the measure of the AOB is x°, then we write m AOB = x°.

5. Angle An angle is the union of two rays with a common initial point. When a ray AB rotates in the plane about the point A and takes a new position AC, an angle BAC is formed. The Symbol of angle is 

Minutes and Seconds Each degree is divided in to 60 equal parts called minutes. Symbolically 1° = 60|. Each minute is divided into 60 equal parts called seconds. Symbolically 1| = 60||.  1° = 60| = 60 × 60n = 3600n. Thus one degree is equal to 3600 seconds.

Formative Worksheet 1.

2.

3.

Write the complement of each of the following : i) 54° ii) 90° iii) 47°22| iv) 65°10|20|| v) 87°59|| Through what angle does the minute hand of a clock turn in i) 5 minutes ii) 21 minutes a) Two supplementary angles are in the ratio 4 : 5. Find the angles. b) If the angles (2a – 10)° and (a – 11)° are complementary angles, then find ‘a’. c) Two supplementary angles differ by 32°. Find the angles.

Conceptive Worksheet 1.

2. 3.

Write the supplement of each of the following : i) 50° ii) 90° iii) 132°45| iv) 56°20|48|| v) 112° 48|| How many degrees are there in an angle which equals one-third of its supplement? The earth makes a complete rotation on its axis in 24 hours. Through what angle will it turn in 3 hours 20 minutes and how long will it take to turn through 130°?

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  The angle formed by the two rays AB and AC is  denoted by BCA or CAB. The two rays AB  and AC are called the arms, and the common initial point A is called the vertex.

Bisector of an angle A line which divides an angle into two equal parts is called the Bisector of the angle. Example In the figure below, the line OP divides AOB into two equal parts i.e., ?AOP = ?POB = x° The line OP is called the bisector of AOB.

6. Types of Angles There are six types of angles. i) Acute Angle ii) iii) Obtuse Angle iv) v) Reflex Angle vi)

They are: Right Angle Straight Angle Complete Angle

Acute Angle: An angle which is less than 90° is called an acute angle.

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163

Right Angle: An angle which is equal to 90° is 7. Pair of Angles called a right angle. Complementary Angles: When the sum of the measures of two angles is equal to 90°, then the angles are said to be complementary angles.

Obtuse Angle: An angle which is greater than 90° but less than 180° is called an obtuse angle. Here x + y = 90°, therefore x and y are complementary angles.

Supplementary Angles: When the sum of the measures of two angles is equal to 180°, then the angles are said to be supplementary angles.

Straight Angle: An angle which is equal to 180° is called a straight angle. Here x + y = 180°, therefore x and y are supplementary angles.

Reflex Angle: An angle which is greater than 180° but less than 360° is called a reflex angle.

Note: An angle is said to be a zero angle when two rays coincide and the measure is 0°.

Adjacent Angles: Angles having the same vertex and a common side, and which lie on the opposite sides of the common side are called adjacent angles.

Complete Angle: An angle which is exactly

Angles AOB and COB with common vertex O and common side OB are adjacent angles.

equal to 360° is called a complete angle.

Linear Pair of Angles: Two adjacent angles are said to form a linear pair of angles if they lie on the same straight line. The two non–common arms are opposite rays.

In the above figure, AOC and BOC are adjacent angles. OA and OB are the rays which are non–common arms and are collinear. www.betoppers.com

8th Class Mathematics

164 The sum of the two adjacent angles of a linear pair of angles is 180°; hence they are also supplementary. Two adjacent angles are a linear pair of angles if, and only if, they are supplementary. Note: If p + q = 180°, then AB is a straight line.

Note: Two angles are said to be a pair of vertically opposite angles if their arms form two pairs of vertically opposite rays . Angles forming a pair of vertically opposite angles are congruent (or equal).

Position of points with respect to Angle Interior of an Angle: The interior of an angle is the set of all points P in its plane, which lie on the same side of line AB as C and also on the same side of line AC as B. Linear pair of angles

Adjacent angles

are always   Adjacent angles

are always   Linear pair of angles

Perpendicular Lines: Any two straight lines

Angle addition Axiom: If P is an interior point

lying in the same plane, that meet each other at right angles are called perpendicular lines. Perpendicular lines are denoted by the symbol ^.

of any angle Ð ABC, then m ÐABC = m ÐABP + m ÐPBC

Let AB be a straight line and CD be a line on AB such that BDC = CDA = 90°; then we say that CD  AB.

Exterior of an Angle

Congruent Angles and An axiom Two congruent angles have the same measure and, conversely, two angles of equal measure are congruent. Consider two angles, ABC and DEF. If the two angles are congruent, then we write ABC  DEF

The exterior of an angle is the set of all points Q in its plane, which do not lie on the angle or in its interior.

Angles around a Point When a number of lines meet at a single point, they form angles around that point. The sum of the measures of all these angles formed around the point is equal to 360° i.e., four right angles are formed.

Vertically Opposite Angles: If two lines AB and CD intersect at a point O, then the pair of angles ÐAOC and ÐBOD is said to be a pair of vertically opposite angles. Vertically opposite angles (also vertical angles) are those which have a common vertex and the sides of one are the extended sides of the other. OA and OB are opposite rays. OC and OD are also opposite rays. Also AOD and BOC form another pair of vertically opposite angles.

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i.e., measures of (a + b + c + d + e + f + g + h) = 360° ma + mb + mc + md + me + mf + mg + mh = 360° The sum of the angles around a point is 360°.

Plane Geometry

Angles in a Transversal: When a transversal AB intersects two lines CD and EF at two distinct points G and H, then the three lines determine eight angles, four angles at point G, and four angles at point H.

165 In the above figure ‘l’ is the transversal intersecting the parallel lines ‘m’ and ‘n’. Then the following pairs of corresponding angles are equal. i) a = e ii) b = f iii) c = g and iv) d = h

Alternate Angles: Two angles are said to be a

The angles at the point G are 1, 2, 3 and 4 and the angles at the point H are 5, 6, 7 and 8.

pair of alternate angles if the angles are interior angles, which lie on either side of the transversal and are not adjacent angles. In the above figure, the pairs of alternate angles formed are (i) d and e (ii) c and f.

Results of a Transversal Intersecting Parallel Lines

Interior Angles: Angles inside the parallel lines and on both sides of the transversal are called interior angles. The angles Ð3, Ð4, Ð5 and Ð6 are called interior angles.

Co-interior Angles: Two angles are said to be co-interior angles or interior opposite angles if they are interior angles and lie on the same side of the transversal. In the above figure, the following pairs of angles are co-interior angles: i) 3 and  6

ii) 4 and 5

Exterior Angles: Angles lined outside the parallel lines on both sides of the transversal are called exterior angles. Corresponding Angles Two angles are said to be a pair of corresponding angles if they are on the same side of the transversal with one angle interior and the other angle exterior and the angles are not adjacent angles.

Axiom of corresponding Angles: If a transversal intersects two coplanar parallel lines, then the corresponding angles are equal. Conversely, if the a transversal intersects two coplanar lines, making a pair of corresponding angles equal, then the lines are parallel.

i)

When a transversal intersects two parallel lines, eight angles are formed. ii) The sum of the four interior angles is 360°. iii) The sum of the co-interior angles is 180°. iv) The sum of the four exterior angles is 360°. v) The alternate angles are equal. vi) The corresponding angles are equal. vii) If two lines are cut by a transversal such that two corresponding angles are equal, then the two lines are parallel. viii) If two lines are cut by a transversal such that two alternate angles are equal, then the two lines are parallel. ix) Alternate exterior angles are equal. In the figure 1 and 7, 2 and 8 are alternate exterior angles. x) The sum of the exterior angles that are on the same side of the transversal is 180°.

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8th Class Mathematics

166

8. Basic Theorems on Angles Theorem Theorem The adjacent angles formed when one straight line stands over another, are together equal to two right angles i.e., 180 0.

Proof: Consider a straight line AB.

Let the straight line OC make the adjacent angles AOC and COB with this line. It is required to prove that AOC and COB are together equal to 180°. Suppose OD is perpendicular to AB. i.e., OD is at right angles to AB. Then, AOC + COB = AOC + COD + DOB –––––––– (1) Also, AOD + DOB = AOC + COD + DOB –––––––– (2) From (1) and (2) AOC + COB = AOD + DOB ––––– (3) But AOD + DOB = 90° + 90° = 180° Hence (3) becomes AOC + COB = 180° Hence the adjacent angles f ormed when one straight line stands over another are together equal to 180°. Theorem If two adjacent angles are supplementary, then their outer arms are in a straight line.

Proof:

Let two adjacent angles ACD and DCB be supplementary i.e., ACD + DCB = 180°. It is required to prove that A, C, B are collinear. Assume that A, C, B are not collinear. Produce AC to E such that A, C, E are collinear. www.betoppers.com

 ACD + DCE = 180°__________ (2) ( A, C, E are collinear and CD stands on AE) From (1) and (2) ACD + DCB = ACD + DCE  DCB = DCE But, DCB DCE ( DCE is a part of DCE) A contradiction to our assumption that A, C, B are not collinear. Hence the points A, C, B are collinear. i.e., AC, CB are in a straight line. Hence the theorem. Theorem If two straight lines intersect each other, the four angles so formed are together equal to four right angles i.e., 3600 .

Proof: Let AB and CD be the two straight lines which intersect each another at ‘O’. Let the line OD make the adjacent angles BOC and COA with the straight line AB.

It is required to prove that AOD +  DOB + BOC + COA = 360° We know that “the adjacent angles formed when one straight line stands over another straight line on one side of it, are together equal to 180°”.  AOD + DOB = 180° _______ (1) and BOC + COA = 180° _______ (2) Adding (1) and (2), we get AOD +DOB+BOC+COA = 180° + 180° = 360° Therefore when two straight lines intersect each other, the four angles so formed are together equal to four right angles i.e., 360°.

Plane Geometry

167 OA is a straight line, revolving about O, and turning in successively through the angles AOB, BOC, COD, DOE and EOA and returning to its original position,, will have made one complete revolutions and turned through 360°, equal to four right angles. Therefore when a number of straight lines meet at a point, the sum of the angles so formed at that point is equal to four right angles i.e., 360°.

Theorem If two straight lines intersect each other, then the vertically opposite angles are equal.

Proof:

Properties of Angles

Let the straight lines AB and CD intersect each other at ‘O’. It is required to prove that i) AOC = BOD and ii) AOD = BOC OA is a straight line on CD.  AOC + AOD = 180° ––––––– (1) ( linear pair of angles are supplementary) and OD meets the straight line AB.  AOD + DOB = 180° ––––––– (2) From (1) and (2), AOC + AOD = AOD + DOB  AOC = BOD Again, OC meets AB,  AOC + BOC = 180° ––––––– (3) and OA meets CD,  AOC + AOD = 180° ––––––– (4) From (3) and (4) AOC + BOC = AOC + AOD  AOD = BOC Hence when two straight lines intersect each other, the vertically opposite angles are equal. Theorem When a number of straight lines meet at a point, the sum of all angles so formed at that point is equal to four right angles i.e., 3600 .

The adjacent angles formed when one straight line stands over another, are together equal to two right angles i.e., 180°. If two adjacent angles are supplementary, then their outer arms are in a straight line. If two straight lines intersect one another, the four angles so formed are together equal to four right angles i.e., 360°. If two straight lines intersect one another, the vertically opposite angles are equal. When a number of straight lines meet at a point, the sum of all angles so formed at that point is equal to four right angles i.e., 360°.

Formative Worksheet 4.

In the figure below, POR and QOR form a linear pair. Determine the value of t.

5.

If OP is a ray standing on a line QR such that POQ = POR, then show that POQ = 90°. In the figure below, OX, OY are opposite rays. i) If a = 75°, what is the value of b? ii) If b = 110°, what is the value of a?

6.

Proof:

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8th Class Mathematics

168 7.

In the figure, find t. Further find QOS, ROS and POR.

7.

In the figure, find  ROS, when  POR +  QOS = 85°. R

8.

AB and CD are two coplanar lines and XY is the transversal intersecting the lines at P and Q. If APX = 125° and DQP = 55°, then are the lines AB and CD parallel? If so, why?

S

O

P

8.

Q

In the figure,  XOZ and  YOZ form a linear pair. If p – q = 80°, find the values of ‘p’ and ‘q’. Z

p q O

X

9. 9.

In the figure below, PQ and RS are two coplanar lines, and XY is the transversal intersecting the two lines. Find the value of a°. Is PQ | | RS? If so, why?

Y

In the figure below, two lines PQ and AB are parallel. XY is a transversal cutting these two lines.  1 = 70°. Show that the pairs of alternate interior angles are equal. X L P

2 3

M 6

5

A

Conceptive Worksheet 4.

In the figure, if POQ is a straight line, then a = ______

a +30° P

5.

a +10° a +20° O

Q

In the figure below, ON bisects mm POQ and OM bisects QOR and ON  OM. Show that P, O, R are collinear.

7

1 4

Q

B

8

Y

9. Basic Theorems on Parallel Lines Theorem If a transversal intersects two coplanar lines in such a way that a pair of interior angles on the same side of the transversal are supplementary, then the two lines are parallel.

Proof:

Q

N

M P

6.

O

R

In the figure, determine the value of ‘t’. U 5t° P

S O

2t° Q

5t°

R V

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Let ‘l ’ and ‘m’ be the two coplanar lines and ‘n’ be the transversal. Then 1 + 3 = 180° ––––––– (1) ( linear pair of angles)

Plane Geometry 1 + 2 = 180° –––––––(2) (given) Equating (1) and (2), we get 1 + 3 = 1 + 2 Cancelling ?1 on both sides, we get 3 = 2, which are corresponding angles. According to the corresponding angles axiom, if any pair of corresponding angles are equal, then the two lines are parallel. Hence ‘l ’ is parallel to ‘m’.

169

Theorem If a transversal intersects a pair of coplanar lines in such a way that a pair of alternate angles are equal, then the two lines are parallel. Proof:

Theorem If a transversal intersects a pair of parallel lines, then the interior angles on the same side of the transversal are supplementary. Proof:

Let ‘l ’ and ‘m’ be the parallel lines and ‘n’ the transversal which intersects ‘l ’ and ‘m’ in such a way that 2 and 6 are one pair of interior angles on the same side of the transversal and the other such pair formed is 1 and 5. It is required to prove that 2 + 6 = 180° and 1 + 5 = 180°. We have 1 + 3 = 180° ––––––– (1) ( linear pair of angles) But 3 = 6 ( corresponding angles) Substituting the value of 3 in equation (1), we get 2 + 6 = 180° ––––––– (2) Again 1 + 4 = 180° ––––––– (3) ( linear pair of angles) But 4 = 5 ( corresponding angles) Substituting the value of 4 in equation (3), we get, 1 + 5 = 180° ––––––– (4) From (2) and (4), we get 2 + 6 = 1 + 5 = 180° Hence proved .

Let the two coplanar lines be ‘l ’ and ‘m’ and ‘n’ be the transversal intersecting ‘l ’ and ‘m’ in such a way that a pair of alternate angles are equal i.e., 1 = 2. It is required to prove that l || m. We have 1 = 3 ––––––– (1) ( vertically opposite angles) But given that 1= 2 ––––––– (2) From (1) and (2), we get 2 = 3, which are corresponding angles. According to the corresponding angles axiom, if the two lines are parallel, Hence ‘l’ is parallel to ‘m’. Theorem If a transversal intersects a pair of parallel lines, then each pair of alternate interior angles are equal.

Proof:

Given AB and CD are parallel lines and EF is the transversal intersecting AB and CD at ‘X’ and ‘Y’. 1, 3 are one pair of alternate interior angles and 2, 4 are another pair of alternate interior angles. It is required to prove that 1 = 3 and 2 = 4. We have 1 = ?5 ( corresponding angles) ––––––– (1) and 5 = 3 ( vertically opposite angles) ––––––– (2) www.betoppers.com

8th Class Mathematics

170 From (1) and (2) 1 = 5 = 3  1 = 3 Since XY is a ray on line AB  1 + 2 = 180° ( linear pair of angles) Also YX is a ray on line CD  3 + 4 = 180° ( linear pair of angles)  1 + 2 = 3 + 4 But 1 = 3  2 = 4 Hence if a transversal intersects two parallel lines, then each pair of alternate interior angles are equal. Theorem Theorem Straight lines which are parallel to the same straight line are parallel to one another.

Proof:

Let the straight lines AB, CD be parallel to the straight line EF. It is required to prove that AB and CD are parallel to each other. Draw a line, GH, cutting AB, CD and EF at P,Q and R. Since AB || EF, and GH meets them.  APR = PRF ––––––– (1) ( alternate angles) Also CD || EF and GH meets them,  CQR = PRF ––––––– (2) ( alternate angles)  From (1) and (2) APR = CQR i.e., the corresponding angles are equal. Hence AB and CD are parallel.  The lines which are parallel to the same straight line are parallel to one another.

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Theorem Theorem It there are three or more parallel straight lines, and the intercepts made by them on any transversal are equal, then the corresponding intercepts on any other transversal are also equal. This is called Intercept Theorem.

Proof:

Let the parallel lines AB, CD and EF cut equal intercepts PQ and QR from the transversal PQR. Let XY, YZ be the corresponding intercepts cut from another transversal XYZ. It is required to prove that XY = YZ. Through Y draw a line MN parallel to PR and touching the lines AB and EF. Since PM || QY and PQ || MY, therefore PMY Q is a parallelogram ––––––– (1) Similarly QYNR is also a parallelogram –––– (2) From (1) and (2) PQ = MY (opposite sides of a parallelogram) and QR = YN (opposite sides of a parallelogram) But PQ = QR  MY = YN In ’s XYM and NYZ, YM = NY XYM = NYZ (vertically opposite angles) XMY = YNZ (alternate interior angles)   XYM   NYZ (by ASA congruent rule). Hence XY = YZ Note: Converse of Intercept Theorem does not exist. For example, consider the quadrilateral ABCD, where AB is not parallel to CD. Let P, Q be the midpoints of AD and BC.

Now AP = PD and BQ = QC. But AB, PQ and DC are not parallel. But in case AB || DC, and if P and Q are midpoints of AD and BC then AB, PQ and DC are parallels.

Plane Geometry

171

Properties of Parallel Lines 1.

2.

3. 4.

If a transversal intersects two coplanar lines in such a way that a) A pair of alternate angles are equal, then the two lines are parallel. b) A pair of interior angles on the same side of the transversal are supplementary, then the two lines are parallel.. If a transversal intersects a pair of parallel lines, then a) The interior angles on the same side of the transversal are supplementary. b) Each pair of alternate interior angles are equal. Straight lines which are parallel to the same straight lines are parallel to one another. If there are three or more parallel straight lines, and the intercepts made by them on any transversal are equal, then the corresponding i n t e r c e p t s made by them on any other transversal are also equal.

Conceptive Worksheet 10. Straight lines which are perpendicular to the same straight line are parallel to one another.     11. In the figure below, OP || XY and OQ || XZ , then show that A  B .11. In the figure below,     OP || XY and OQ || XZ , then show that A  B . P Y

Formative Worksheet 10. A || B and C || D. If 1 = 75°, prove that 3 = 1

A O

Q B

1 + of a right angle. 3 11. If two parallel lines are intersected by a transversal, show that the bisectors of any two alternate angles are parallel. 12. AC and DF are parallel. AB and DE are parallel lines. x + y = ___________ .

C

T

X

Z

12. In figure, p || q and q || r. If  1 = 120°, then find 2 . p 2 q 1

r

13. In figure, r  q and r  p. If  1 = 65°, then find 2= 13. In the following figure PQ is parallel to RS and they cut XY and ZY at E, F and G, H respectively. Given that ?XEQ = 80°, ?YHS = 120° and ?XYZ = a, find the value of ‘a’, No proof is required. Only the essential steps of working must be given.

p

q

3

4 r

1 2 t

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8th Class Mathematics

172

10. Triangle and its Angles A triangle is a closed figure bounded by three straight lines. A triangle is denoted by the symbol . In the figure, XYZ is a triangle with XY, YZ and ZX as sides and X, Y, Z are called the vertices of the triangle XYZ.

3.

Obtuse–angled Triangle: A triangle which has one of its angles greater than 900 and less than 1800 is called an obtuse– angled triangle.

1, 2 and 3 are the interior angles The sum of the three interior angles of a triangle is 180°. 1 and 2 are the interior opposite angles and 3 is the interior adjacent angle with respect to 4. 90° A < 180°, ?B and ?C are necessarily acute.

Types of Triangles with respect to their Angles 1.

Types of Triangles with respect to their sides

Right-angled Triangle:

A triangle which has one of its angles as a right angle i.e. 90 0, is called a right–angled triangle or simply a right triangle.

1.

Equilateral Triangle: A triangle which has all its sides equal is called an equilateral triangle.

The side opposite to the right angle is called the hypotenuse and it is the longest side. The sum of the other two angles is equal to 90°. When the other two angles are equal, then the other two sides are also equal. P = 90°, Q = and R are necessarily acute.

When all the three sides are equal, then all the three angles are also equal. Since all the three angles are equal, this triangle is also called an equiangular triangle. Note: Here each angle is 60°. Note: The sides forming the right angle are called the legs of the right angle.

2.

Acute-angled Triangle: A triangle which has all its angles less than 900 is called an acute –angled triangle.

When all the three angles are equal, then all the three sides are also equal.

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2.

Isosceles Triangle: A triangle in which two sides are equal is called an isosceles triangle.

Plane Geometry AC = BC  AB. The angles opposite to these two sides are also equal.  A = B Note: Opposite angles of equal sides are equal.

3.

Scalene Triangle: T he triangle in which all the three sides are of different lengths is called a scalene triangle. i.e., AB  BC  CA.

173 If the side AC is extended to E to form a ray AE, then BCE is an exterior angle of ABC. From above, we can conclude that at each vertex of a triangle there exist two exterior angles. In the above figure in ABC, the exterior angles ACD and BCE are equal as they are vertically opposite angles i.e., ACD = BCE. The two interior opposite angles will be the same for the two exterior angles formed at a particular vertex. Interior Angle Theorem The sum of the three interior angles of a triangle is 180 0 .

Proof:

When all the three sides are different , the three angles are also different. In any triangle the maximum number of acute angles that can be drawn are three, and minimum are two.

Acute Angles are necessary in a Triangle. Only one right angle (or) one obtuse angle is possible in any triangle.

Exterior Angle: If the side BC of a  ABC is extended to D to form a ray BD, then ACD is called an exterior angle of ABC angled at C and is denoted by the exterior angle, ACD.

Interior Angle: The angles A and B are called the remote interior angles or the interior opposite angles of the exterior angle ACD. The sum of the interior opposite angles is equal to the exterior angle formed. A + B = ACD  ACD > A and ACD > B. The exterior angle is greater than each of the interior opposite angles.

Given ABC is a triangle. The interior angles of this triangle are p, q and r. It is required to prove that the sum of the three interior angles is 180°. i.e. p + q + r = 180° Through A, draw a line XY parallel to BC. s and t are the exterior angles on the line XY at point A. Since XY is parallel to BC.  q = s and r = t ( alternate interior angles) Adding p on both the sides, we get p + q + r = s + t + p But p + s + t = 180° (by linear pair axiom)  p + q + r = p + s + t = 180°  p + q + r = 180° Hence the sum of the three interior angles of a triangle is 180°.

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8th Class Mathematics

174 Exterior Angle Theorem If one side of a triangle is extended, then the exterior angle so formed is equal to the sum of the two interior opposite angles.

Proof:

Consider a ABC, with angles p, q and r. Extend the side BC to D. Let the exterior angle at ‘C’ be ‘s’. It is required to proved that s = p + q From Interior angle theorem, we have p + q + r = 180° ______ (1) Also r + s = 180° ______ (2) ( linear pair) From (1) and (2), we have p + q + r = r + s  p + q = s or s = p + q Hence the exterior angle of a triangle is equal to the sum of the two interior opposite angles.

4.

In a triangle, the measure of an exterior angle is equal to the sum of the measures of the interior opposite angles.

Formative Worksheet 14. Show that sum of the three interior angles of a triangle is 180°. 15. If one side of a triangle is extended, then the exterior angle so formed is equal to the sum of the two interior opposite angles. 16. In the figure below, ‘l’ and ‘m’ are two plane mirrors which are parallel to each other. Show that the





incident ray CA is parallel to the reflected ray BD .

17. In the figure below, the side HG is produced both sides. Prove that x° + y° > 180°.

Importance of Interior and Exterior Angle Theorems 1.

The interior angle theorem involves the three angles inside a triangle. It states that together they make 180°. 2. Making use of alternate angles and linear pair of onceptive orksheet angles made by a transversal on straight lines, we prove that the sum of the three angles in a triangle 14. In the adjacent figure, straight lines AB and CD is 180°. intersect at O. If the magnitude of  is four times 3. The exterior angle theorem involves an external that  then  = angle formed by the extension of one side of the C triangle in addition to the two interior angles opposite B to it.  4. We make use of the interior angle theorem and the  exterior angle is equal to the sum of the two interior O opposite angles. Exterior angles = Sum of the two interior opposite A D angles. 15. If CE is parallel to DB in the given figure, then the value of x will be ________ Properties of a Triangle

C

1. 2. 3.

The sum of the three interior angles of a triangle is 180°. In a triangle, the sides opposite to the angles of equal measure are of equal length. The angles opposite to the sides of equal length are of equal measure, in a triangle.

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W

A

B 110°

75° D

E

30°

60°

x C

Plane Geometry

175

16. What are the values of p and q in the figure below:

Here, A  D, B  E and C  F Two triangles ABC and DEF are said to be congruent if, and only if, there is a correspondence between their vertices such that the corresponding sides and the corresponding angles are equal. Congruent triangles are similar but similar triangles are not congruent always.

A p°

110° q° D

C

30° B

17. In a triangle with angles 90°, 45° and 45°, what is the ratio of the legs of the right angle? 18. The sum of two angles of a triangle is 80°, and their difference is 20°. Find all the angles of the triangle.

11. Congruent Triangles

1.

2.

are Congruent triangles   Similar triangles always

not Similar triangles   Congruent triangles always

Properties of Congruent Relation

Two geometrical figures are said to be congruent if they have exactly the same shape and size. 1. Congruency is denoted by the symbol . Two line–segments are said to be congruent, if and 2. only if their lengths are equal. Eg: 3.

From the above example ABC  DEF Congruence Relation is Reflexive: ABC  ABC Congruence Relation is Symmetric: If ABC  DEF, then  DEF  ABC Congruence Relation is Transitive: Suppose there is a third triangle XYZ which is congruent to DEF.

AB  CD Two angles are said to be congruent if, and only if, their measures are equal. Eg: If ABC  DEF and DEF  XYZ then ABC  XYZ.

Axioms

3.

1. ABC DEF  mABC = mDEF Two triangles are said to be equal in all respects, when one is placed on the other to exactly coincide with it, in which each part of the first triangle is equal to the corresponding part of the other. Also the triangles are equal in area. The method of testing equality is known as the method of superimposition. The triangles which are made to coincide by superimposition are said to be congruent or identically equal.

SAS (side – angle – side) Congruence Axiom: The axiom states that two triangles are congruent if any two sides and the included angle of one triangle are equal to the corresponding two sides and the included angle of the other triangle.

If AB = XY B = Y and BC = YZ then ABC  XYZ. Note: The included angle must be the angle formed by the two equal sides, otherwise the triangles may not be congruent.

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8th Class Mathematics

176

2.

ASA (angle – side – angle) Congruence Axiom: In this axiom, two triangles are congruent if two angles and the included side of one triangle are equal to the corresponding two angles and the included side of the other triangle.

If A = D = 90°; AB = DE and BC = EF then ABC  DEF. Note: AAA is not a criterion for congruence of triangles, since the lengths of the sides need not be equal in triangles of equal angles.

12. Congruent Triangles - Theorems If A = D; AC = DF and C = F then ABC  DEF. Note: The included side is the side edged by the two angles.

3.

AAS (angle – angle – side) Congruence Axiom (corollary of ASA Axiom): If any two angles and a non – included side of one triangle are equal to the corresponding two angles and the non – included side of the other triangle, then the two triangles are congruent.

If A = D; B = E; AC = DF then  ABC  DEF.

4.

SSS (side – side – side) Congruence Axiom: In this axiom, two triangles are congruent if the three sides of one triangle are equal to the corresponding three sides of the other triangle.

If AB = DE; BC = EF; AC = DF then  ABC   DEF.

5.

RHS (right angle – hypotenuse – side) Congruence Axiom: In this axiom, two right triangles are congruent if the hypotenuse and one side of a triangle are respectively equal to the hypotenuse and the corresponding side of the other triangle.

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ASA congruence theorem Two triangles are congruent if two angles and the included side of one triangle are equal to the corresponding two angles and the included side of the other triangle.

Proof:

Let the two triangles be  ABC and PQR. Given that, B = Q, BC = QR and C = R It is required to prove that ABC  PQR. Case (i): If AB = PQ, we also have BC = QR and B = Q, then by SAS congruence axiom  ABC  PQR Case (ii): Suppose AB  PQ. Let AB < PQ. Take a point S on PQ such that QS = AB. Join RS. Now in  ABC and  SQR, AB = SQ; B = Q and BC = QR By SAS axiom  ABC   SQR.  ACB = SRQ, but ACB = PRQ  PRQ = SRQ This is impossible unless RS coincides with RP or S coincides with P.  AB must be equal to PQ.

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177

Case (iii): If we suppose that AB > PQ, then a similar argument applies and ABC   PQR Hence, in all the cases  ABC   PQR. AAS congruence theorem If any two angles and a non – included side of one triangle are equal to the corresponding two angles and the non – included side of another triangle, then the two triangles are congruent.

Proof:

Let the two triangles be  ABC and  XYZ. Given that A = X, B = Y and BC = YZ It is required to prove that  ABC   XYZ Since the sum of the three angles of a triangle is 180°, A + B + C = 180° and X + Y + Z = 180°  A + B + C = X + Y + Z But, A = X and B = Y  C = Z In ’s ABC and XYZ B = Y, C = Z and BC = YZ  By ASA congruence theorem,  ABC = XYZ.

SSS congruence theorem Two triangles are congruent if three sides of one triangle are equal to the corresponding three sides of the other triangle. Proof:

Let the two triangles be ABC and DEF. Given that AB = DE, BC = EF and CA = FD It is required to prove that  ABC   DEF Suppose BC is the longest side. Draw EG from E such that EG = AB Join FG and DG FEG = ABC

In ’s ABC and GEF AB = GE ABC = FEG and BC = EF  By SAS axiom,  ABC  GEF  A = G and AC = GF Now AB = DE and AB = GE  DE = GE Similarly, we can prove that DF = GF Now in  EDG, DE = EG  1 = 4 ( angles opposite to equal sides of a triangle are equal) And in  DFG, DF = GF  2 = 3 ( angles opposite to equal sides are equal)  1 + 2 = 3 + 4 i.e., D = G But G = A  D = A Now in  ABC and  DEF, AB = DE AC = DF and A = D  By SAS congruence axiom  ABC   DEF. RHS congruence theorem Two right triangles are congruent if the hypotenuse and one side of a triangle are respectively equal to the hypotenuse and the corresponding side of another triangle. Then the two triangles are congruent.

Let the two triangles be ABC and DEF. Given in ’s ABC and DEF, B = E = 90° www.betoppers.com

8th Class Mathematics

178 AC = DF and BC = EF. It is required to prove that  ABC   DEF. Produce DE to P such that EP = AB. Join FP. Now in ’s ABC and PEF, AB = EP, BC = EF and  ABC = PEF   BC  PEF (By SAS axiom)  A = P and AC = PF Also AC = DF  PF = DF Now in  DFP, DF = PF  D = P ( angles opposite to equal sides are equal) But A = P  A = D In ’s ABC and DEF A = D; B = E  C = F Now in ’s ABC and DEF, we have BC = EF C = F and AC = DF  By SAS axiom  ABC  DEF.

Formative Worksheet 18. Show that the given triangles are congruent.

19. Given is a figure of two triangles ABC and ADE. Prove that they are congruent.

20. In  ABC and PQR, AB = PQ and the angles at A, B, P and R are respectively 40°, 60°, 40° and 80°. Prove that the triangles are

Important Concepts 1. 2. 3.

Two triangles are congruent if they have exactly the same shape and size. Congruence relation is equivalent i.e., reflexive, symmetric and transitive. Congruence axioms are: i) SAS (side – angle - side) : Two sides and included angle of  ABC = Corresponding two sides and included angle of  PQR. ii) ASA (angle – side – angle) : Two angles and included side of  ABC = Corresponding two angles and included side of  PQR. iii) AAS (angle – angle – side) : Two angles and non-included side of  ABC = Corresponding two angles and non-included side of DPQR. iv) SSS (side – side – side) : Three sides of ABC = Corresponding three sides of DPQR. v) RHS (right angle – hypotenuse – side) : Hypotenuse and oneside of  ABC = Hypotenuse and corresponding side of  PQR.

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21. Prove that in the figure given below ABC and ACD are congruent to each other.

22. In the figure below, PQR is an isosceles triangle in which PQ  PR PS is the bisector of P. Show that i) PQS  PRS

ii) PS  QR

23. In the adjoining figure ABC is an isosceles triangle in which AB  AC AD is the bisector of A. Show that (i)  ABD   ACD

(ii) AD  BC

Plane Geometry

179 24. In the adjoining figure AB  CD . BC and AD are 25. In the adjoining figure ABC is an Isosceles triangles in which transversals. AB = AC. AD  BC. Show that AD bisects  A. Intersecting ‘O’, such that OA  OD A Show that OAB  OCD

Conceptive Worksheet 19. Two triangles ABC and DEF are such that AB = 6 cm,  BAC = 50°, and  ABC = 80°, DE = 6 cm,  EDF = 50° and  DEF = 80°. Prove that the triangle are congruent. 20. Show that the given triangle are congruent. 21. Show that the triangles ABC and DEF given below are congruent. A

B

D

C

13. Particular Type of Triangles Equilateral triangle: A triangle is called an equilateral triangle if and only if all its three sides are equal.

D

C

B

E

F

22. Two triangles ABC and CDE are such that AC = CE, BC = CD,  E = 60°,  DCE = 30° and  B = 90°. Show that the triangles are congruent. (see the figure below) E

A

90°

30° D

In ABC, AB  BC  AC Isosceles triangle: A triangle is called an isosceles triangle if and only if at least two of its sides are equal. Angles opposite to the equal sides are also equal. The equal angles are called the base angles and the other angle is called the vertical angle. In an isosceles triangle, the unequal side is called the base of the triangle.

B

C

23. In the figure below, PR = QR,  SRP =  TRQ and  SQR and  TRP are congruent and hence SR = TR. S

T

P

Q

R

24. In the figure AB || XY . BX and AY are the transversals intersecting at ‘O’, such that

In ABC, AB  AC An equilateral triangle is also an isosceles triangle. Whereas an isosceles triangle is not an equilateral triangle. Theorem In a triangle, if two sides are equal, then the angles opposite to them are equal.

Proof:

OA  OY . Show that  OAB   OXY.. X

Y

O A

B

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8th Class Mathematics

180 Given ABC is a triangle in which AB = AC. It is required to prove that B = C Draw AD  BC Consider the triangles AB D and ACD. We have ADB = ADC = 90° AB = AC (given) and AD is the common side By R.H.S. congruence theorem, ABD  ACD. Hence B = C Hence angles opposite to the equal sides of a triangle are equal. ( corresponding angles of congruent triangles are equal.)

Since the sum of the three interior angles of a triangle is 180°, A + B + C = 180° _______ (3) But from (1) and (2) we get A = B = C _______ (4)

 From (3) and (4) A = 180°  A = 60°  A = B = C = 60°. Hence proved. An isosceles triangle is a triangle with atleast two of its sides equal. If two sides are equal, then the angles opposite to them are equal. Also, if two of its angles are equal, then the sides opposite to them are equal. An equilateral triangle has all its sides equal. Moreover, all the three angles are equal and each angle is 60°.

Theorem If two angles of a triangle are equal, then the sides opposite to them are equal.

Proof: Given ABC is a triangle and B = C. It is required to prove that AB = AC Draw AD  BC. In the triangles ABD and ACD, we have B = C (Given) ADB = ADC = 90° and AD is the common side  By R.H.S. congruence theorem. ABD  ACD. Hence AB = AC.

Important Concepts 1. 2. 3. 4. 5.

Theorem In an equilateral triangle, all the three angles are equal and each is 60 0 . Proof: Given that in ABC, AB  BC  CA It is required to prove that A = B = C = 60° Let us apply Theorem 18. In ABC, AB  CA  B = C ––––––––– (1)

Also AB  BC  A = C ––––––––– (2) www.betoppers.com

A triangle with all its sides equal is an equilateral triangle. A triangle with atleast two sides equal is an isosceles triangle. Equilateral triangle can be an Isosceles triangle Isosceles triangle need not be an Equilateral triangle In an equilateral triangle all the three angles are equal and each is 60°. In an isosceles triangle, Two sides are equal  angles opposite are equal Two angles are equal  sides opposite are equal

Plane Geometry

181

14. Concurrent lines in a Triangle If a number of straight lines intersect at one common point, they are called concurrent lines.

Altitude In a triangle, the perpendicular drawn from a vertex to its opposite side is called an altitude.

In the above figure, AD, BE and CF are the three altitudes of ABC. The three altitudes of a triangle always meet at a single common point called the orthocentre, denoted by ‘O’ which can lie inside or outside the triangle. In a right – angled triangle it lies at the vertex containing the right angle. In an isosceles triangle it lies on the altitude drawn to the base.

Incentre The point through which all the internal bisectors pass is called the incentre of the triangle and is denoted by ‘I’. It always lies inside the triangle. In an isosceles triangle it lies on the altitude of the base. It is always equidistant from the sides of the triangle. The circle drawn with ‘I’ as centre and touching all the three sides of the triangle is called the incircle. The radius of the circle is called inradius and is denoted by ‘r’.

Excentre The external bisectors of B and C and internal bisector of A are concurrent at I1 called Excentre.

Median The straight line joining any vertex to the midpoint of the opposite side of a triangle is known as a median. In the figure below, AF, BE and CD are the medians of ABC.

.

The excentre is opposite to A. Similarly, there are another two excentres I2 and I3 opposite to B and C.

Perpendicular Bisectors

The three medians meet at a common point called the centroid of the triangle and is denoted by G. Centroid is the centre of gravity of all the triangles. Centroid divides each median in the ratio of 2 : 1. A median divides a triangle into two triangles of equal area.

Angular bisectors The line-segments which bisect the angles of a triangle are called the angular bisectors. The linesegments which bisect the angles of a triangle internally are called the internal bisectors. In the figure below, AD, BE and CF are the angular bisectors of ABC.

The lines that are perpendicular and passing through the midpoint of the sides of a triangle are called the perpendicular bisectors of the sides. Perpendicular bisectors need not pass through the opposite vertex. The point through which the perpendicular bisectors pass is called the circumcentre and is denoted by ‘S’. It can be inside or outside the triangle. In an obtuseangled triangle, the circumcentre lines outside of it. In a right-angled triangle it lies on the midpoint of the hypotenuse. In an isosceles triangle it lies on the altitude of the base. By taking ‘S’ as centre, we can draw a circle which passes through all the three vertices. Such a circle is known as the ‘circumcircle’. Radius of the circumcircle is called the ‘circumradius’ i.e., denoted by ‘R’.

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182

Nine point circle A circle passes through the nine points given below which are very specific to the triangle. They are: From the figure: 1. Midpoints of sides BC, AC and AB are X, Y and Z. 2. Feet of the altitudes are D, E and F (altitudes). 3. Midpoints of line-segments joining the vertex and the orthocentre(O) are P, Q and R.

 APQ  QRC  AP = CR But AP = PB ( P is the midpoint of AB)  PB = CR Also since  APQ  QRC  PAQ = ACR But these are alternate angles  BP || CR Since BP is equal and parallel to CR, BCRP is a parallelogram. PR || BC ( opposite sides of a parallelogram are parallel)  PQ || BC and PR = BC ( opposite sides of a parallelogram are equal)

PQ 

Properties: 1. 2. 3. 4.

The medians of a triangle are concurrent. The perpendicular bisectors of the sides of a triangle are concurrent. The altitudes of a triangle are concurrent. The angular bisectors of a triangle are concurrent.

1 1 PR  PQ  BC  PR  BC  2 2

Thus the line-segment joining the midpoints of any two sides of a triangle is parallel to the third side and equal to half of it. Theorem If two altitudes of a triangle are equal, then the triangle is isosceles.

Theorem The line-segment joining the midpoints of any two sides of a triangle is parallel to the third side and equal to half of it.

Proof: In ABC, P and Q are the midpoints of the sides AB and AC. It is required to prove that PQ || BC and

PQ 

1 BC 2

Extend PQ to R, such that PQ = QR. Join CR. Now in ’s APQ and QRC, AQ = QC ( Q is the midpoint of AC) AQP = CQR ( vertically opposite angles) and PQ = QR  By SAS congruence axiom, www.betoppers.com

Proof: Given, PQR is a triangle and altitudes PL and QM are equal. It is required to prove that PQR is isosceles. Consider right–angled triangles PLR and QMR. PL = QM (given) R is the common angle PLR = QMR = 90° Using AAS congruence axiom, PLR  QMR. Hence PR = QR  PQR is an isosceles triangle.

Plane Geometry Theorem The medians on equal sides of an isosceles triangle are equal.

Proof: Let ABC be a triangle with AB = AC . It is required to prove that BD = CE. Consider CEB and BCD.

183

Key Concepts 1.

2. 3. 4. 5. 6.

Here BC is the common side BE = DC (halves of equal sides) BCD = CBE (  In a triangle, angles opposite to equal sides are equal)  By SAS congruence axiom. BDC  CEB Thus BD = CE.

7. 8. 9.

A line-segment which joins the midpoints of two sides of a triangle is parallel to the third side and is equal to half of it. A triangle is isosceles if two altitudes are equal. In an isosceles triangle the medians on the equal sides are equal. Concurrent lines are a number of straight lines that intersect at one common point. Altitude is the perpendicular drawn from a vertex to its opposite side in any triangle. Ortho centre is the point at which the three altitudes of a triangle meet. Median is a straight line drawn from a vertex to the midpoint on its opposite side, in any triangle. Centroid is the point at which the three medians of a triangle meet. The line-segments which bisect the angles of a triangle are called angular bisectors. They are internal, external are perpendicular. Note: In a right triangle the median drawn to the hypotenuse is equal to half of the hypotenuse.

Summary of Triangles

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8th Class Mathematics

184

Important Concepts 1.

If a line stands on another line the sum of the two adjacent angles, called a linear pair, is equal to 180° i.e., p + q = 180° i.e., p + q = 180°. Conversely if p + q = 180°, AB is a straight line.

2.

The bisectors of the adjacent angles of a linear pair enclose a right angle POQ = 90°.

3.

When a number of straight lines meet at a point the sum of the angles soformed at the point is equal to 360°.

6.

Conversely, if a transversal intersects two straight lines such that (i) a pair of corresponding angles are equal or (ii) a pair of alternate angles are equal or (iii) a pair of interior angles on the same side of the transversal are supplementary, then the two lines are parallel. 7. If there are three or more parallel straight lines and the intercepts made by them on any transversal are equal, then the corresponding intercepts on any other transversal are also equal. 8. The sum of the three angles of a triangle is two right angles. 9. The exterior angle of a triangle is equal to the sum of the two interior opposite angles. 10. In a right triangle the median drawn to the hypotenuse is half ofthe hypotenuse. In the figure B = 90°, BD is the median. AD = BD = DC.

11. Conversely, in a triangle, if a median drawn to a side equals half of that side, then the triangle is right -angled. 12. Two triangles are congruent, if and only if, all the sides and angles of the one are equal to the corresponding sides and angles of the other. 13. SAS axiom: If two sides and the included angle of one triangle are equal to the corresponding sides and the included angle of the other, then the triangles are congruent.

4.

If two straight lines intersect one another, then vertically opposite angles are equal. In the figure p = r and q = s.

5.

If two parallel lines are intersected by a transversal then (i) Corresponding angles are equal (ii) Alternate angles are equal (iii) Interior angles on the same side of the transversal are supplementary. 14. ASA axiom: if any two angles and the included c1 = c2 (corresponding angles) side of one triangle are equal to the a1 = a2 (alternate angles) corresponding angles and the included side of a1 + c2 = 180° the other triangles then the triangles are congruent.

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Plane Geometry

185

15. ASA axiom: If the three sides of a triangle are 27. ABC is an isosceles triangle where AB = AC and equal to the corresponding sides of another triangle, B = 2A, then A = ______. then the two triangles are congruent.

28. 16. RHS axiom: In two right triangles if the hypotenuse and a side of one triangle are equal to the hypotenuse and the corresponding side of the 29. other then the two triangles are congruent.

The line-segment joining the midpoints of any two sides of a triangle is parallel to the third side and equal to half of it. M is the midpoint of the hypotenuse AB of a right angled triangle ABC. Prove Cm 

17. In an isosceles triangle; (i) The angles opposite to equal sides are equal. (ii) The sides opposite to equal angles are equal.

1 AB 2

30.  ABC is right angled at A. Prove that the three midpoints of the sides and the foot of the altitude through A are equidistant from N, the midpoint of the median through A.

Conceptive Worksheet 26. In an isosceles triangle, the length of the equal sides is greater than the length of the unequal side. Show that the angle opposite to the unequal side is less than 60°.

18. In an equilateral triangle all thethree angles are equal and each is 60°.

27. In a  ABC, AD bisects  BAC and AD = DC. If  BDA = 70°, then  ABD = _____. A x

x

70° B

D

C

28. In a triangle ABC, the external bisector of vertical angle A is parallel to BC. Prove that the  ABC is isosceles. 19. In an isosceles triangle, the centroid, circumcentre, incentre, orthocentreare collinear whereas in an equilateraltriangle these points coincide.

Formative Worksheet 25. One of the base angles of an isosceles triangle is 40°. Calculate the vertical angle. 26. In an isosceles triangle if the vertical angle is twice the sum of the base angles, calculate the angles of the triangle.

29. X is the orthocentre of  PQR. Show that P is the orthocentre of  XQR.

15. Circles A circle is a simple, closed curve consisting of all points in a plane which are at a fixed distance (say ‘r’) from a fixed point (‘O’) inside it. Consider a circle and a line on a plane.

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8th Class Mathematics

186 There are three possible arrangements of each with respect to the other. a) The line does not touch the circle. b) The line intersects the circle at two points. c) The line touches the circle at one point. Results: 1. A line intersecting the circle at two distinct points is called the secant of the circle. 2. A line touching the circle at only one point is called the tangent of the circle.

Tangent of a Circle The tangent of a circle is a line which touches the circle only at one point. This point is known as the point of contact. The figure shows that ‘’ is a tangent at ‘P’ drawn to a circle with centre ‘O’. The radius of a circle is  to the tangent at the point of contact.

Theorem The tangent at any point of a circle is perpendicular to the radius drawn to the point of contact.

Theorem A line drawn through the end point of the radius and perpendicular to it, is a tangent to the circle.

Proof : Given, a circle with centre O, OX is a radius. A line AB is drawn through X such that AB  OX at X. It is required to prove that AB is a tangent to the circle. Consider any other point Y on AB. Since OX  AB, OY is the hypotenuse of  OXY.. Hence OY > OX.  Y lies outside the circle (since OY > radius) Similarly, with any other point taken on the line, except point ‘X’, we have a right-angled triangle whose hypotenuse will be greater than OX, which is the radius of the circle. Hence there is only one  point ‘X’ on AB which lies on the circle. Therefore  AB is a tangent at point ‘X’. Theorem From a given external point, two tangents can be drawn to a circle.

Proof : Draw a tangent AX at the point A. Here the point of contact is A. It is required to prove that OA  AX. Consider any point B on the tangent AX. Join OB. Since AX is a tangent to the circle at the point A, the point A lies on the circle. Other points are outside the circle. Consider any other point ‘C’ between A and B. Join OC… OC is also greater than the radius OA.This is true of any other point on the tangent. Hence OA is the shortest distance. The shortest distance between a point and a line is the  distance. Hence OA  AX. But OA is the radius. Hence, the tangent at any point of a circle is perpendicular to the radius drawn to the point of contact.

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Proof : Consider a circle with centre O. Let T be an external point. It is required to prove that two tangents can be drawn from T. Join ‘OT’. Taking this as diameter, draw a circle. This circle cuts the circle with centre O at only two points, P and Q. Join TP, TQ, OP and OQ. TPO = 90° ( angle in semicircle)  TP is perpendicular to OP. Hence TP is a tangent to this circle. TQO = 90° ( angle in semicircle)  TQ is perpendicular to OQ. Hence TQ is a tangent to this circle. Hence we have two tangents, TP and TQ, from an external point T.

Plane Geometry

187

Length of the Tangent PA is a tangent of the circle with centre ‘O’.  OA  AP  OAP is a right triangle.

2

2

2

By Pythagoras theorem, OP = PA + OA  PA2 – OP2

36. In the adjoining figure, there are two concentric circle with centre ‘O’. Chord AD of the bigger circle intersects the smaller circle at B and C. Show that AB = CD.

37. In the adjoining figure, chords AB and CD of circle with centre ‘O’ intersect at E. If OE bisects BED. Show that chord AB = chord CD.

 PA  OP 2  OA2

If ‘’ is the length of the tangent of a circle. r – radius of the circle. d – distance between external point and the centre. Then  =

d 2  r2

Formative Worksheet 31. The tangent at any point of a circle is perpendicular to the radius drawn to the point of contact.

Conceptive Worksheet 30. In a figure ‘O’ is the centre of the circle and PQ is a tangent if PQ = 12cm and OQ = 4cm, find PO. Q P

32. In the figure shown, ‘O’ is the centre of the circle and ‘PQ’ is a tangent. If PO = 17 cm and OQ = 7 cm, find PQ.

33. A line drawn through the end point of radius and perpendicular to it, is a tangent to the circle. 34. In the figure below, radius of the circle is 4.8cm, length of the chord, AB is 4.8cm. Find the distance of the chord from the centre of the circle and the area of the ? AOB.

O

31. From a given external point, two tangents can be drawn to a circle. 32. From a point 10cm away from the centre, two tangents are drawn to a circle of radius 6cm. Find the length of the tangents. 33. Find the length of a chord which is at a distance of 8cm from the centre of a circle having radius 10cm.

Summative Worksheet 1.

‘p’ and ‘q’ are complementary angles. Given that

2.

1 th of  q, calculate the value of  q in 6 degrees. An angle is 42° more than one half of its complement. Find the angle in degrees. If two adjacent angles are supplementary, then their outer arms are in a straight line.  p is

35. Prove the converse theorem for chords which are at the same distance of 12 cm from the centre of a circle of radius 13cm.

3.

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8th Class Mathematics

188 4.

If ray OT stands on line RS such that  ROT =  SOT, show that  ROT = 90°.

10. In the figure below,  1 = 45° and  2 =

1 of a 2

right angle. Prove that M || N. T

P M 1

R

5.

O

S

2

In the figure, AB is an incident ray and BC the reflected ray, if  ABC = 124°,  CBX = _____. A

C

N

3

11. In the figure shown, AB is parallel to CD. Find the values of ‘p’ and ‘q’. E

124°

t

t

X

Y

65°

G

A

B

45°

B

6.

In the figure below, ‘l’ and ‘m’ are two lines intersected by a transversal ‘n’. Is l || m? If so, why?

P°

O

40° q°

C

D

n

F 130°

a

l c

b

t E 1

d

e

l

m

50°

f

7.

12. Prove that the two lines which are both parallel to the same line are parallel to one another.

2

F

In the figure below,  1 = 70°. Show that each pair of consecutive interior angles are supplementary.

G

m

3

n

X P

A

3

Q

C

7

2

1

B

13. In the given figure, PQ and RS are two parallel lines. Giving reasons, find the values of x, y and z.

4

R

6

5

(3x)° (4x)°

D

8

Y

8.

9.

75°

If a transversal intersects a pair of parallel lines, then the interior angles on the same side of the transversal are supplementary. In the figure below, PQ, RS and TU are three coplanar lines. AB is the transversal intersecting the lines at X,Y and Z. If  PXA = 115°,  UZX = 65° and  UZX +  SYZ = 180°, then show that PQ || RS and RS || TU. Q

115°

X

S

R Y 65°

T

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Z

B

U

y°

P

z° C

B

Q

14. Find the value of ‘x’ in the following figure. A x

A

P

S

A

30° B

70° C

15. Find the third angle of a triangle whose two angles are 45° and 25°.

Plane Geometry

189

16. Angles of a triangle are (x + 10°), (x + 40°), and (2x – 30°). The value of x is ________. 17. In the figure below:

(a)

 ABC : AB = 3, BC = 4,  B = 90°  DEF : DE = 3, EF = 4,  E = 90°

(b)

b is _______;  c is _______;

 ABC : AC = 2, BC = 3,  A = 90°  DEF : DE = 2, DF = 3,  D = 90°

a is ______

(c)

A

 ABC : AB = 3, AC = 5, BC = 6  PQR : PQ = 6, QR = 5, PR = 3

a

(d)

 ABC and  DEF, BC = EF,,  A = 90°,  B = 50°,  F = 40°

b c

35° D

B

23.

46° C

In the figure below,  XYZ and  AYZ are two triangles such that XY  AY and XZ  AZ .

E

18. In the figure below, ABC is a triangle in which B  50 and C  70 . Sides AB and AC are produced. If z is the angle between the bisectors of the exterior angles formed, then the value of z = ?

Show that  XYZ   AYZ. YZ. X

A

Z

Y

A

B x°

50° x°

24. In a parallelogram, show that (i) the opposite sides are equal (ii) the opposite angles are equal (iii) the diagonals bisect each other.

70° C y° y° z°

D

D

19. In the figure given below, t he bisectors of B and C of  ABC intersect at D. Show that

D  90 

z 2

O B

A

25.

In  ABC the sides are in the ratio 1: 3 : 2 then the angles of the triangle are _____. 26. In  ABC,  A = 120°, AD, BE, CE are bisectors of  A,  B and  C.

A z

x x

C

D y

C

y

B

C

20. Show that the given triangles are congruent. B

D

E

C

A O D

B

F

(i) Prove that E is excentre of  ADB. (ii) Prove that F is excentre of  ADC.

27. In  ABC if  A +  B =  C then orthocentre is 21. In triangles ABC and ACD, ACB  90 , (A) A (B) B (C) C ACD  90 and AB = AD. Prove that the triangles are congruent. (D) midpoint of AB (E) none of these 22. In the following pairs of triangles, find which pairs are congruent. The lengths of the sides are measured in cm. A

its

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8th Class Mathematics

190 28. In the figure below, the sides AB and AC of a triangle ABC are produced to X and Y respectively. The bisectors of  XBC and  YCB intersect at O.

2.

In the given figure, AB || CD. Find the value of x. C B

A

F

110°

1 Prove that  BOC = 90° –  BAC. 2

D

120°

x°

A

E

3. B

C

In the figure given below, find the values of x, y, z, a, b and c. b° a°

O X

x°

c°

Y

29. Prove that the angle between the internal bisector of base angle and the external bisector of the other is equal to one half of the vertical angle.

z°

y°

110°

65°

A E

B

4.

E

2

1

C

G

D

30. A chord of length 10 cm is at a distance of 12 cm from the centre of a circle. What is the radius of the circle? 31. What is the length of the tangent of a circle drawn from a point at a distance of 13cm from the centre? Given radius of the circle is 5cm. 32. The Perpendicular bisector of a chord of a circle passes through the centre of the circle. 33. The diameter which is perpendicular to a chord of a circle bisects the chord.

A

P

Q

R

S

C F

5.

6.

H

C AC C= +B

A

x°

7.

120°

D

C 45°

x°

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B

Using information given in the adjoining figure, find the values of x and y. A

A

C

60 80 D

90

B

105°

(ii)

D

In a triangle, if the median drawn to a side is equal to half of that side, then prove that triangle is right angled. The base of a triangle is 80 and one of the base angles is 60. The sum of the lengths of the other two sides is 90. The shortest side is

In each of the following figures, AB || CD. Find the value of x in each case. (i) A E

B (3x + 5)°

4x°

HOTS Worksheet 1.

In the adjoining figure, lines EF and GH are parallel. Find the value of x for which the lines AB and CD will be parallel.

B

x° x°

O 50° D

100°

65°

B

C

y°

D

Plane Geometry 8.

191

Using information given in the adjoining figure, find 13. ABC is a triangle. D and E are two points on AB the values of x and y. AB AC A  and AC such that = 2m for 01) is also a rational number.. i) Let an odd number be (2p + 1) and its Odd perfect square = (2p + 1)2 = 4p2 + 4p + 1 = 4p(p + 1) + 1 = 4k + 1 where k = p(p + 1). Therefore an odd perfect square will be odd of the form (4k + 1). ii) Let an even number be 2k and its Even perfect square = (2k) 2 = 4k 2 = 4p where p = k2. Therefore an even perfect square will be even.

8th Class Mathematics

194 We know that the sum of two even numbers is even and also the sum of two odd numbers is even. i.e. a + b is even when a, b are even. Given a and b are two natural numbers. (a + b) is even implies either a and b are both even or a and b are both odd. Case1: Suppose a and b are both even. Then let a = 2n and b = 2m  a – b = 2n – 2m = 2(n – m) which is even. Hence the difference is also even. Case 2: Suppose a and b are odd. Then let a = 2p + 1 and b = 2q + 1 Their difference = (a – b) = (2p + 1) – (2q + 1) = 2p + 1 – 2q – 1 = 2p – 2q = 2(p – q) which is even. Hence in either of the cases (a – b) is even. 7. Now x, y and z are odd. i) So, x2 is odd, also y2, z2 is odd. So x2 + y2 + z2 + 12 is odd Choice (1) is wrong ii) Also x2y2, y2z2 are odd and (z – x)2 is even. So even – odd is odd. Choice (2) is correct. iii) y2 + z2, (y – z)2, (x – y)2 is even So x2[y2 + z2 – (y – z)2 – (x – y)2] is even So even + odd is odd Choice (3) is wrong. iv) Also x3, y3, z3, (y – x)3, (z – x)3 is odd x3(y3 + z3 – (y – x)3 – (x – x)3 + 73 odd [ odd + odd – odd – odd] + odd = even . choice (4) is correct Therefore the correct answer is (2) and (4). 8. Let x = 4, y = 2 Then 42 – 22 = 16 – 4 = 12(even). 9. By trial and error method, let us find all the factors of 12. They are 1, 2, 3, 4, 6 and 12. The set of these factors is represented as F(12) = {1, 2, 3, 4, 6, 12} Similarly, by trial and error method, F(36) = {1, 2, 3, 4, 6, 9, 12, 18, 36} 10. The multiples of any number are obtained by multiplying the number with the natural numbers starting from 1. Then multiples of 20 are 20 × 1 = 20 20 × 2 = 40 20 × 3 = 60 ........ Now the multiples of 20 are 20, 40, 60, 80, 100,...., 200,....... And the multiples of 25 are 25 × 1 = 25 25 × 2 = 50 25 × 3 = 75........

Now the multiples of 25 are 25, 50, 75, 100,........., 200,........

6.

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M  20   M  25  

{20, 40, 60, 80, 100,.., 200,..} {25, 50, 75, 100,.., 200,..}

{100, 200,.....} M  24   M  42  

{24, 48, 72, 96, 120, 144, 168,....336..} {42, 84, 126, 168, 210,..336,..} ={168, 336, ........} 11. Given, n3 – 1 is a prime number. n3 – 1 = (n – 1) (n2 + n + 1). So n – 1 must be equal to 1.  n2 + n + 1 is prime. 12. For n = 1,

1(1  1) 2   1 , which is not a prime 2 2

number.

2(2  1)  3 , which is a prime number 2 and for all other numbers (n > 2), we get an even For n = 2,

n( n  1) . 2 13. When n = 41, (n2–n+41) = 412 which is composite.  square of a prime is not a prime 14. Let n = 1 then, 21 – 1= 1 Let n = 2 then, 22 – 1= 3(Prime) Let n = 3 then, 23 – 1= 7(Prime) Let n = 4 then, 24 – 1= 15(Not Prime)  For all n = prime number 15. a3 – b3 = (b + 2)3 – b3 ( a > b) [  a and b are twin primes ] = 6b2 + 12b + 8 number for

16. Suppose x is the number to be added to

then

5  5  5 5 + x =  x =   9  8  8 9

 x=

5 5 + 9 8

 x=

5× 8 + 5 × 9 40 + 45 85 = = 72 72 72



85 -5 5 must be added to to get . 72 8 9

5 8

,

Number System Solutions 17. Concept: To arrange the same, we need to first convert the given fractions into decimals. Given numbers are Now,

3 7 1 2006 413 , , , , 4 9 2 2007 1115

3  0.75 4 7  0.777.......... 9 1  0.5 2

2006  0.99 (by considering two 2007 decimal places) 413  0.37 1115 From the above, it is clear that: 0.37 < 0.5 0 if x > b   i   if x < b b bx

check if it is negative.



221

a ax  b b- x

a ax  b bx

a b  x  b  a  x x b  a = b b  x  b b  x

a Now, we know that a > b, since is of greater b inequality.

a  a b   21 16  21       21: 26 c  b c   32 13  26 20. i) Compound ratio of ii) a : c 

2 :1,3: 5, 20 : 9 is 

2 3 20 2  45 2 2     1 9 3 5 5 3

 The compound ratio is 2 2 : 3 . ii) Compound ratio of x : y, y : z and z : p is

x y z x  ×   y z p p  The compound ratio is x : p. iii) Compound ratio of 2x : 3y, a : 2b and 3p : 4q is



2 x a 3 p axp × × = 3 y 2b 4q 4 ybq

 The compound ratio is axp : 4ybq

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8th Class Mathematics

222 21. Duplicate ratio of 2

a :b

2

b

2

 =a:b



2





3

  ab  2 a  b   x 2 a  b

 3 : 5 3: 5  5  3:5

x 8 52  2 2  x  16  8



288 144  14 7

144 7 24. i) The sub-duplicate ratio of 16p4x6 : 9q6y4

 x

 16 p 4 x 6 : 9q 6 y 4 = 4p2x3 : 3q3y2

ii) The sub-duplicate ratio of 9p2 : 16q4  9 p 2 : 16q 4  3 p : 4 q 2

iii) The sub-duplicate ratio of 729x12 : 4096x6 6 3 3  729 x12 : 4096 x 6 = 27x : 64x = 27x : 64  Sub duplicate ratio = 27x3 : 64

25. The sub-duplicate ratio of a : b is Given: (2a – x) : (b – 2x) = a



b





a  b  2 x   b  2a  x 



ab  2a b  2 x a  x b



a b b 2 b a a





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a: b.

a: b

2a  x b  2x

x 2 a b

 x   ab  x2 = ab Hence proved. 26. i) The triplicate ratio of

 p  q

7

3

: p  q 3

x  8 25  2 50   x  16 64 64  64 (x + 8) = 50 (x + 16)  64x + 512 = 50x + 800  64x – 50x = 800 – 512  14x = 288  x



7

3

2

23. Duplicate ratio of 5 : 8 is 52 : 82 . Given, (x + 8) : 2 (x + 16) is the duplicate ratio of 5 : 8. 



4

22. i) Duplicate ratio of 2p : 4q = (2p)2 : (4q)2 = (2p × 2p) : (4q × 4q) = 4p2 : 16q2 ii) Duplicate ratio of 3: 5 



ab  b  2 a   x 2 a  b

 2 2

 a  : b    a  a  : b 

2

7 7   p  q  3  :  p  q  3      = (p + q)7 : (p – q)7 ii) The triplicate ratio of

3

3

3

1 2 2 1 is   a 3 b 3  :  a 3 b 3      2 2 = ab : a b = b : a 27. First we have to find the sub-duplicate ratio of 4096a6 : 729b12 The sub duplicate ratio of 4096a6 : 729b12 is

a

1

3

2

b 3 :a

2

3

b

1

3

 4096a 6 : 729b12 = 64a3 : 27b6 The sub triplicate ratio of 64a3 : 27b6 is  3 64 a 3 : 3 27b 6 = 4a : 3b2  The sub-triplicate ratio of the sub duplicate ratio of 4096a6 : 729b12 is 4a : 3b2. 28. The ratio of 1.25 litres to 500ml is 1.25 litre : 500ml But they are of different units So, change 1.25 litre to ml i.e., 1.25 × 1000 ml = 1250 ml Now, the ratio is 1250 ml : 500 ml = 5 : 2 So, we can express the ratio of 1.25 litres to 500 ml in the form of a : b as 5 : 2. Therefore, the ratio of 1.25 litres : 500 ml is commensurable. ( 5, 2 are two integers) 29. i) We know that four quantities a, b, c, d are said to be in proportion, if a : b = c : d. Hence, if 40, 30, 60, 45 are to be in proportion 40 : 30 should be equal to 60 : 45

40 4 60 4  and  30 3 45 3  40 : 30 = 60 : 45. Hence 40, 30, 60, 45 are in proportion. ii) We know that four quantities a, b, c, d are said to be in proportion if a : b = c : d Hence if 5, 6, 20, 18 are to be in proportion 5 : 6 should be equal to 20 : 18. Now,

Ratio, proportions and Variations Solutions

5 5 20 10  and  6 6 18 9  5 : 6  20 :18 Hence, 5, 6, 20, 18 are not in proportion. 30. Since 2,4,6 and x are proportion, we have, 2 : 4 :: 6 : x Product of means = Product of extremes. 4×6=2×x x = 12 31. Since x, 2x, y and y2 are in proportion, we have, x : 2x : : y : y2 Product of means = Product of extremes. x × y2 = 2xy  y = 2 and x can take any positive value.

223 37. As a, b, c, d are in continued proportion

Now,

a c e g 32. As = and = (given) b d f h



a e c g ae cg × = ×  = b f d h bf dh

 ae : bf = cg : gh. 33. Let the mean proportional be x. Then, x2 = 7 × 63  x = 21 34. Let c be the third proportional. We know that 8, 24 and c are in continued proportion if 8: 24: : 24 : c

8 24  or 24 c

24  24  72 8 Hence the third proportional is 3. 35. Let ‘c’ be the third proportional. We know that (p 2 – q 2 ) (p + q) and c are in continued proportion if (p2 – q2) : (p + q) : : (p + q) : c c

or

p2  q 2 p + q = p+q c

c

 p + q  p + q  2

p q

2



p+q p-q

p+q Hence, the third proportional is p  q . 36. Let it be x. Then, 1: 2 : : 3 : x 6=x  x=6

we have

a b c = = . b c d

a b c = = =k b c d Then, a = bk; b = ck, c = dk i.e, a = bk = ck × k = dk × k2 = dk3 b = ck = dk × k = dk2 L.H.S is 2a + 3d : 3a – 4d Substituting the value of a, we get 2dk3 + 3d : 3dk3 – 4d Now, let



d  2k 3  3 d  3k 3  4 



2k 3  3 3k 3  4

________

(1)

R.H.S is 2a3 + 3b3 : 3a3 – 4b3, substituting the values of a, we get 2b3k3 + 3b3 : 3b3k3 – 4b3 



b 3  2k 3  3  b3  3k 3  4 



2k 3  3 3k 3  4

2b3 k 3  3b3 2b3 k 3  4b3

________

(2)

From (1) and (2), we have 2a + 3d : 3a – 4d = 2a3 + 3b3 : 3a3 – 4b3. 38. Let a, b, c, d be the four proportionals.

a c.  ad = bc. = b d Also, given a + d = 21, b + c = 19, a2 + b2 + c2 + d2 = 442  (a + d)2 + (b + c)2 = (21)2 + (19)2 = 441 + 361  a2 + d2 + b2 + c2 + 2ad + 2bc = 441 + 361  442 + 2ad + 2bc = 441 + 361  442 + 4ad = 441 + 361 ( ad = bc)  ad = 90 and bc = 90 (a – d)2 = (a + d)2 – 4ad = (21)2 – 4 × 90 = 81 Then,

 a  d  9 If a + d = 21 and a – d = 9  a = 15, d = 6 If a + d = 21 and a – d = – 9  a = 6 and d = 15 (b – c)2 = (b + c)2 – 4bc = (19)2 – 4 × 90 = 1  b  c  1 If b + c = 19, b – c = 1  b = 10, c = 9 If b + c = 19 and b – c = –1  b = 9 and c = 10  The four proportionals a, b, c, d are 6, 9, 10, 15 or 15, 10, 9, 6. www.betoppers.com

8th Class Mathematics

224 39. Given: a + c = 2b  c = 2b – a –––––– (1) and

1 1 2 + =  c(d + b) = 2bd b d c

 c

2bd –––––– (2) d +b

2bd d +b (2b – a)(d + b) = 2bd 2bd + 2b2 – ad – ab = 2bd 2b2 – ab = ad  b(2b – a) = ad bc = ad ( c = 2b – a)

From (1) and (2), 2b  a 

   

c a   a : b :: c : d . d b 40. Let the three numbers be a, b and c Given they are in continued proportion, then

41. By Componendo, we have x 2  x  2  (2  x ) x 2  2 x  3  (2 x  3)  2 x 2x  3 x2 x2  2  x 2x  3  2x – 3 = 2 – x  3x = 5 

 x

5 3

3a  6b  c  2d 3a  6b  c  2d  3a  6b  c  2d 3a  6b  c  2d By componendo and dividendo rule, we have

42. We have,



 3a  6b  c  2d    3a  6b  c  2d   3a  6b  c  2d    3a  6b  c  2d 

a b = = k (say) i.e., a = bk ; b = ck b c a = bk = ck (k) = ck2 b = ck Given that sum of the first and third is 130. i.e.,a  ck2 + c = 130  c(k2 + 1) = 130



130 __________ (1) k 2 1 Also, the middle one is 16.  b = ck = 16

 3a  6b  c  2d   (3a  6b  c  2d )  3a  6b  c  2d    3a  6b  c  2d 



6a  2c 6a  2c  12b  4d 12b  4d



2(3a  c ) 2(3a  c )  2(6b  2d ) 2(6b  2d )

 c

16 __________ (2) k From (1) and (2) c

c

    

130 16  k 2 1 k 130k = 16k2 + 16 16k2 – 130k + 16 = 0 16k2 – 128k – 2k + 16 = 0 16k (k – 8) – 2(k – 8) = 0 (16k – 2) (k – 8) = 0

2 or k = 8 8 If k = 8 k 

16 16  2 k 8 b = ck = 2(8) = 16 a = bk = 16(8) = 128 Hence, the numbers are 2, 16, 128. then, c 

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3a  c 3a  c 3a  c 6b  2d    6b  2d 6b  2d 3a  c 6b  2d Applying Componendo and Dividendo rule again. We get, 

 3a  c    3a  c    6b  2d    6b  2d   3a  c    3a  c   6b  2d    6b  2d  6a 12b a b a c      2c 4d c d b d  a,b,c,d are in proportion. Hence it is proved. 

9 x  7 y 9a  7b 43. We have, 9 x  7 y  9a  7b By componendo and dividendo, we get

 9 x  7 y    9 x  7 y    9 a  7b    9 a  7 b   9 x  7 y    9 x  7 y   9 a  7b    9 a  7b  

18 x 18a x a    14 y 14b y b

Therefore, x : y = a : b

Ratio, proportions and Variations Solutions

225

44: To show that,

 a  1   a  1  a  1   a  1

2 p + q + 4r + 2s 2 p + q - 4r  2 s = 2 p - q + 4r  2 s 2 p  q - 4r + 2s p r = q s (given)



multiplying both sides by 2, we get 

2 p 2r  q s

Applying componendo and dividendo, we get,

2 p  q 2r  s  2 p  q 2r  s Applying alternendo, we get



2

 8a  1  16 a 2  24a  9 

2

 8a  1  16a 2  24 a  9 

2a 32a 2  16a  10  2 32a  8

16a 2  8a  5 16a  4 2  16a – 4a = 16a2 – 8a + 5  a

 4a = 5  a 

2p  q 2p  q  2r  s 2r  s



16a 16a

 a

Now, multiplying both sides by

1 , we get, 2

2p  q 2p q  4r  2s 4r  2s Now, applying componendo and dividendo, we get

5 4 4 6

46. Given x 

2 3



2 p + q + 4r + 2s 2 p  q + 4r  2 s = 2 p + q  4r  2s 2 p  q  4r + 2 s Now, apply alternendo



Hence proved. Using componendo and dividendo,

  

2 2

x 2 2



x2 2

a +1  a  1

x2 2

a +1 +

a +1 



 a  1

 4a  1  2  4a  1  2



2 a 1



4a  1  4a  3

a 1



4a  1 4a  3

2 3





2 3 2 3



2 3 

a  1 16a 2  8a  1  a  1 16a 2  24a  9 Again, using componendo and dividendo

we have,

2 3 2 3 2 3 2 3

3 3 2

_______

3 2 x



(1) 2 2

2 3 2 3 Applying componendo and dividendo we get, x2 3 2 2 2 3  x2 3 2 2  2 3

2

 a 1   4a  1    2  a  1   4a  3

i.e.,



2 3

Similarly,

2 a 1 a 1 Now, squaring both sides, we get, 2

2 2

6  2 3

a +1 + a  1 +

  a  1  

4 6



Applying componendo and dividendo to

2 p + q + 4r + 2s 2 p + q  4r  2 s  = 2 p  q + 4r  2s 2 p  q  4r + 2 s 45

x



5 . 4



3 2 3

__________ (2) 2 3 From (1) and (2) we get

x2 2 x2 2



x2 3 x2 3



3 3 2 3 2



3 2 3 2 3

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8th Class Mathematics

226

3  





3 2



   2  2  3 

2 3  3 2 3



3

3 2



3 6 9  2  6 3 6 6  3 6 6 3 2  6

4 6  10 2 6 5 x2 2 x2 2



2 2 6 5 



2 6 5

x2 3

 2

k3  k  k 2  1 sk 3

____________

(2)

From (1) and (2)  1 1   1 1   p  q  p  r        pqr p s q r

x y z 49. Let p  q  r  k i.e., x = pk ; y = qk ; z = rk Substituting the values of x, y and z in L.H.S, we get

2

x2 3



a c = =k b d  a = bk and c = dk

x 2  p2 y2  q 2 z 2  r 2    L.H.S xp yq zr

a 2  ac  c 2 b2 k 2  bdk 2  d 2 k 2   2 a  ac  c 2 b 2 k 2  bdk 2  d 2 k 2

 pk  

47. Let

2



2

2

k b  bd  d  b  bd  d = 2 2 k 2  b2  bd + d 2  b  bd  d 2

 1 1  1 1 we get         p s q r

 sk  sk  sk  sk  sk sk  sk  3

sk 2  k  1 sk  k 2  1

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 q2

 rk  

qk  q

2

3



2

 r2

rk  r

p 2  k 2  1 p  k  1 2



q 2  k 2  1 q  k  1

 1  q  p  r 



________

3

2

 k  1  k 2  1 sk 3

r 2  k 2  1 r  k  1

(1)

Again substituting the values of x, y and z in R.H.S., we get, 2

R.H.S 

 x  y  z   q  p  r   x  y  z   q  p  r  2

1 1  k 3  k  k 2     ________ (1) s k3  Also substituting the values of p, q and r in R.H.S, we get

s3 k 6





1  1 1 1 1  1 1  1   3     2     3 1 2   s   sk sk  s  k k k  sk



2

k 1

i.e., p = qk ; q = rk; r = sk p = qk = rk × k = sk × k2 = sk3 q = rk = sk × k = sk2 r = sk Substituting the values of p, q and r in L.H.S

pqr

 qk  

pk  p

k

p q r  = = =k q r s



 p2

2

Hence, it is proved. 48. Given p, q, r, s are in proportion

 p  q  p  r 

2

 pk  qk  rk    q  p  r   pk  qk  rk    q  p  r 

2

2

2

 q  p  r   k 2  1   q  p  r  k  1 

 q  p  r   k 2  1

___________

k 1

Hence from (1) and (2) x 2  p2 y2  q2 z 2  r 2   xp yq zr 2

 x  y  z   q  p  r    x  y  z  q  p  r

2

(2)

Ratio, proportions and Variations Solutions

227 54.

a c e   k b d f  a = bk, c = dk, e = fk

50. Let

 k1a  k 2 c  k 3e  Consider,  k b n  k d n  k f n    1 2 3 

3 2 9 7 2  K. 7 or K     1 9 3 14 3 3 14

 k1  bk n  k 2  dk n  k 3  fk n   k1 b n  k 2 d n  k 3 f n 

 48  1.

n

n

n

   

1 1  k a n  k 2 cn  k 3e n  n n n  1 n  k   k n n   k1b  k 2 d  k 3 f 

3

3

CONCEPTIVE WORKSHEET

3

x 3 y3 z3  ak   bk   ck   3 3 3 3  3  3 a b c a b c

1.



3

x 3 y3 z 3 3xyz    a 3 b3 c3 abc 52: If x  y , then x = ky i.e., k = x/y __________(1)

So,

x = 8, y = 15 Now x 

8 1  10  5 15 3

 k

( from (1), x = ky)

x 2  y3 or x2 = ky3 or k = x2/y3 But x = 3, y = 4

__________

2 3.

8 15

2A = 3B = 4C

2A 3B 4C A B C      k 12 12 12 6 4 3  A = 6k, B = 4k, C = 3k  A : B : C = 6k : 4k : 3k = 6 : 4 : 3.

3xyz 3.ak.bk.ck   3k 3 = k + k + k = 3k and abc abc 3

Q = 60.

(where N is some other constant) Hence, x2 + y2 = N (x2 – y2) or y2 + x2  x2 – y2.

x y z   k a b c So x = ak, y = bk, z = ck

51: Here,

3

or

2 2 x 2  y 2 y  k  1 k 2  1  2 2 2 2  2 N x y y  k  1 k  1

a c e    (Hence proved). b d f

3

Q 75

 Q = 60 when P  48 and R  75 55. If x  y then x = ky  x2 + y2 = k2y2 + y2 = y2 (1 + k2) ( By substituting x = ky) Again x2 – y2 = k2y2 – y2 = y2(k2 – 1) ( By substituting x = ky)

 k b n  k 2 d n  k 3f n   kn  1 n  kn n n  k b  k d  k f  1 2 3 

53.

P  Q and P  1/ R  P = K(Q/R); where K is any constant.

4.

(1)

 k = 9/64

A A B 3 5 15       A:C = 15 : 8 C B C 2 4 8 B = 3A C = 2B  C = 2 × 3A = 6A  A : B : C = A : 3A : 6A = 1 : 3 : 6 x y = k 5 2 x = 5k, y = 2k 

2

 1  9 3   .y ( by substituting in (1) ) 64  3



or y  3

64 4 1 or y   1 27 3 3

 y 1

1 1 when x  3 3

5.

8 x  9 y 40k  18k 58 29    8 x  2 y 40k  4k 44 22

4 x 2  3 y 2 12  2 x 2  5 y 2 19 19(4x2 – 3y2) = 12 (2x2 + 5y2)  76x2 – 57y2 = 24x2 + 60y2  52x2 = 117y2  4x2 = 9y2 

x 3 x2 9    2 y 2 y 4

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8th Class Mathematics

228 6.

4 x  3 y 11  6 x  5 y 17 Dividing the numerator and the denominator by y,

x 4 3 11 y   x 6  5 17 y  68 

x x  51  66   55 y y

 68 

x x 2x  66   55  51  4 y y y



7.

 4x   6x   17   3   11  5  y y    

x 4 x 2     x : y  2 :1 . y 2 y 1

x 3  y 5 Dividing the numerator and denominator by y,

x:y=3:5 

x 8   3 8  3  3 y 8x  3 y 39 5      3 5x  2 y x 25 . 5   2 5  2 5  y

8.

 x 3  Substituting y  5     the ratio (8x + 3y) : (5x + 2y) = 39 : 25 Let the numbers be a and b.

a 8  b 5 8b  5a  8b  a  _________ (1) 5

a:b= 8:5 

Given the difference of squares of the two numbers is 156.  a2 – b2 = 156  (a – b) (a + b) = 156 _________ (2) Substituting the value of a from (1) in (2)

 8b  8b     b   b   156  5  5 

 3b  13b       156  5  5   b2 

156  25 12  25  39 3

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 b2 = 4 × 25 8 10  16 . 5 ( by substituting b = 10 in (1))  The two numbers are 16, 10 9. a : b = 2 : 3 ––––––––– (1), b : c = 4 : 5 ––––––––– (2) Let us make the magnitude of b, equal in the two given ratios. The value of b is ‘3’ in (1) and ‘4’ in (2) LCM of 3 and 4 is 12. To make b = 12, we multiply the first ratio by 4 and the second ratio by 3.  a : b = 2 : 3  2 × 4 : 3 × 4 = 8 : 12 –––– (3) b : c = 4 : 5  4 × 3 : 5 × 3 = 12 : 15 –––– (4)  a : b : c = 8 : 12 : 15 (From (3) and (4)) a : c = 8 : 15. 10. Let the two lengths of sides of the triangles be 3x and 4x. Let the hypotenuse be y. Given the perimeter of the triangle is 24cm.  3x + 4x + y = 24  7x + y = 24 ____________ (1)

 b = 10

a

Also in a right triangle,

2

 3x    4 x 

2

y

( According to pythagoras theorem, In the right angle triangle (hyp)2 = (base)2 + (height)2 )

 y  5x  25 x 2 Substituting y = 5x in (1) we get 7x + 5x = 24  12 x  24  x2 The lengths of sides of the triangle are 3x = 3 × 2 = 6 and 4x = 4 × 2 = 8 The hypotenuse is 5x = 5 × 2 = 10 Hence, the sides of the right triangle are 6, 8 and 10. 11. Work done by (a + 7) workmen in (a – 8) days = (a + 7)(a – 8). Similarly, work done by (a – 8) workmen in (a + 9) days = (a – 8)(a + 9).

 a  7  a  8   5  a  8 a  9  6

a7 5  a 9 6  6(a + 7) = 5(a + 9)  6a + 42 = 5a + 45  6a – 5a = 45 – 42  a = 3.  The value of a = 3. 12. a : b = 3 : 7 _________ (1) b : c = 6 : 11 _________ (2) Let us make the magnitude of b equal in the two given ratios. 



Ratio, proportions and Variations Solutions

229 L.C.M. of denominators 5 and 8 is 40.

The value of b is ‘7’ in (1) and ‘6’ in (2) L.C.M. of 7 and 6 is 42. To make a = 42, we multiply the first ratio by 6 and the second ratio by 7. a : b = 3 : 7 = 3 × 6 : 7 × 6 = 18 : 42 ___________ (3) b : c = 6 : 11 = 6 × 7 : 11 × 7 = 42 : 77 __________ (4)  a : b : c = 18 : 42 : 77 [ from (3) and (4)] 13. Let

a be the ratio. b

4 4  8 32 7 7  5 35    ;  5 5  8 40 8 8  5 40 35 32   7 :8  4 : 5 40 40 Hence, 7 : 8 is greater than 4 : 5. 15.

a 1, b i.e., a > b. Let us add x to both the numerator and

7 2 4 ; 2:9  ; 4:5  3 9 5 L.C.M. of 3, 9, 5 is 45 7 :3 

As the ratio is of greater inequality,

the denominator. The new ratio is

7 15 105  3 15 45

 

a+ x . b+ x

2 5 10 4 9 36   ;   9 5 45 5 9 45

a+ x a is smaller than . b+ x b Concept: If p > q, (p – q) > 0. In other words, (p – q) is positive. We have to show that

a a+ x >0 b b+ x

Let us take the difference of

a a+ x and and b b+ x

a a + x a  b+ x  - b  a + x  = b b+ x b  b+ x  ab + ax - ab - bx x  a - b  = b b + x  b  b+ x 

Now,

x a - b will be positive, if (a – b) > 0 b b + x 

We know that a > b, since ratio is of greater inequality.

a a+ x  a a+ x 0   b b+ x  b b+ x  14: Convert the consequent of each ratio to the same value and compare the antecedent.

 -

4:5 

4 7 and 7 :8  . 5 8

4 7 28 6 5 30   &   5 7 35 7 5 35

30 28  . 35 35  The descending order is 6 : 7 > 4 : 5 17. Compound ratio of 2 : 3 and 3 : 7 is 2 × 3 : 3 × 7 = 6 : 21 = 2 : 7  The compound ratio is 2 : 7. 18. Compound ratio It’s clearly indicates that

check if it is positive.

=

4 6 & 6:7  5 7 The L.C.M of denominators 5, 7 is 35. 4:5 



 a a + x  is positive. i.e.,    b b+ x 

105 36 10   45 45 45

7:3>4:5>2:9 16.

a a+ x Similarly, to show > , b b+ x we show that

It clearly indicates that

2 2 p  x p2  x2  p  x     p  x  p  x 2 p4  x4

 p  x   p 2  x2  p 2  x 2    p  x  p  x 2  p 4  x4  

2

2

 p  x   p 2  x 2   p  x  2  p  x 2  p  x  p  x  2  p 2  x2  p 2  x 2  2

2

 p  x   p 2  x2   p  x   p  x   2  p  x  p  x   p  x  p  x   p 2  x 2  1   1:1 . 1 www.betoppers.com

8th Class Mathematics

230 19. Duplicate ratio of 2

3: 6 





3

2

 3 :  6  3: 6  6 = 3 : 6

20. Duplicate ratio of 6 : 10 is 62 : 102 Given: (x + 2) : (3x + 4) is the duplicate ratio of 6 : 10. 2

x2 36 x2 6   2  3x  4 100 3x  4 10  100(x + 2) = 36(3x + 4)  100x + 200 = 108x + 144  100x – 108x = 144 – 200  –8x = –56  x = 7. x = 7.  21. Duplicate ratio of (p + r) : (q + r) is given to be p : q. The duplicate ratio of (p + r) : (q + r) is (p + r)2: (q + r)2. 

2

 (p + q)2 : (q + r)2 =

 p  r 2 q  r



p q

p 2 + r 2 + 2pr p = q 2 + r 2 + 2qr q

 q(p2 + r2 + 2pr) = p(q2 + r2 + 2qr)  p2q + qr2 + 2pqr = pq2 + pr2 + 2pqr  p2q – pq2 = pr2 – qr2  pq(p – q) = r2(p – q)  r2 = pq.  Hence proved. 22. i) The sub-duplicate ratio of 9x2a4 : 25y6 =

9 x 2 a 4 : 25 y 6 = 3xa2 : 5y3.

ii) The sub-duplicate ratio of 64a6 : 729x4 64a 6 : 729 x 4  8a 3 : 27 x 2

23. i) The sub-triplicate ratio of 64a3 : 27b6 is  3 64 a 3 : 3 27b 6  4a : 3b 2 ii) The sub-triplicate ratio of a2 : b2 is 2

 3 a2 : 3 b2  a 3 : b

2

1.8  1.8  2.7 1.2 ii) Let the third proportional be x. Then 225 : 75 : 75 : x  x

75  75  25 225 28. Let it be x. Then 6 : 8 : : 9 : x  8 × 9 = 6x  x = 12 29. Let a, b, c are in continued proportion, then given a = 16, c = 4, b = ? (a : b = b : c  b2 = ac)  x

 b  ac  16  4  64  8 Therefore the mean proportional is 8. 30. Let a, b, c are in continued proportion, then given a = 16, b = 8; c = ?

a : b : : b : c or ac = b or c  2

b2 82 8  8   4 a 16 16

c4.  3rd proportional = 4 31. As q is the mean proportional between p and r,  q2 = pr

 q  r  p  q   q2      2  2  pq + q 2 + pr + qr 4  4q2 = q2 + pq + qr + rp  3q2 = pq + qr + rp  3pr = pq + qr + rp  pq + qr = 2pr Dividing throughout by pqr we have,  q2 

3

1 1 24. The reciprocal ratio a : b is : or b : a a b 1 1 The reciprocal ratio of 4 :  : 4 9 9 The reciprocal ratio of www.betoppers.com

25. Let the mean proportional be x. Then, x2 = 13 × 637  x = 91 26. Let it be x. Then 16 : 4 : : 4 : x  16 = 16 x  x=1 27. i) Let the third proportional be x. Then 1.2 : 1.8 :: 1.8 : x

a b b a :  : 3 5 5 3

1 1 2 pq qr 2pr    + = p r q pqr pqr pqr Hence proved. 32. Let r be the mean proportional We know that if ‘q’ is the mean proportional of p and r then

p q = . q r

Ratio, proportions and Variations Solutions x2

Similarly,

231 2 xy a 2y 35. Given , a    x+ y x x y Applying componendo and dividendo, we get

r 4 pq  p r qy 2

2

a  x 2y  x  y 3y  x   a  x 2y  x  y yx

2

x x p x r   2  r2  2 2  r  2qy 4 pq qy 4q y 2

 The mean proportional 

a 2x  y x y Applying componendo and dividendo, we get

x 2qy

a  y 2 x  x  y 3x  y __________   (2) a  y 2x  x  y x y Subtracting (1) from (2), we get

6a  7b 6c  7 d  6a  7b 6c  7d By componendo and dividendo, we get 6a  7b  6a  7b 6c  7 d  6c  7 d  6a  7b  6a  7b 6c  7 d  6c  7 d

12a 12c a c     a:b  c:d 14b 14d b d a, b, c, d are in proportion.  34. Let the quantity added to the ratio (p + q) : (p – q) be x. 



 3 y  x  3x  y a x a y     ax a y  yx  x y



  3 y  x  x  y    3x  y  y  x   y  x  x  y 



3xy  3 y 2  x2  xy  3xy  3 x2  y 2  xy xy  y 2  x 2  xy



4  x2  y 2  4  x  y  4 y2  4x2   2  x 2  2 xy  y 2 x y  x  y



2

p+ q+ x  p  q Then p - q + x  2  p  q

36.

Applying componendo and dividendo, we have 2

2

p + q + x+ p - q + x  p  q    p  q   p + q + x - p + q - x  p  q 2   p  q  2

a  x a  y 4 x  y    ax a y x y

p q r 2 p  3q  5r    m 2 3 4 k  p = 2m ; q = 3m ; r = 4m 2 p  3q  5r m k

2 2 2  p  x 2 p  q  p  x p2  q 2     2q 4 pq q 2 pq



2  2m   3  3m   5  4m  m k

by cross-multiplying, we get 2qp(p + x) = q(p2 + q2) 2p(p + x) = p2 + q2 2p2 + 2px = p2 + q2 2px = q2 – p2



m  4  9  20  m k



m 15   m  k  15 k

q2 - p2 x= 2p



q2  p2 must be added to the terms of ratio 2p

(p + q) : (p – q) to make it equal to (p + q)2 : (p – q)2.

(1)

Also,

6a  7b 6a  7b  33. Given, 6c  7d 6c  7 d Let us bring a, b terms to L.H.S. and c, d terms to R.H.S. By alternendo, we get



__________

 k  15 . 4 p  9q 4 p  9q  4r  9s 4r  9s Let us bring p, q terms to L.H.S and r, s terms to R.H.S.

37. Given,

By alternendo, we get

4 p  9q 4r  9 s  4 p  9q 4 r  9 s www.betoppers.com

8th Class Mathematics

232 By componendo and dividendo, we get 4 p  9 q  4 p  9 q 4r  9 s  4 r  9 s  4 p  9 q  4 p  9 q 4r  9 s  4 r  9 s 

8p 8r p r    18q 18s q s

38. Given, p : q = r : s 

p r  c q s

i.e., p = qc; r = sc Substituting the value of p and r in L.H.S

p  q qc  q  q  k  1  q  s  k  1 s r  s sc  s

we get

R.H.S 



q c2  1 2

p 2  q2 r 2  s2 



q 2 c2  q 2 s 2c2  s 2

q s

s c 1 Since L.H.S = R.H.S p+q:r+s 

39. We know that

p2  q 2 : r 2  s 2

a c e = = b d f

a c e = = = k (say). b d f Then a = bk, c = dk and e = fk  Let

Then

Also

a + c - e bk + dk - fk k(b+ d - f) = = =k b+ d - f b+ d - f (b + d - f)

41. Let the sums of money in the bags be Rs. x, Rs. y, Rs. z respectively. As these are on the ratio 4 : 3: 2, we get x = 4k, y = 3k, z = 2k. Rs. 50 is added to each bag. So, the sums are now 4k + 50, 3k + 50, 2k + 50 respectively. But these are in the ratio 14 : 13 : 12.  4k + 50 = 14p ___________ (1) 3k + 50 = 13p ___________ (2) 2k + 50 = 12p ___________ (3) Subtracting equation (2) from equation (1) we get, k = p  4k + 50 = 14p  4k +50 = 14k  10k = 50  k = 5 ( p = k)  The sums in the bags are, Rs 4 × 5, Rs. 3 × 5, Rs. 2 × 5 i.e., Rs. 20, Rs. 15, Rs, 10. 42. a, b, c, d are in continued proportion

a b c    k  say  b c d Then, c = dk, b = ck = (dk) (k) = dk2 a = bk = (dk2) k = dk3 

3





a c = = k (say) b d then a = bk, c = dk 

2 2 2 2 ab + cd bk.b+ dk.d k (b  d ) b  d   = 2 2 2 2 ab - cd bk.b - dk.d k (b  d ) b  d

a 2 c2  k2 k2 a 2  c2 = a2 c 2 = 2 a  c2  k2 k2

Hence it is proved. www.betoppers.com

3

md 3 k 9  nd 3 k 6  rd 3 k 3 md 3 k 6  nd 3 k 3  rd 3 d 3 k 3  mk 6  nk 3  r 



40. If

3

3 2 ma 3  nb3  rc 3 m  dk   n  dk   r  dk   3 3 mb3  nc3  rd 3 m  dk 2   n  dk   rd 3

a + c + e bk + dk + fk k(b+ d + f) = = =k b+ d + f b+ d + f (b + d + f)

a+c - e a+c+e = b+ d - f b+ d + f Hence it is proved.

a dk 3   k3 d d



d 3  mk 6  nk 3  r 

 k3

a ma 3  nb3  rc 3  d mb3  nc 3  rd 3  a : d : :(ma3 + nb3 – rcc3) : (mb3 + nc3 – rd3 )



x y  k . 5 8 Then, x = 5k and y = 8k.

43. Let



x  5 5k  5 5  k  1 5    y  8 8k  8 8  k  1 8

 (x + 5) : (y + 8) = 5 : 8.

Ratio, proportions and Variations Solutions

233

44: Given that x  y .  x = ky –––––––– (1). i.e. = k =

k

x . ; given x = 18, y = 7 y

18  21 ( By Substituting k & y in (1)) 7

8  21  54 . 7  x = 54 when y = 21 45. Given that x  y ,  x = ky ( Where k is a constant) –––– (1)

k

x y

2 3

i.e., 3a + 7b = k(3a + 13b) ––––––– (1) where k is a constant 3(5) + 7(3) = k(3(5)+13(3)) 15 + 21 = k(15 + 39)

 k

36 = k(54)

2 3

2 3

 3a  7b    3a  13b  ( By substituting k 

(  x = 2, y = 3 given )

From (1), we get, y = y

1 when y = 5. 5

48. Given that,  3a  7b    3a  13b 

x

k=

1 ( by substituting y = 5, k = 1) 5

 x

18 ( By Substitution) 7  x

x

x ––––––– (2) k

18 54   27  2 / 3 2

2 ) 3

 y = 27 when x = 18. 46. Given that x 

1 y

1 x  k   –––– (1) (where k is a constant) y  k = xy  k = 4 × 15 = 60 ( x = 15, y = 4 given)

From (1), we get, y =

k ––––––– (2) x

60  10 (by substituting k of x = 6) 6  y = 10 when x = 6. y

1 1  y  k   –––––––– (1) x x where k is a constant k = (x) (y) ( from (1)) k = (1) (1)  k = 1.

47. Given that y 

From (1), we get x =

9a + 21b = 6a + 26b 3a – 5b = 0  3a = 5b. 49. Given x 

y

( by substituting x = 18, k 

k ––––––– (2) y

2 ) 3

1 1  x  k1   ––––––––– (1) y  y

1 1  y  k 2   ––––––––– (2) z z

dividing (1) by (2), we get

x  k1  1/ y  x  k   z     1  y  k2  1/ z  y  k2   y  k  x   1  z  x  z where  k1  is a constant.  k2   k2   z  x hence proved. 50. Given that

a3b, i.e. a1/2 = kb1/3 ––––––––– (1) ( where k is a constant) 1/ 2 41/ 2 2 From (1), k  a  k   1 b1/ 3 81/ 3 2  k = 1. a1/2 = (1)b1/3 ( by substituting in (1)) Squaring on both the sides, we get a = b2/3  again cubing on both sides, we get a3 = b2.  Equation between a & b is a3 = b2.

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8th Class Mathematics

234

SUMM ATIVE WORKSHEET 1.

a b c   k, 3 4 7 then, a = 3k, b = 4k and c = 7k 

2.

4.

Let the number to be added be x.

7 x 2  13  x 3  21 + 3x = 26 + 2x  x=5 Let the numbers be 3x and 4x. We know that product of two numbers = L.C.M.  H.C.F.. Here L.C.M.=180 and H.C.F. is clearly x  180 × x = 3x × 4x  x = 15  The numbers are 45 and 60.  Smallest number is 45.

504  42Kmph 12  Required ratio = 56 : 42 = 4 : 3 p r   k  p  qk , r  sk q s

Considering LHS and substituting ‘p’ by ‘qk’. 2 p  3q 2 qk  3q q  2k  3  2k  3    –(1) 2 p  3q 2qk  3q q  2 k  3  2 k  3

Considering R.H.S, and substituting ‘r’ by ‘sk’. 2r  3q 2 sk  3q s  2k  3 2k  3    2r  3q 2sk  3q s  2k  3 2k  3 –(2)

From (1) and (2), LHS = RHS 2 p  3q 2r  3q  . 2 p  3q 2r  3q To find ratio, both the terms should be of same units.

 6.

Ratio of length to width

10m 2.4cm

As 1m = 100cm.

 Ratio of length to width

10000 1250  24 3

Thus, the required ratio = length : width = 1250:3 www.betoppers.com

1 1 Given, a : b  6 : 4 2 2

13 9 :  13 : 9 ______________ (1) 2 2 b : c = 8.5 : 4.5 

85 45 :  85 : 45  17 : 9 ___________ (2) 10 10 Let us make the magnitude of b equal in the two ratios. The value of b is ‘9’ in (1) and 17 in (2) and L.C.M of 9 and 17 is 153. To make b = 153, we multiply the first ratio by 17 and the second ratio by 9. a : b = 13 : 9 = 13 × 17 : 9 × 17 = 221 : 153 b : c = 17 : 9 = 17 × 9 : 9 × 9 = 153 : 81  a : c = 221 : 81. Duplicate ratio of 3 : 5 is 32 : 52 Given (3a + 3) : (9a + 7) is the duplicate ratio of 3 : 5. 

448  56Kmph 8

The speed of the second train =

Let

8.

Total distance covered Timetaken

 The speed of the first train =

5.

312 78  4   4 x 4 78 78 The numbers are 6x = 6 × 4 = 24 13x = 13 × 4 = 52 Hence, the two numbers are 24 and 52.

a +b + c 3k  4k  7k 14   2 c 7k 7

Speed =

Let the two numbers be 6x and 13x. Now, L.C.M of 6x and 13x is 78x. Given, 78x = 312

x

Then, 3.

7.

9.

3a  3 9 3a  3 32   2  9a  7 25 9a  7 5  25 (3a + 3) = 9 (9a + 7)  75a + 75 = 81a + 63  81a – 75a = 75 – 63  6a = 12  a = 2 10. Let the numbers be denoted by 5x and 8x. 

5x  9 8  8 x  9 11 Hence, the numbers are 15 and 24. 11. (a + 4) : (a + 12) = (a –1) : (a + 5) Then,

a  4 a 1  a  12 a  5  (a + 4) (a + 5) = (a – 1) (a + 12)  a2 + 5a + 4a + 20 = a2 + 12a – a – 12  9a + 20 = 111a – 12  2a = 32  a = 16.  a = 16. 

Ratio, proportions and Variations Solutions

235

2

A Ax 12. By the given condition,    B  B x  2 2  B(A + x) = A(B + x) A2B + 2ABx + Bx2 = AB2 + 2ABx + Ax2 x2 (A – B) = AB (A – B);  x2 = AB since A – B is not zero. 13. If p, q, r are in continued proportion p q we have, = = k q r Then p = qk ; q = rk p = qk = rk × k = rk2 q = rk L.H.S. is (p + q + r) (p – q + r) = (rk2 + rk + r) (rk2 – rk + r) = r2 (r2 + k + 1) (k2 – k + 1) = r2 [k4 – k3 + k2 + k3 – k2 + k + k2 – k + 1] = r2 [k4 + k2 + 1) ________ (1) R.H.S is r2 + q2 + r2 = (rk2)2 + (rk)2 + r2 = r2k4 + r2k2 + r2 = r2(k4 + k2 + 1) _________ (2) From (1) and (2) we get (p + q + r) (p – q + r) = p2 + q2 + r2 Hence proved.

1 14. Given, p  4ab multiplying both sides with 2a a b 

p 4ab 2b   2a 2a  a +b  a +b

applying componendo and dividendo, to

p 2b  we have, 2a a  b p  2a 2b  a  b 3b  a   p  2a 2b   a  b  b  a –––––––(1)

p 2a  . 2b a  b Applying componendo and dividendo, we have, Similarly,

p  2b 2a   a  b  3a  b   p  2b 2a   a  b   a  b  –––––––(2)

adding (1) and (2), we get p  2a p  2b 3b  a 3a  b    p  2a p  2b b  a a b



3b  a  3a  b 2a  2b 2  a  b    2 ab  a  b a  b

15.

y 2  xy  y 2  xy

a  1

y 2  xy  y 2  xy

By componendo and dividendo, we get, a 1  a 1

 

   xy   

y 2  xy  y 2  xy  2

y  xy  y

2

2

y  xy  y

a  1 2 y 2  xy a 1    a  1 2 y 2  xy a 1



  xy 

y 2  xy  y 2  xy 2

y 2  xy y 2  xy

Squaring both the sides, we get 2

 a  1  y 2  xy 2 .  a  1 y 2  xy



a 2  2a  1 y 2  xy  a 2  2a  1 y 2  xy

Again applying componendo and dividendo, we have,

a a

2

 2a  1   a 2  2a  1

2

 2a  1   a 2  2a  1

y  y 

2

 xy    y 2  xy 

2

 xy    y 2  xy 

2  a 2  1 y 2 2a 2  2 2 y 2    4a 2 xy 4a xy

a2  1 y   x  a 2  1  2ay 2a x 2  xa – 2ay + x = 0 16. We know that each of the given fractions 



sum of the numerators sum of the denominators

x yz q  r  p  r  p  q   p  q  r  x yz ––––––––– (1) pqr Now, multiplying and dividing each fraction by (y + z), (z + x) and (x + y), respectively, we get 

x y  z x yz   ;  q  r  p  y  z  y  z  q  r  p  y  z  x y  z  x   ;  r  p  q   z  x   z  x  r  p  q 

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8th Class Mathematics

236 zx  y z  x  y   p  q  r  x  y   p  q  r  x  y  ––––(2)

18. Let c be the third proportional to 5  2 3 and

37  20 3 .

From (2) 5 2 3

sum of the numerators  sum of the deno min ators



x y  z y  z  x    y  z  q  r  p   z  x  r  p  q 

 37  20 3  c

37  20 3



37  20 3 c 2

5 2 3



2569  1480 3 5 2 3



z x  y  x  y  p  q  r 



 2569  1480 3  5  2 3   5  2 3  5  2 3 



x y  z  y  z  x  z  x  y ––––– (3) 2  px  qy  rz 



12845  5138 3  7400 3  8880 25  12



3965  2262 3 3965 2262 3   13 13 13

From (1) and (3) x  y  z x y  z  y z  x  z  x  y  pqr 2  px  qy  rz 

 305  174 3 a 2  a  7  a 1 1 a  7  a 1

17. Given

a 



a  7  a 1

Hence, the third proportional is 305  174 3 .

x y yz zx 19. Given px  qy  py  qz  pz  qx

  2

by adding all the terms a  7  a 1

2  a  7  a 1 a

1 a  7  a 1 Applying componendo and dividendo, we get, a  7  a 1  a  7  a 1 a  7  a  1  a  7  a 1 2  a   a  7  2 a  1 2  a  a 1 Squaring both the sides, we get 

2 a7



 x  y   y  z    z  x  px  qy    py  qz    pz  qx 

2a 2a



 2a    2a





If

2 x  y  z  p x  y  z  q x  y  z 2 x  y  z 

 x  y  z  p  q   x  y  z  0

___________

(1)

then equation (1) becomes

2

a  7 4  a  4a  a  1 4  a 2  4a Again, applying componendo and dividendo we get,

a  7  a  1 4  a 2  4a  4  a 2  4a  a  7  a  1 4  a 2  4a  4  a 2  4a 



2 2a  6 2 4  a  8 8a

 a3 a

4  a2 a

4 3

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

2 pq Hence proved. 20. Since a, b, c are three proportionals,

a b  or b2 = ac b c  (b2 + bc + c2) (ac – bc + c2) = (b2 + bc + c2) (a2 – bc + c2) (  ac = b2) = [(b2 + c2) + bc][(b2 + c2) – bc] = (b2 + c2)2 – b2c2 = b4 + 2b2c2 + c4– b2c2 = b4 + c4 + b2c2 = b4 + c4 + (ac)c2 (  b2 = ac) = b4 + c4 + ac3 Hence it is proved. We have,



2  a  3 8



 a 2  3a  4  a 2 a 

4 3



2 4  a2 8a



Ratio, proportions and Variations Solutions 21.

237

a c e = = = k (say) b d f

 120 – y +

Then, a = bk, c = dk and e = fk

5a  7c  3e 5bk  7 dk  3 fk  5b  7 d  3 f 5b  7d  3 f 

3.

1862  49  x = 7 38  The numbers are 21,14 and 35. 2  38x2 = 1862  x 

A A B C 3 8 15 15 5         D B C D 4 9 16 24 8 24. Clearly, men and time taken are inversely proportional.  30 × 27 = 18 × x

4.

23.

x 2   3 13  y  18 x 3   4  y

30  27  45 days. 18 But the work is thrice as great, hence 18 men would take 3 × 45 = 135 days.

 2x   3x   18   3  13   4  y  y 

p q r = = b-c c-a a-b we have to show that p + q + r = 0. p q r = = =k Let b-c c- a a -b Then p = (b – c)k, q = (c – a)k, r = (a – b)k p + q + r = (b – c)k + (c – a)k + (a – b)k = k[b – c + c – a + a – b] = k . 0 = 0 Hence it is proved.

HOTS WORKSHEET 1.

2.

Let the number of apples be 5x and the number of mangoes be 4x. Given the number of apples is 210.  5 x  210  x  42 The number of mangoes in basket is  4x = 4 × 42 = 168. Let the rupee coins be x and the ten paise coins be y. Then x + y = 120 x = 120 – y

10 y Also, x   y  57  x   57 100 10

2 x  3 y 13  3 x  4 y 18 Dividing the numerator and the denominator by y, we get

 x

25.

y 9y = 63  = 63 10 10  y = 70  x = 120 – 70 = 50  The number of one rupee coins in the purse are 50. Let male be 5x and female be 3x Now 5x – 3x = 40  x = 20  Total number of workers = 5x + 3x = 8x = 8 × 20 = 160  y–

k (5b  7 d  3 f ) c k (5b  7d  3 f ) d

Hence it is proved. 22. Let the numbers be 3x, 2x and 5x. Then (3x)2 + (2x)2 + (5x)2 = 1862

y = 57 (  x = 120 – y) 10

5.



36 x 39 x  54   52 y y



39 x 36 x 3x   54  52  2 y y y



x 2   x: y  2:3 y 3

Duplicate ratio of p : q is p2 : q2 Given (p – x) : (q – x) is the duplicate ratio of p : q. 

p  x p2  q2 (p – x) = p2 (q – x) q  x q2 

    

pq2 – q2x = p2q – p2x p2x – q2x = p2q – pq2 (p2 – q2)x = pq (p – q) (p + q) (p – q) x = pq (p – q) (p + q) x = pq



pq 1 p q 1 1 1 1        pq x pq pq x x p q www.betoppers.com

8th Class Mathematics

238 6.

7.

p q r   k b c c  a a b i.e., p = (b – c)k ; q = (c – a)k ; r = (a – b)k  ap + bq + cr = a(b – c) k + b (c – a) k + c(a – b)k = k[ab – ac + bc – ab + ac – bc] = k (0) = 0  ap + bq + cr = 0 Also p + q + r = (b – c)k + (c – a) k + (a – b)k = k(b – c + c – a + a – b) = k(0) = 0 Let

Given





ay  bx cx  az bz  cy = = c b a

c  ay  bx  c

2

c  ay  bx 

=

b  cx  az  b

2

b  cx  az  2

=

=

a  bz  cy  a2

12 ____________ (1) k2 Also, b + c = 30  ek3 + ek2 = 30 [ c = ek2] Substituting (1) in (2), we get  e

c  ay  bx  +b  cx  az  + a  bz  cy 

_____

(2)

12 3 12 2  k  2  k  30 k2 k  12k + 12 = 30  12k = 30 – 12

 12k = 18  k  Now, e 

a  bz  cy 

c b a2 Adding numerators and denominators 

2

=

 e2k4 = 144  ek 2  144  12

d  ek 

18 3 3  k 12 2 2

12 12 4 16   12   k2 9/ 4 9 3

16 3  8  d = 8 3 2

c  ek 2 

16 9   12  c = 12 3 4

b  ek 3 

16  3      18  b = 18 3 2

c 2 +b 2 + a 2

3

acy  bcx + bcx  abz + abz  acy  c 2 + b2 + a 2 

4

0 0 2 c + b2 + a 2

16  3  a  ek      27  a = 27 3 2 4

ay  bx cx  az bz  cy = = 0 c b a  ay – bx = 0; cx – az = 0; bz – cy = 0  ay = bx ; cx = az ; bz = cy 

 8. 9.

x y x z y z = ; = ; = a b a c b c



x y z = = a b c

2a 6ab c 4a    3b 5c 2 a 5c Let the five numbers be a, b, c, d and e. Given that they are in continued proportion. The required ratio 

a b c d Then, = = = = k (say) b c d e Let us convert a, b, c and d in terms of e we have d = ek, c = dk = ek2, b = ck = ek2 k = ek3  b = ek3 a = bk = ek3k = ek4 Given that the product of first and the fifth numbers is 144. i.e., a × e = 144 Also the sum of second and the third number is 30, i.e., b + c = 30 ae = 144  (ek4)e = 144 www.betoppers.com

 The five numbers are 27, 18, 12, 8 and 16/3. 10. i) As p, q, r, s are in continued proportion,

we have

p q r = = =k q r s

Then p = qk; q = rk; r = sk i.e., p = qk = rk × k = sk × k2 = sk3 q = rk = sk × k = sk2 r = sk L.H.S is p : q + s Substituting the values of p and q, we have

p sk 3 k3  2  2 qs sk  s k  1 3 2 3 R.H.S is r : r s + s Substituting the value of r, we have p:q  s 

3

 sk  r3  r :rs+s  2 3 2 r ss  sk  s  s3 3

2

3

s3k 3 k 3 ___________  (2) s3k 2  s 3 k 2  1 From (1) and (2) we have p : q + s = r3 : r2 s + s3 

_____

(1)

Ratio, proportions and Variations Solutions ii)

As p, q, r, s are in continued proportion, we have 

p q r = = =k q r s Then, p = qk, p = rk, r = sk i.e., p = qk = rk × k = sk × k2 = sk3 q = rk = sk × k = sk2 r = sk L.H.S is (p2 + q2 + r2) (q2 + r2 + s2) Substituting the values of p, q and r, we have (p2 + q2 + r2) (q2 + r2 + s2) = [(sk3)2 + (sk2)2 + (sk)2] × [(sk2)2 + (sk)2 + s2] = (s2k6 + s2k4 + s2k2) (s2k4 + s2k2 + s2) = s2k2 (k4 + k2 + 1) s2 (k4 + k2 + 1) = s4k2 (k4 + k2 + 1)2 ___________ (1) R.H.S = (pq + qr + rs)2 = [(sk3) (sk2) + (sk2) (sk) + (sk) (s)]2 = [s2k5 + s2k3 + s2k]2 = s4k2 [k4 + k2 + 1]2 _________ (2) From (1) and (2) we have, (p2 + q2 + r2) (q2 + r2 + s2) = (pq + qr + rs)2 11.

239



k 3b 4  2d 2 k 2 fk  3bkk 2 f 2 f b 4  2d 2 f  3bf 3 k 3  b 4  2d 2 f  3bf 3 

b

4

 2d 2 f  3bf 3 

a c e ace = k3 = k × k × k = b × d × f = bdf Hence, proved. 13. Step1: Using concept of equal ratios.  a c e a + c + e + .....   i.e., = = =  , we have b d f b + d + f + .....  

q+ r  p r + p  q p+ q  r = = y+ z  x z+ x  y x+ y  z 

q + r  p+ r + p  q+  p + q  r   y + z  x +  z + x  y +  x + y  z

p q r = = =k q+r - p r+ p - q p+q - r



p+q+r x + y + z –––––––––– (1)

 p = k(q + r – p) ; q = k(r + p – q); r = k (p + q – r) –––––––––––– (A) Now p + q + r = k[(q + r – p) + (r + p – q) + (p + q – r)] = k[q + r – p + r + p – q + p + q – r] or p + q + r = k (p + q + r)  k = 1 Using k = 1 in (A), we have p = q + r – p or 2p = q + r. Adding p on both the sides,  p + q + r = 3p –––––––– (1) 2q = r + p, adding q on both the sides,  p + q + r = 3q –––––––– (2) 2r = p + q, adding r on both the sides,  p + q + r = 3r –––––––– (3) L.H.S. of (1), (2), (3) are same.  R.H.S. is same  3p = 3q = 3r  p = q = r.

q+r  p p+q+r Step2: Now, y + z  x  x + y + z

a c e 12. Let = = = k b d f

3

 bk  

Applying dividendo to both the sides,

q + r  p   p+ q + r  p+q+r 

 y  z  x   x  y  z



 2

b  2  dk  fk  3bk  fk  f b4  2d 2 f  3bf 3

q+r  p y+ z  x  p+q+r x+ y+ z

xyz

2 p 2 x  p+q+r x+ y+ z 2 p p  q  r p pqr    2 x x  y  z x xyz

q r p+ q+r Similarly , y  z  x + y + z

a3b  2c 2 e  3ae 2 f , b 4  2d 2 f  3bf 3 2





so that a = bk; c = dk; e = fk L.H.S. 

By applying alternendo

p+q+r p q r    x + y + z x y z –––––––– (2)

Equating (1) and (2), we have www.betoppers.com

8th Class Mathematics

240

q+ r  p r + p  q p+ q  r = = y+ z  x z+ x  y x+ y  z =

p q r p+ q+ r = = = x y z x+ y+ z 1

14.

y z q r   rc  pa  qb pa  qb  rc y z + q r  rc  pa  qb  pa  qb  rc

1

y  a  1 3   a  1 3  1  a  1 13   a  1 13

y 1 By componendo and dividendo we get, y 1 1

1

1

1

 a  1 3   a  1 3   a  1 3   a  1 3 1 1 1 1  a  1 3   a  1 3   a  1 3   a  1 3 2  a  1

1

Cubing both sides,

 y  13 3  y  1



a 1 y3  1  3y2  3y a  1  3  a 1 y  1  3y2  3y a  1

Again applying componendo and dividendo we get y3  1  3 y2  3 y  y3  1  3y 2  3y y 3  1  3 y 2  3 y  y3  1  3 y 2  3 y



x y p q Similarly,  qb + rc  pa rc + pa  qb

1

 a  1 3   1 1 2  a  1 3  a  1 3 3

y z + q r  ry  qz    2 pa qr  2 pa

a 1  a 1 a 1  a 1

2 y 3  6 y 2a y3  3y   a  a 2 2  6 y2 1  3y2

 y3 + 3y = a + 3ay2  y3 – 3ay2 + 3y – a = 0 x y 15. Given p qb + rc - pa = q rc + pa - qb     z r  pa + qb - rc 

x y + p q  qb + rc - pa + rc + pa - qb x y + p q qx  yp   2rc pq  2rc x z p r Similarly,   qb + rc - pa pa + qb - rc x z + rx + zp p r   2qb pr× 2qb

ry + qz qx+ yp rx + zp = = a c b Multiply the first of these fractions above and below by ‘x’, second by ‘z’, and the third by ‘y’ then 

rxy + qxz ryx + pyz qzx+ ypz = = ax by cz

consider



y z  q  rc  pa  qb  r  pa  qb  rc  . Dividing the

rxy + pyz + qxz + pyz - rxy - qxz by + cz - ax



2 pyz by + cz - ax

numerator and the denominator of LHS by q, and the numerator and the denominator of RHS by r we get,

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Similarly, we also get

2qxz 2rxy and cz + ax - by ax +by - cz

Ratio, proportions and Variations Solutions



241

2 pyz 2qxz 2rxy = = by + cz - ax cz + ax - by ax + by - cz



Dividing the numerator by xyz. p q = x  by + cz - ax  y  cz + ax - by 

d 3 k 6  2k 3  3  d 3 k 6  3k 3  4 



2k 3  3 3k 3  4 –––––––– (2)

(2a + 3a) : (3a – 4d) = (2a3 + 2b3) : (3a3 – 4b2) Hence proved. 5.

r  z  ax + by - cz 





Mean proportion between 6  3 3 and

 6  4 3  is  6  3 3 8  4 3 

Hence proved

 48  24 3  24 3  12  3

IIT JEE WORKSHEET 1.

Given

a c =  ad = bc –––––––– (1) b d

 48  36  12  4  3  2 3 6.

b d and x = y

 by = dx ––––––––– (2) Now, ad : dx = bc : by 

2. 3.

7.

ad bc = dx by



i.e. a : x = c : y The mean proportion of 16 and 4 is

16  4  64  8 . Let the third proportional to 16 and 8 be x then 16 : 8 :: 8 : x  16 × x = 8 × 8 64  x 4 16

4.

x  4 x 1  x  12 x  5  (x + 4) (x + 5) = (x + 12) (x – 1)  x2 + 9x + 20 = x2 + 111x – 12  9x – 111x = –12 – 20  –2x = –32  x = 16 p + r = 2q ––––––– (1) Given

 1 1  r    2 q s

 r

a b c = = =k b c d i.e., a = bk, b = ck, c = dk  a = bk = (ck) k = ck2 = (dk).k2 = dk3 b = ck = (dk)k = dk2



d  3k 3  4 

2k 3  3 3k 3  4 ––––––––(1)

R.H.S 3

3

3 2 2a 3  3b3 2  dk   3  dk     3a  4b3 3  dk 3 3  4  dk 2 3



2qs –––––––– (3) qs

2qs  2q qs dividing by ‘q’



p 2qs  2 q qq  s



p 2q p 2s 2q  2 s  2s    2  q q s q qs qs

2a + 3d 2dk + 3d = 3a  4d 3dk 3  4d 

1 1    q s



 p

3

d  2k 3  3

2

Now substitute (3) in (1) we get

Let

L.H.S. 

1 1 2   ––––––––– (2) q s r

2d 3k 9  3d 3k 6 3d 3k 9  4d 3 k 6

p pr  [ From (1) 2q = p + r] q qs   pq + ps = pq + rqq  ps = rqq 

p r =  p:q=r:s q s Hence proved. 

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8th Class Mathematics

242 8.

Since b is the mean proportional between a and c we have b2 = ac

a 2  ac + c 2 a 2  b2 + c2 a 2  ac+ c 2 = 2 = 1 1 1 1 1 1 a  ac + c 2  +  + a2 b2 c 2 a 2 ac c 2 a 2c 2

a

2

 ac + c 2  a 2c 2

a 9.

2

 ac + c 2 

a2c2 = (ac)2 = (b2)2 = b4 Let the required number be x then 6 + x, 10 + x, 14 + x, 22 + x, are proportional

6  x 14  x  10  x 22  x  (6 + x)(22 + x) = (14 + x)(10 + x)  132 + 28x + x2 = 140 + 24x + x2  132 – 140 = 24x – 28x  – 8 = – 4x 

 x

8 2 4

10. Mean proportional between









27  18 and



27  18 is



27  18





27  18

2

27

   18 



a  b

a

2

 27  18  9  3

2

 b2  2

 x

a  b  a  b  a  b  a  b  a  b

 27 × x = 9 × 9  x 

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30  30  x = 75 12 mean proportional to 9 and 25 is  x

 3  3  5  5  3  5  15  Ratio of the third proportion to mean proportion is 75 : 15 = 5 : 1 15. (a + b + c) (a – b + c) = a2 + b2 + c2  (a + b + c) (a – c – b) = a2 + b2 + c2  (a + c)2 – b2 = a2 + b2 + c2  a2 + c2 + 2ac – b2 = a2 + b2 +c2 a b = b c a, b, c are in continued proportion

 2ac = 2b2  ac = b.b 

 

12. 27 : 9 :: 9 : x

 x3

 a  c  15876 a – c = 126 ––––––– (4) Solving (2) and (4) a + c = 130 a – c = 126 ––––––––––––––––––  2a = 256  a = 128  From (4) a – c = 126  128 – c = 126  c=2  The three numbers are 128, 16, 2 14. Third proportional to 12 and 30 is 12 : 30 : : 30 : x 12 × x = 30 × 30

2

11. Let the third proportional be x then (a2 – b2): (a + b) : : (a + b) : x  (a2 – b2) × x = (a + b) (a + b)

 x

a b   b 2  ac ––––––––– (1) b c Given middle number i.e, b = 16. and sum of other two numbers = 130 i.e a + c = 130 ––––––––– (2) From (1) we have (16)2 = a × c a × c = 256 ––––––––– (3) Now (a – c)2 = (a + c)2 – 4ac  (a – c)2 = (130)2 – 4 × 256 ( from (2) and (3) a + c = 130, ac = 256)  (a – c)2 = 16900 – 1024  (a – c)2 = 15876 

a2  b2 + c2 L.H.S. = 2 2 2 a  b +c

=

13. Let a, b, c are in continued proportion

81 27

4. COMPARING QUANTITIES SOLUTIONS 5.

FORMATIVE WORKSHEET 1. (a) Ratio of the speed of cycle to the speed of

15  1: 2 30 (b) Since 1 km = 1000 m, scooter 

5m 5m   1: 2000 10km 10  1000m (c) Since Re 1 = 100 paise, Required ratio 

Required ratio 

6.

50paise 50paise   1:10 Rs5 500paise

2.

a

3:4 

2 100 2 200     100%  % 3 100 3 3

 30   50  lakh=15 lakh    100  Number of people who like other games

 66  3  2   % 3  

3.

28  25  7 100 Thus, 7 students are not good in mathematics. Let the total number of matches played by the team be x. It is given that the team won 10 matches and the winning percentage of the team was 40%.

7.

Therefore,

40  x  10 100

100  x = 25 40 Thus, the team played 25 matches. x  10 

 10    50  lakh =5 lakh  100  Let the original salary be x. It is given that the new salary is Rs 1,54,000. Original salary + Increment = New salary However, it is given that the increment is 10% of the original salary. Therefore, x 

mathematics  4.

100   25  x  Rs 600  x  Rs  600   25  100  Rs 2400 Thus, she had Rs 2400 in the beginning. Percentage of people who like other games= (100 – 60 – 30)% = (100 – 90)% = 10 % Total number of people= 50 lakh Therefore, number of people who like cricket  60    50  lakh  30 lakh  100  Number of people who like football

3 3 100 3 100     4 4 100 4 100

2  66 % 3 It is given that 72% of 25 students are good in mathematics. Therefore, Percentage of students who are not good in mathematics= (100 – 72)% = 28%  Number of students who are not good in

Let the amount of money which Chameli had in the beginning be x. It is given that after spending 75% of Rs x, she was left with Rs 600. Therefore, (100 –75)% of x = Rs 600 Or, 25 % of x = Rs 600

8.

10  x  154000 100

100  110x   154000  x  154000   110  100   x = 140000 Thus, the original salary was Rs 1,40,000. It is given that on Sunday, 845 people went to the zoo and on Monday, 169 people went. Decrease in the number of people= 845 –169 = 676 Percentage decrease  Decrease in the number of people × 100   %  Number of people who went to zoo on sunday 

8th Class Mathematics

244 9.

It is given that the shopkeeper buys 80 articles for Rs 2,400. Cost of one article =Rs

2400  Rs 30 80

Profit percent= 16

Profit Percent 

Proft Proft  100  16   100 C.P. Rs 30

 16  30  Profit  Rs    Rs 4.80  100  Selling price of one article = C.P. + Profit = Rs (30 + 4.80) = Rs 34.80 10. Total cost of an article= Cost + Overhead expenses = Rs 15500 + Rs 450 = Rs 15950 Profit%  15 

Profit  100 C.P.

Proft  100 Rs 15950

 15950  15  Profit  Rs    Rs 2392.50  100   Selling price of the article= C.P. + Profit = Rs (15950 + 2392.50) = Rs 18342.50 11. C.P. of a VCR= Rs 8000 The shopkeeper made a loss of 4 % on VCR. This means if C.P. is Rs 100, then S.P. is Rs 96.  96   8000  = When C.P. is Rs 8000, S.P. = Rs   100  Rs 7680 C.P. of a TV = Rs 8000 The shopkeeper made a profit of 8 % on TV. This means that if C.P. is Rs 100, then S.P. is Rs 108.  108   8000  = When C.P. is Rs 8000, S.P. =Rs   100  Rs 8640 Total S.P. = Rs 7680 + Rs 8640= Rs 16320 Total C.P. = Rs 8000 + Rs 8000= Rs 16000 Since total S.P.> total C.P., there was a profit. Profit = Rs 16320 “ Rs 16000 = Rs 320 Profit 320  100   100  2% Profit%  C.P. 16000 Therefore, the shopkeeper had a gain of 2% on the whole transaction. www.betoppers.com

12. Total marked price= Rs (1,450 + 2 × 850) = Rs (1,450 +1,700) = Rs 3,150 Given that, discount % = 10%

 10   3150   Rs 315 Discount =Rs   100  Rs 315= Rs 3150 “ Sale price  Sale price= Rs (3150 – 315)= Rs 2835 Thus, the customer will have to pay Rs 2,835. 13. S.P. of each buffalo= Rs 20000 The milkman made a gain of 5% while selling one buffalo. This means if C.P. is Rs 100, then S.P. is Rs 105. 100   C.P. of one buffalo = Rs  20000   = Rs 105   19,047.62 Also, the second buffalo was sold at a loss of 10%. This means if C.P. is Rs 100, then S.P. is Rs 90. 100    C.P. of other buffalo = Rs  20000  90    = Rs 22222.22 Total C.P. = Rs 19047.62 + Rs 22222.22 = Rs 41269.84 Total S.P. = Rs 20000 + Rs 20000 = Rs 40000 Loss= Rs 41269.84 “ Rs 40000 = Rs 1269.84 Thus, the overall loss of milkman was Rs 1,269.84. 14. On Rs 100, the tax to be paid= Rs 12 On Rs 13000, the tax to be paid will  12   13000  = Rs 1560 be  Rs   100  Required amount = Cost + Sales Tax = Rs 13000 + Rs 1560 = Rs 14560 Thus, Vinod will have to pay Rs 14,560 for the T.V. 15. Let the marked price be x.

Discount percent 

20 

Discount  100 Marked price

Discount  100 x

Discount 

20 1 x  x 100 5

Comparing Quantities Solutions Also, Discount = Marked price – Sale price

1 x  x  Rs 1600 5 1 x  x  Rs 1600 5 4 x  Rs 1600 5

5  x  Rs 1600    Rs 2000 4  Thus, the marked price was Rs 2000. 16. The price includes VAT. Thus, 8% VAT means that if the price without VAT is Rs 100, then price including VAT will be Rs 108. When price including VAT is Rs 108, original price = Rs 100 When price including VAT is Rs 5400, original price  100   Rs   5400  = Rs 5000  108  Thus, the price of the hair-dryer before the addition of VAT was Rs 5,000. 17. (a) Principal (P) = Rs 10, 800

1 25 Rate (R)  12 %  %  annual  2 2 Number of years (n) = 3 R   Amount, A = P  1    100 

n

3

 25     Rs 10800 1     200    3   225    Rs 10800     200   

225 225 225    Rs 10800×    200 200 200   = Rs 15377.34375 = Rs 15377.34 (approximately) C.I. = A – P = Rs (15377.34 – 10800) = Rs 4,577.34

245 (b) Principal (P) = Rs 18,000 Rate (R) = 10% annual

 Number of years (n) = 2 years 2 The amount for 2 years and 6 months can be calculated by first calculating the amount for 2 years using the compound interest formula, and then calculating the simple interest for 6 months on the amount obtained at the end of 2 years. Firstly, the amount for 2 years has to be calculated. 2  1   A  Rs 18000 1     10   

11 11    Rs 18000     Rs21780 10 10   By taking Rs 21780 as principal, the S.I. for the next    

1 year will be calculated. 2

1    21780  2  10  S.I  Rs    Rs 1089 100      Interest for the first 2 years= Rs (21780 – 18000) = Rs 3780

1 year= Rs 1089 2  Total C.I. = Rs 3780 + Rs 1089= Rs 4,869 A = P + C.I. = Rs 18000 + Rs 4869= Rs 22,869 (c) Principal (P)= Rs 62,500 Rate= 8% per annum or 4% per half year and interest for the next

Number of years  1

1 2

1 There will be 3 half years in 1 years. 2 3 n R   Rs 62500 1  4     A  P 1   100         100 

26 26 26    Rs  62500     = Rs 70304 25 25 25   C.I. = A – P= Rs 70304 – Rs 62500 = Rs 7,804

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8th Class Mathematics

246 (d) Principal (P) = Rs 8000 Rate of interest= 9% per annum

9 % per half year 2 Number of years= 1 year There will be 2 half years in 1 year. or

2 n R   Rs 8000 1+ 9     A  P 1   200         100  2   209    Rs 8000     Rs 8,736.20  200   

C.I. = A – P= Rs 8736.20 “ Rs 8000= Rs 736.20 (e) Principal (P)= Rs 10,000 Rate= 8% per annum or 4% per half year Number of years= 1 year There are 2 half years in 1 year. R   A  P 1    100 

n

2 2   4   1      Rs 10000 1+  Rs 10000 1      25    100       

26 26    Rs 10000×    Rs 10,816 25 25   C.I. = A – P= Rs 10816 – Rs 10000 = Rs 816 18. Principal (P)= Rs 26,400 Rate (R)= 15% per annum 4 years 12 The amount for 2 years and 4 months can be calculated by first calculating the amount for 2 years using the compound interest formula, and then calculating the simple interest for 4 months on the amount obtained at the end of 2 years. Firstly, the amount for 2 years has to be calculated. Number of years (n)  2

2  15    A  Rs 26400 1     100    2  3     Rs  26400 1     20   

23 23    Rs  26400     Rs 34,914 20 20   www.betoppers.com

By taking Rs 34,914 as principal, the S.I. for the next

1 years will be calculated. 3

1    34914  3  15  S.I  Rs    Rs 1,745,70 100    

Interest for the first two years = Rs (34914 – 26400) = Rs 8,514

1 year = Rs 3

And interest for the next

1,745.70 Total C.I. = Rs (8514 + Rs 1745.70) = Rs 10,259.70 Amount = P + C.I. = Rs 26400 + Rs 10259.70 = Rs 36,659.70 19. Interest paid by Fabina 

PR T 100

 12500  12  3   Rs    Rs 4,500 100   Amount paid by Radha at the end of 3 years R   A  P 1    100 

n

3  10    A  Rs 12500 1     100   

110 110 110    Rs 12500     100 100 100   = Rs 16, 637.50 C.I. = A – P = Rs 16637.50 – Rs 12500= Rs 4,137.50 The interest paid by Fabina is Rs 4,500 and by Radha is Rs 4,137.50. Thus, Fabina pays more interest. Rs 4500 – Rs 4137.50= Rs 362.50 Hence, Fabina will have to pay Rs 362.50 more. 20. P= Rs 12000 R= 6% per annum T= 2 years  12000  6  2  PR T  Rs   = Rs 1,440 100 100   To find the compound interest, the amount (A) has to be calculated. S.I. 

Comparing Quantities Solutions

247

2 n R   Rs 12000 1  6     A  P 1   100         100 

11    Rs  80000   = Rs 88,000 10   By taking Rs 88,000 as principal, the SI for the

2  3    Rs 12000 1     50   

next 

53 53    Rs 12000    50 50   = Rs 13,483.20  C.I. = A – P= Rs 13483.20 – Rs 12000 = Rs 1,483.20 C.I. – S.I. = Rs 1,483.20 – Rs 1,440 = Rs 43.20 Thus, the extra amount to be paid is Rs 43.20. 21. (i) P = Rs 60,000 Rate = 12% per annum = 6% per half year n = 6 months = 1 half year 1 n R   Rs 60000 1  6     A  P 1   100        100  

106    Rs  60000    Rs 63,600 100   (ii) There are 2 half years in 1 year. n = 2 2  6    A  Rs 60000 1     100   

106 106    Rs  60000    100 100  

 Rs 67, 416 22. (i) P = Rs 80,000 R = 10% per annum

1 n =  1 years 2 The amount for 1 year and 6 months can be calculated by first calculating the amount for 1 year using the compound interest formula, and then calculating the simple interest for 6 months on the amount obtained at the end of 1 year. Firstly, the amount for 1 year has to be calculated. 1  10      1  A  Rs 80000 1    Rs 80000 1+     100    10    

1 year will be calculated. 2

1   88000  10  2  P  R  T  Rs   = Rs 4,400 S.I.  100   100  

Interest for the first year= Rs 88000 “ Rs 80000 = Rs 8,000

1 year = Rs 4,400 2 Total C.I. = Rs 8000 + Rs 4,400 = Rs 1,2400 A = P + C.I. = Rs (80000 + 12400) = Rs 92,400 (ii) The interest is compounded half yearly. Rate = 10% per annum = 5% per half year And interest for the next

1 There will be three half years in 1 years. 2 3 3   5   1     A  Rs 80000 1   Rs 80000 1      20    100       

21 21 21    Rs  80000     = Rs 92,610 20 20 20   Difference between the amounts = Rs 92,610 – Rs 92,400 = Rs 210 23. (i) P = Rs 8,000 R = 5% per annum n = 2 years 2 2   5   1     A  Rs 8000 1+  Rs 8000 1      20    100       

21 21    Rs  8000     Rs 8,820 20 20   (ii) The interest for the next one year, i.e. the third year, has to be calculated. By taking Rs 8,820 as principal, the S.I. for the next year will be calculated.

 8820  5  1  S.I.  Rs    Rs 441 100   www.betoppers.com

8th Class Mathematics

248 24. P = Rs 10,000 Rate = 10% per annum = 5% per half year

1 n = 1 years 2 1 There will be 3 half years in 1 years. 2 3

 5    A = Rs 10000 1+    100    3  1    = Rs 10000 1+    20   

21 21 21    Rs 10000     = Rs 11, 576.25 20 20 20   C.I. = A – P = Rs 11576.25 – Rs 10000 = Rs 1,576.25 The amount for 1 year and 6 months can be calculated by first calculating the amount for 1 year using the compound interest formula, and then calculating the simple interest for 6 months on the amount obtained at the end of 1 year. The amount for the first year has to be calculated first. 1  10     1   A = Rs 10000 1+   Rs 10000 1      100    10    

11    Rs 10000   = Rs 11,000 10   By taking Rs 11,000 as the principal, the S.I. for the next

1 year will be calculated. 2

1   11000  10  2  S.I.  Rs    Rs 550 100      

Interest for the first year= Rs 11000 – Rs 10000 = Rs 1,000 Total compound interest= Rs 1000 + Rs 550 = Rs 1,550 Therefore, the interest would be more when compounded half yearly than the interest when compounded annually.

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25. P = Rs 4,096

1 25 % per half year R = 12 % per annum = 2 4 n = 18 months There will be 3 half years in 18 months. 3  25    Therefore, A  Rs  4096 1  400       3  1    Rs  4096 1     16   

17 17 17    Rs  4096     =Rs 4,913. 16 16 16   Thus, the required amount is Rs 4,913. 26. We start by finding the interest received by Sumit after three years. Principal (P) = Rs 125000 Rate of interest (R) = 12% per annum Number of years (n) = 3 R   Amount (A)  P  1    100  12   = Rs 125000  1    100  3   = Rs 125000  1    25   28  = Rs 125000    25 

n

3

3

3

28  28  28 = Rs 175616 25  25  25  C.I. = A – P = Rs (175616 – 125000) = Rs 50616 = Rs 125000×

S.I. 

PR T 100

125000  12  3 100 = Rs 45000 Thus, the difference between the interest received by Sumit and Sarita is (C.I – S.I.) = Rs (50616 – 45000) = Rs 5616. Therefore, Sumit will receive Rs 5616 more than Sarita as the interest. Thus, statement A is  correct.  Rs

Comparing Quantities Solutions

249

CONCEPTIVE WORKSHEET 1.

4.

Number of chocolates eaten by Rohit= 7 Total number of chocolates in the bag= 25

7 25 Therefore, percentage of chocolates eaten by  Fraction of chocolates eaten by Rohit 

7  100  28 25 Thus, Rohit ate 28% chocolates in the bag. Total number of cricket bats = 200 Number of bats of brand A= 65 Number of bats of brand B= 90  Number of bats of brand C= 200 – (65 + 90) = 200 – 155 = 45

 Rs 1000  month   Rs 5000  100  %  20%   Thus, 20% of Vicky’s monthly salary constitutes his monthly savings.

Rohit  2.

Thus, fraction of bats of brand A 

Vicky’s monthly expenditure= Rs 4000 Vicky’s monthly savings= Rs 1000  Vicky’s monthly income= Rs (4000 + 1000) = Rs 5000  Percentage of money saved by Vicky every

5. a.

To find the ratio of girls to boys. Ashima and John came up with the following answers. They needed to know the number of boys and also the total number of students.

65 200

Similarly, fraction of bats of brand B 

90 200

Similarly, fraction of bats of brand C 

45 200

 Percentage of bats of brand  65   100  %  32.5%  200 

A

 Percentage of bats of brand

 90   100  %  45% B 200  

OR

 Percentage of bats of brand

3.

 45   100  %  22.5% C  200  Number of day scholars in the school= 325 Total number of students in the school= 500 Thus, number of hostlers in the school= 500 – 325 = 175 Thus, percentage of students who are  175   100  %  35% hostlers    500  Number of day scholars in the school= 325 Total number of students in the school= 500 Percentage of students who are day  325  100  %  65%  500 

scholars  

Thus, percentage of students who are hostlers= 100% –65% = 35%

So, the number of boys = 30 – 18 = 12. Hence, ratio of the number of girls to the number of boys is 18 : 12 or

18 3  . 12 2

3 is written as 3 : 2 and read as 3 is to 2. 2 www.betoppers.com

8th Class Mathematics

250 b.

To find the cost per person. Transportation charge = Distance both ways × Rate = Rs (55 × 2) × 12 = Rs 110 × 12 = Rs 1320 Total expenses = Refreshment charge + Transportation charge = Rs 4280 + Rs 1320 = Rs 5600 Total number of persons =18 girls + 12 boys + 2 teachers = 32 persons Ashima and John then used unitary method to find the cost per head. For 32 persons, amount spent would be Rs 5600. The amount spent for 1 person = Rs

c.

6.

Let the original length and breadth of the rectangle be l and b respectively.  Original area of the rectangle = l × b = lb It is given that the length and breadth are respectively increased by 20% and 25%.  New length =

20 l 6l l l  100 5 5 New breadth = b + 25% of b l + 20% of  l  l 

b

25 b 5b b b  100 4 4

6l 5b 3lb   5 4 2  Percentage change in the area of the rectangle  New area of the rectangle 

5600 = Rs. 32

175. The distance of the place where first stop was made = 22 km. To find the percentage of distance:



Originalarea  New area  100 Original area

3lb lb  lb 1  2  100  2  100  2  100  50% lb lb

7.

OR

Thus, the area of the rectangle increased by 50%. Let the number of votes that Khushboo received be x. Then, number of votes that Meenakshi received= x + 20%  of x

20 x 6x x  x  100 5 5 Now, number of votes received by Meenakshi and Khushboo = Total number of votes casted x



6x  x = 111100 5



Both came out with the same answer that the distance from their school of the place where they stopped at was 40% of the total distance they had to travel. Therefore, the percent distance left to be travelled = 100% – 40% = 60%.

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11x  111100 5

111100  5 Þ x = 50500 5  Difference between the numbers of votes received by the two contestants  x



6x x 50500 x    10100 5 5 5

Comparing Quantities Solutions 8.

251

It is given that the cost price of 15 cloth pieces equals the selling price of 10 cloth pieces. Let the cost price of 15 cloth pieces= Selling price of 10 cloth pieces = x  Cost price of 1 cloth piece 





x 15

x Selling price of 1 cloth piece  10  Profit made by the merchant on each cloth piece= Selling price – Cost price

3x x 4  100  4  100 1 1   100  33 % 3x 3x 3 3 4 4

x

Thus, Meenakshi’s expenditure must increase

1 33 % . 3 so that her savings remain the same. Let the cost price of the shirt be Rs x. It is given that the shopkeeper suffered a loss of 10%. This implies that his selling price was Rs (x – 10% of x) i.e., Rs 450. by

10.

x x x   10 15 30

Profit

 x

x  30  100 x 15

 x

x 15   100  50% 30 x Thus, the merchant makes a profit percent of 50%. Let Meenakshi’s original salary be x. Then, original expenditure

450  10  500 9 Thus, the shirt had cost the shopkeeper Rs 500. Discount = Marked Price – Sale Price = Rs 840 – Rs 714 = Rs 126 Since discount is on marked price, we will have to use marked price as the base. On marked price of Rs 840, the discount is Rs 126. On MP of Rs 100, how much will the discount be?





9.

10 x  450 100

 Profit percent  Cost price  100

11.

75 3 x x 100 4 Original savings = Salary • Expenditure 

3 1 =x  x  x 4 4 It is known that Meenakshi’s salary increased by 25%.  New salary = x + 25% 25 1 5 xx x x 100 4 4 Since her savings remained the same: 1 New savings = Original savings  x 4 New expenditure = New salary “ New  5 1 savings  x  x  x 4 4 Percentage increase in expenditure 



New expenditure  Original expenditure Original exp enditure

9x  450 10

 x

= 75% of x =

of x = x 

x  450 10

126  100  15% 840 We can also find discount when discount % is given. Marked price is same as the list price. 20% discount means that on Rs 100 (MP), the discount is Rs 20. By unitary method, on Re 1 the discount will be Discount 

12.

Rs

20 100

20  220  Rs 44 100 The sale price = (Rs 220 – Rs 44) or Rs 176 Rehana found the sale price like this — A discount of 20% means for a MP of Rs 100,

On Rs 220, discount = Rs

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8th Class Mathematics

252 discount is Rs 20. Hence the sale price is Rs 80. Using unitary method, when MP is Rs 100, sale price is Rs 80;

17. (i)

80 . 100 Hence when MP is Rs 220, sale price = Rs

When MP is Re 1, sale price is Rs

13.

80  220  Rs 176. 100 Cost price of 200 bulbs = Rs 200 × 10 = Rs 2000 5 bulbs were fused. Hence, number of bulbs left = 200 – 5 = 195 These were sold at Rs 12 each. The SP of 195 bulbs = Rs 195 × 12 = Rs 2340 He obviously made a profit (as SP > CP). Profit = Rs 2340 – Rs 2000 = Rs 340 On Rs 2000, the profit is Rs 340. How much profit is made on Rs 100?

It is given that, population in the year 2003 = 54,000 Therefore, 54000 5   = (Population in 2001)  1    100 

Population in 2001  54000 

2

20 20  = 21 21

48979.59 Thus, the population in the year 2001 was approximately 48,980. (ii)

340  100  17% 2000 Overall CP of each fan = Rs 1200. One is sold at a loss of 5%. This means if CP is Rs 100, SP is Rs 95. Therefore, when CP is Rs 1200,

5   Population in 2005 = 54000  1    100 

2

Profit 

14.

95  1200  Rs 1140 100 Also second fan is sold at a profit of 10%. It means, if CP i0s Rs 100, SP is Rs 110. Therefore, when CP is Rs 1200, then SP = Rs

then SP = Rs

15.

110  1200  Rs 1320 100 We need to find the combined CP and SP to say whether there was an overall profit or loss. Total CP = Rs 1200 + Rs 1200 = Rs 2400 Total SP = Rs 1140 + Rs 1320 = Rs 2460 Since total SP > total CP, a profit of Rs (2460 – 2400) or Rs 60 has been made. On Rs 100, the tax paid was Rs 5. On Rs 450, the tax paid would be 5  450 = Rs 22.50 100 Bill amount = Cost of item + Sales tax = Rs 450 + Rs 22.50 = Rs 472.50. The price includes the VAT, i.e., the value added tax is10%. Total cost = actual cost + sales tax

= Rs

16.

10   3300   x  x   100   

3000 = x  Original cost = 3000 www.betoppers.com

33000 = 11x

2

1  21 21   54000  1    54000   =59,535. 20 20 20  

Thus, the population in the year 2005 would be 59,535. 18.

The initial count of bacteria is given as 5,06,000. Bacteria at the end of 2 2

2.5  1    hours  506000  1    506000  1    100   40 

 506000 

41 41  = 531616.25 40 40

= 5,31,616 (approx.) Thus, the count of bacteria at the end of 2 hours will be 5,31,616 (approx.). 19.

Principal = Cost price of the scooter = Rs 42,000 Depreciation= 8% of Rs 42,000 per year

 42000  8  1   Rs   = Rs 3, 360 100   Value after 1 year = Rs 42000 – Rs 3360 = Rs 38,640

2

Comparing Quantities Solutions 20.

253

Let the sum invested be P. Rate of interest, R = 8% p.a. Time T for the sum invested in simple interest = 1 year The simple interest received (S.I.) is given by: S.I. 

 2000 

It is also given that compound interest on the same sum for the same time period at the same rate of interest compounded annually is Rs 2160.

PRT 100

 Rs 640 

 P  Rs

R    A  P 1    100 

P  8 1 100

 1000  A  P 1   P  

The amount received (A) is given by: n

P  2160

 Rs 8000  

21.

108  108 = Rs 1331.20 100  100

Compound interest, C.I. = Rs (9331.20 – 8000) = Rs 1331.20 Let the original sum be P. It is given that rate of interest, R = 40% p.a. Let the time period T be n years. Amount, A is given by:

n



22.

3

P

1000000 =6250 160

Substituting the value of P in equation (i): R = 16 Thus, the required rate of interest is 16% per annum. On Rs 100, interest charged for 1 year is Rs 15. So, on Rs 10,000, interest charged



15  10000  Rs 1500 100

Interest for 2 years = Rs 1500 × 2 = Rs 3000 Amount to be paid at the end of 2 years = Principal + Interest = Rs 10000 + Rs 3000 = Rs 13000

n

 n=3 Thus, the required time period is 3 years. Let the sum be Rs P and the rate of interest be R%. It is given that: Time, T = 2 years Simple interest, S.I. = Rs 2000 PR2 100

23.

n

1331  110   11   11         1000  100   10   10 

 S.I. 

2

2

P

n

2

P  + 2160P = P  + 1000000 + 2000P 160 P = 1000000

2

R  1331 10    P 1  P  P 1     1000  100   100 

2

 P  1000  

2

Here, n (number of years) = 2 8   A  Rs 8000 1    100 

2

 100000  A  P 1   P  100  

640  100  Rs 8000 8

R   P 1    100 

PR 100000  R ____ (1) 50 P

n

24.

R   We have, A = P  1   , where Principal (P)  100  = Rs 12600, Rate (R) = 10, Number of years (n) = 2 2

10    11   Rs 12600  1    Rs 12600    100   10   Rs 12600 

2

11 11   Rs 15246 10 10

CI = A – P = Rs 15246 – Rs 12600 = Rs 2646 www.betoppers.com

8th Class Mathematics

254 25.

26.

Mayuri first converted the time in years. 3 1 1 year 3 months = 1 year  1 years 12 4 Mayuri tried putting the values in the known formula and came up with: 1

1

17  4  A  Rs 10000  1    200  Now she was stuck. She asked her teacher how would she find a power which is fractional? The teacher then gave her a hint: Find the amount for the whole part, i.e., 1 year in this case. Then use this as principal to get simple

interest for

1 year more. Thus, 4

17   A  Rs 10000  1    200 

 Rs 10000 

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217  Rs 10,850 200

Now this would act as principal for the next We find the SI on Rs 10,850 for

1 year.. 4

1 year 4

1 10850  17 4 SI  Rs 100  2 10850  1 17  Rs 230.56 800 Interest for first year = Rs 10850 – Rs 10000 = Rs 850  Rs

And, interest for the next

1 year = Rs 230.56 4

Therefore, total compound Interest = 850 + 230.56 = Rs 1080.56.

Comparing Quantities Solutions 27.

255

Principal = Rs 21,000 Reduction = 5% of Rs 21000 per year 21000  5  1  Rs  Rs 1050 100 value at the end of 1 year = Rs 21000 – Rs 1050 = Rs 19,950 Alternately, We may directly get this as follows:

SUMM ATIVE WORKSHEET 1.

Discount = 20% of Rs 1850  Rs

19  Rs 19, 950 20 Let R be the rate at which Deepak borrowed the loan. Principal (P) = Rs 16000 C.I = Rs 1640  A = P + C.I. = Rs 16000 + Rs 1640 = Rs 17640 Number of years (n) = 2

= Rs 21000 

  1480 = x +

15 x  100

  1480 = x +

3x   20

n

R   We know that, A  P  1      100  R    Rs 17640 = Rs 16000  1    100  

17640  R   1   16000  100 

  1480 =  2

2.

2

2

R    1   = (1.05)2  100 

1480  20  1286.96  1287 23 Thus, the cost price of the article is approximately Rs 1287. The correct answer is C. Principal = Rs 16000 Rate of depreciation = 5% per year Therefore, the value of the washing machine at the end of two years is: 5   = Rs 16000  1    100 

2

R    1   = (1 + 0.05)2  100  2

23x   20

 x

2

R    1.1025 =  1    100 

2

2

2

R   5    1    1    100   100   R = 5% Thus, Deepak borrowed the loan at the rate of 5% per annum. The correct answer is B.

20  1850 = 100

Rs 370 S.P. = M.P. – Discount = Rs (1850 – 370) = Rs 1480 Let Rs x be the cost price of the article. It is given that the shopkeeper can earn a profit of 15% on the sale of the article. Profit = 15% of Rs x It is known that S.P. = C.P. + Profit Rs 1480 = Rs x + 15% of Rs x

5   Value at the end of 1 year = Rs 21000  1   100  

28.

M.P. of the article = Rs 1850 Discount percent = 20

3.

2

1    19  = Rs 16000  1   = Rs 16000 ×    20   20  = Rs 14440 Thus, the cost of the washing machine after two years will be Rs14440. The correct answer is B. Population in the year 2007 = 625000 Rate of increase = 8% p.a. Time period = 2 years Therefore, population at the end of the year 2009 is: 2

8   = 625000  1   = 625000  100 

2   1    25 

2

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8th Class Mathematics

256 2

4.

 27  = 625000   = 729000  25  Thus, at the end of the year 2009, the population of the district will be 729000. The correct answer is B. The value of the ornament after three years can be calculated using the compound interest formula. Principal = Rs 20000 Rate of appreciation = 5% p.a. Therefore, the value of the ornament after 3 years: 5   = Rs 20000  1    100   21  = Rs 20000×    20 

 Discount = 20% of x 

   x 



3

7.

3

2

6.

 20   Rs   1600   100  = Rs 320  S.P. = Rs (1600 + 320) = Rs 1920 Discount allowed on jacket = 20% Let x be the marked price of the jacket. www.betoppers.com

5  1920 = Rs 2400 4 Thus, the shopkeeper should mark the jacket at Rs 2400. The correct answer is C. Cost price of the ceiling fan = Rs 1800 Profit percent = 20%

 Discount percent 

Discount  100   M.P.

240  100 2400 = 10% Thus, 10% discount should be allowed by the shopkeeper on the ceiling fan. The correct answer is A. Principal (P) = Rs 75000 Rate of interest (R) = 10% p.a. 

2

 26  = Rs 1562500    25  = Rs 1690000 Thus, the cost of the land will be Rs 1690000 after two years. The correct answer is B. Cost price of jacket = Rs 1600 Profit percent = 20%  Profit = 20% of Rs 1600

4x  Rs 1920 5

 20   Profit  Rs  100  1800   Rs 360   S.P. of the ceiling fan = C.P. + Profit  = Rs (1800 + 360) = Rs 2160 M.P. of the ceiling fan = Rs 2400  Discount allowed = Rs (2400 – 2160) = Rs 240

21 21  21 20  20  20 = Rs 23152.50 Thus, the value of the ornament after 3 years is Rs 23152.50. The correct answer is B. Current value of land = Rs 1562500 Rate of appreciation = 4% per annum The value of land at the end of two years can be calculated using the compound interest formula. Therefore, the value of land after two years will be: 4   = Rs 1562500  1    100 

x  Rs 1920 (S.P. = M.P. – discount) 5

 x = Rs =

= Rs 20000×

5.

x   5

8.

1 1 Number of years (n)  2  2  2 2 We know that amount received (A) is given by n

R   P 1      100  We first calculate the amount received at the end of first two years. Amount received at the end of first 2 years = Rs 10   75000  1    100 

2

2

 11  = Rs. 75000×   = Rs. 90750  10 

Comparing Quantities Solutions

257

1 year can be 2 calculated by taking the amount received after two years as the principal.

8    Rs 72900 = Rs 62500  1    100 

Now, the simple interest for

 S.I. 

72900  2   1    62500  25 

PR T 100

n

n

729  27      625  25 

1 90750  10    2  Rs 100 = Rs 4537.50 Thus, the required amount for the given sum at

1 years is 2 Rs (90750 + 4537.50) = Rs 95287.50. The correct answer is D. Principal (P) = Rs 20000

n

the end of 2

9.

n

11.

6 % for half year 2 Conversion period (n) = 2 We know that amount received (A) is given by

2

 27   27        25   25   n = 2 Thus, Manish borrowed the money for 2 years. The correct answer is B. Let P be the sum that is invested. Rate of interest (R) = 10% p.a. Number of years (n) = 2 C.I. = Rs 4200 We know that amount received (A) is given by

Rate (R) = 6% p.a. =

R   P 1    100 

n

n

R    P + C.I. = P  1      100 

n

R   P 1     100  

10    P + Rs 4200 = P  1   100  

2  6     A =  Rs 20000 1  200    

2

2

  103    Rs 20000    100  

10.

2

103  103    Rs  20000  100  100   = Rs 21218  Compound interest (A – P) = Rs (21218 – 20000) = Rs 1218 Thus, the compound interest received on Rs 20000 after one year is Rs 1218. The correct answer is A. Principal (P) = Rs 62500 Rate (R) = 8% p.a. Amount (A) = Rs 72900 Let n be  number  of  years  for  which  Manish borrowed the money. We know that, R   A  P 1    100 

n



  

  

 11  P + Rs 4200 = P    10  100P + Rs 42000 = 121P 121P – 100P = Rs 420000 21P = Rs 420000

 P = Rs

420000 = Rs 20000 21

Now, simple interest is given by

PR T . 100

20000  10  2 Rs = 4000 100 Thus, simple interest is Rs 4000. The correct answer is B. S.P. of shirt = Rs 702 C.P. of shirt = Rs 650 Let Rs. x be the tax imposed on the shirt.  S.I = Rs

12.

 x 

 

 Tax = x% of Rs 650  Rs  100  650 

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8th Class Mathematics

258 According to question,

6

 x   650  = Rs 702 Rs 650 + Rs   100   Rs 65000 + Rs 650x = Rs 70200  Rs 650x = Rs (70200 – 65000)  Rs 650x = Rs 5200

3.

5200 8 650 Thus, 8% sales tax is imposed on the shirt. The correct answer is B. Let x be the cost of the television set.  x

13.

1 x  Rs16060 10



11x  Rs16060 10

150  128.25 21.75   100 = 14.5% 150 150 The correct answer is C.

14.

n

4.

x  Rs

Here A =  84  100, r =  5, n =  2,  and P =  the required initial investment 2   5   84 100  P 1  1         100  

  21 2   441   841  84 100  P 1    P   P 1      400   400    100  

80 80   25 100 100

2.

4 4    25  Rs.16   5 5 The correct answer is C. Let the principal be Rs.P 6

5.

6

R  R    Given that P  1    2P   1   2  100   100 

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400  84 100 841 P = 400 × 100 P = 40 000 Thus, the original amount invested by Sammy was Rs.40 000. The correct answer is C. The money is invested at compound interest. The interest is compounded semi-annually. The time period is 15 years. The value of the investment in 15 years is given  P

HOTS WORKSHEET The selling price 

n

n   r    P 1  1      100  

 14   Rs   1200   100  = Rs 168 Thus, Rajni needs to spend Rs (1200 +168) = Rs 1368 to buy the sweater. The correct answer is B.

1.

r   CI  P 1    100  Amount = A = P + C.I. r    P  P 1    100 

16060  10  Rs14600 11 Thus, the cost price of the television set is Rs 14600. The correct answer is B. List price of sweater = Rs. 1200 It is given that 14% sales tax is imposed on the sweater.  Sales tax = 14% of Rs 1200



24

discount  

10 x x Tax = 10% of x  100 10 x

n

R  R   R    P 1    16P  1    1    100   100   100   n = 24 years The correct answer is D. Since the purchase value is above Rs.100, the additional discount will be applicable. The cost after a 10% discount = Rs.150 – Rs.15 = Rs.135 Now the cost after 5% discount on the discounted price = Rs.135 – Rs.(135 × 0.05) = Rs.128.25 The effective

r   by,  CI  P 1    100 

n

Comparing Quantities Solutions

259

9 Here, P = 1 000, r = , n = 15 × 2 = 30 2 Thus, the required expression 30

6.

10.

4

R   The amount for Johny = P  1      100 

is

30

9    0.09  1000  1    1000  1     2   100  2   The correct answer is D. Let the cost of the racket, before any sales tax is levied on it be Rs.x After the sales tax is levied on it, the racket costs 10% more  Rs.x + 10 % of Rs.x = Rs.71.50  Rs.x + Rs.0.1x = Rs.71.50  Rs.1.1x = Rs.71.50

71.50  Rs.65 1.1 Therefore, before the sales tax is levied, the tennis racket costs Rs.65. The correct answer is C. Original price of the car = Rs.24 000 Amount to be paid as sales tax = 7.5% of Rs.24 000

5

R   The amount for Tony = P  1    100   The ratio of the amounts for Johny and Tony 4

R   P 1   100   5 is =  R  P 1    100 

 x

7.

8.

1 1 5    R 20 6 6   1 1 100 100 5 The correct answer is A. Let she put Rs.x in her account 

11.

1

4    10 000  x 1    100 

8

8

7.5   Rs.24 000 100

 26   10 000  x    25 

= Rs.1 800 Thus, total amount needed to be paid for the car = Rs.24 000 + Rs.1 800 = Rs.25 800 The correct answer is C. Here P = 20 000

10 000  26 268 Hence, x = Rs.7 306.90 The correct answer is A. Le the annual interest rate be r.

20%  10% 2 n = 2×2 = 4 R

12.

r   4 Theamount will be compounded 4 × t times in t years.  The amount to be returned after t years  =

4

R  0     A  P 1    20000 1    100   100  = 20 000 × (1.1)4 = 29 282 The correct answer is A. Here P = 200 000 R = –25% [The value is being reduced] n = 3 n

8

 x

The quarterly interest rate =

[Half yearly]

n

9.

Let the amount invested = P r = 20%

4t

r     A 1  4     100   

3

R  25       A  P 1    200000  1      100   100  = 84 375 The correct answer is B.

r    A 1    400 

4t

= A × (1.125 508 81)t

4t

r    1   =(1.125 508 81)t  400 

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8th Class Mathematics

260 16.

r    1   = (1.125 508 81)1/4  400  = 1.03  r = 12% Theannual interest rate = 12% The correct answer is D. 13.

17.

8 = 4% 12 The amount will be compounded 3 × 2 = 6 times Therate of the half yearly interest =

t

14.

r   So, the amount to be returned = P   1      100   101 22.55 = P × (1 + 0.04)6  P = 7 999.99  8 000   p = Rs.8 000 The correct answer is C. There will be 5 × 4 quarters The amount to be paid after the kth quarter = theamount  to  be  paid  after  the  (k – 1) th quarter+the interest on the amount to be paid after t h e th (k-1)  quarter

2 A k  A k 1   A k 1 100

 102   A k 1     100  but A0 = Rs. 10 000

IIT JEE WORKSHEET 1.

Let C.P. of the cell phone be x. Loss percent = 8%

 Loss = 8% of x = 

8x 2x  100 25

It is given that if Mahender sold the cell phone at Rs 378 less, then the loss would have been 15%.

102 A1  10 000  100 102 102 A 2  10 000    100 100  102  A k  10 000     100 

18.

Present Value = Depreciated Value (P){1 – Rate/ 100}Number of years  512 = P{1 – 20/100}3  512 = (0.80)3P   P = Rs.1 000 The correct answer is C. Let the investment by Mr. Anthony in scheme A = Rs.P  Investment in scheme B = Rs.(27 000 – P) Given [{P × {(1 + 8/100)2 – 1} + (27 000 – P) {(1 + 9/100)2 – 1}] = 4 818.30   P = 12 000 The correct answer is B. Let Sam’s share = Rs.P  Patrick’s share = Rs.(1 301 – P) Given; P(1 + 4/100)7 = (1 301 – P)(1 + 4/100)9   P = Rs.676  (1 301 – P) = Rs.625 So, Sam’s share = Rs.676 Patrick’s share = Rs.625 The correct answer is A.

2x   Therefore, when the loss is Rs  378   , then 25   the loss percent is 15%.

k

Now, loss percent is given by

Loss  100 . C.P.

20

15.

 102  A 20  10 000    = 14 859.47  100  Theamount to be returned = Rs.148 59.47 The correct answer is C. For effective rate of interest, the factor (1 + R/ 100) is considered. As interest is compounded half-yearly, R = 10/2 =5  (1 + R/100) = (1 + 10/100 × 2)2 = (1 + 5/100)2 = 1.1025 So, effective rate of interest = 10.25% The correct answer is B.

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2x    378  25   15     100 x    

 15x = 37800 + 8x  7x = 37800  x

37800  5400 7

Thus, Mahender bought the cell phone for Rs 5400. The correct answer is B.

Comparing Quantities Solutions 2.

Let x be the M.P. of the bag of rice. S.P. of the bag of 5 kg rice = Rs 117 Discount offered = 10%  Discount = 10% of x 

 x 

3.

261

7.

Thus, the bag of 5 kg rice was marked at Rs 130. The correct answer is B. List price (L.P.) of the microwave oven = Rs 3250 S.P. of the microwave oven = Rs 2470  Discount = L.P. – S.P.. = Rs 3250 – Rs 2470 = Rs 780 Now, discount percent is given by

 discount percent=

5.

 9 13      % of Rs. 9 000 = Rs. 15 2 3 

10 x  Rs117 100

90 11700 x  Rs117  x  Rs  Rs130 100 90

8.

780  100 = 24% 3250

Thus, the discount percent offered by the shopkeeper on the microwave oven is 24%. The correct answer is C. M.P. of the book = Rs 250 Discount = 16%  Discount = 16% of Rs 250

 16   250   = Rs 40 = Rs  100    S.P. of the book = M.P. – discount = Rs (250 – 40) = Rs 210 Thus, the selling price of the book is Rs 210. The correct answer is A. Cost of flat last year = Rs 1800000 Cost of flat this year = Rs 1926000  Increase in cost = Rs 1926000 – Rs 1800000 = Rs 126000  Percentage increase in cost of the flat Increase in cost  100    126000  100 Old cost 1800000 = 7% Thus, the percentage increase in the cost of the flat is 7%. The correct answer is C. 

The required difference

 1   1    4 % of Rs.9000    4 % of Rs.9000   2   3 

10 x 100

Discount  100 L.P.

4.

6.

9.

The correct answer is B. Given P(1 + R/100)15 = 2P   (1 + R/100)15 = 2P/P = 2 _____ (i) Let n = required number of years for the money to become eight times its initial value,   P {1 + R/100}n = 8P   {1 + R/100)n = 23 = {{1 + R/100}15}3 _____ by (i)   {1 + R/100}n = (1 + R/100}45

  n = 45 years The correct answer is D. Let the sum be Rs.x. Compound Interest = x × (1 + 50/3 × 100)3 – x  343x/216 – x = 127x/216  127x/216 = 1270   x = 2160  the sum = Rs.2 160  Simple Interest = (P × R × T)/100 = (2 160 × (50/3) × 3)/100 = Rs.1 080 The correct answer is D. After r years. Share = x(1 + R/100)r Where r = the number of years R = Rate Robert’s share for 9 years = (1 + 5/100)9 Luke’s share for 11 years = (1 + 5/100)11 11

5   1   100   9 Ratio  =  5    1     100 

(1 + 0.05)2 = 441/400  Luke’s share was 441/(400 + 441) × 7569 = 3969 The correct answer is C.

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8th Class Mathematics

262 10.

11.

The difference in the simple interest and the compound interest for 2 years is on account of the interest paid on the first year’s interest, when interest is reckoned using compound interest being compounded annually Hence, 12% of simple interest = 90  simple interest = 90/0.12 = 750 As simple interest for a year = 750 @ 12% per annum, the principal = 750/0.12 = Rs.6250 If P = Rs.6 250 The amount at the end of 3 years = 6250 + 3 × (simple interest on 6 250) + 3 × (interest on simple interest) + 1 × (interest on interest on interest) = 6 250 + 3 × (750) + 3 × (90) + 1 × 10.80 = 8 780.80 The correct answer is C. (B, C) Let C.P. of each article = Rs. 1. Then, C.P. of 12 articles = Rs. 12. S.P. of 12 articles = C.P. of 15 articles = Rs. 15

14.

15.



So, 15% of 

16.

 3   100  %  25%  12 

(B, C) Let original price = Rs. 100 Price after second discount

 80   90   Rs.72 =Rs.   100  Price after third discount

13.



 60   72   Rs.43.20 = Rs.   100   Single discount =(100-43.20)=56.8% or 57% (A, C) Let original price be Rs. x.

 12  Discount =   100  %  15%  80  (C, D) Let the marked price be Rs. x Discount availed by the retailer = 15% of Rs. x C.P of the machine by the retailer = (x–15% of x) = Rs.

 Gain= 

12.

(A, B)

17.

17x 17x =1955– 20 20

51x 17x   1955or x  2000 400 20 Discount recived by retailer = (15% of Rs. 2000) = Rs. 300 (B, D) Rs. P is S.I. on Rs. P for 7 years.

7  Rs. 3P is S.I. on Rs. P for   3P  =21 years P   (C) 135% of 80 135  80  13.5  8  108.0 cm 100 =1.08 metres. (A) 8.4 % of a = 42



18.

8.4  a  42 100 a 19.

42  1000  500 84

(B)

3x C.P. = (x – 25% of x) = 4

P% of 24 = 6m 

3x  33x  3x S.P=   10%of  4  40  4

P

33x  660  x  Rs.800 40 = Rs.

660  40 33

20.

P  24  6 100

6  100  25% 24

(A) Let the amout he had in th begining be x So, x 



20 of x  6400 100

80 x  6400 100

x www.betoppers.com

17x 20

6400  100  Rs.8000 80

Comparing Quantities Solutions 21.

(A) Let the original salary = Rs. x Present salary = Rs. 8025 Increase in salary = 7% of original salary =7% of Rs. x According to the question,

263 27.

28.

7 x  x  8025 100 107 8025  100 x  8025  x   Rs.7500 100 107 (D) Present value of the machine = Rs. 40,000. Decrease in value = 5% of Rs. 40, 000 5  40000  Rs.2000 100 Thus, value of machine after a year = Present value - Decrease in value =Rs. 40000–2000 =Rs. 38000 (B) Original Population =60, 000 Increase in Population = 10% of original =

23.

10  60000  6000 100 Thus, population after a year = 60000+6000 =66000 (C) Let the value year ago = x

29.

 x / 10   100  % Now, Gain% =   x  1  Gain% =10 %  n =1 A  s,B  p,C  r, D  q

(A)

2 2   100%  40% 5 5

10 x  Rs.38700 100

(B)

2 2   100%  28.55% 7 7



90 x  Rs.38700 100

(C)

13  100  65% 20

38700  100  Rs.43000 90

(7) We have, number of students who are good at

72 ×25=18 100 Number of Students who are not good at Mathematics =25–18=7 (4) 50% of ( x –y) = 30% of (x + y) Mathematics = 72% of 25 =

26.

30.

1 x of Rs.x  Rs. 10 10

x–

x 25.

15  50000  7500 100 Thus, value of machine after a year = Present value – Decrease in value = Rs. 50, 000 –Rs. 7, 500 = Rs. 42,500 = Rs. 425 × 102 , Hence n = 2 (1) Let the C.P. of the article be Rs. x. Then, Gain =

=

24.

100  6   2400  S.P. of 80 articles = Rs.   100  =Rs. 2754 (2) Present value of machine = Rs. 50,000 Decrease in value = 15% of Rs. 50, 000 =



22.

(4) C.P. of 80 articles = Rs. 2400, Profit = 16%



(D) 31. (A)

13 52  100   100%  17.33% 75 3 A  r,B  s,C  q, D  p Let the number be a 1 6 %of a is 2 4 25 25 1 %a  2   a  2 4 4 100  a  2  4  4  32 

50  x  y   30%of (x  y) 100

 5  x  y   3  x  y   2x  8y  x  4y

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8th Class Mathematics

264 (B)

M.P. =Rs. 30, Discount % = 15%, Discount = ? 15=

Discount  100 30

15  30  Discount 100 45  Rs.4.50 10 C.P. of the house = Rs. 6500 S.P. of the house =Rs. 7150 Gain = 7150–6500=Rs.650 Cost price of the goods = Rs. 8000 Profit % = 7 S.P =?  Discount 

(C)

(D)

 100  Pr ofit%  S.P=    C.P 100    100  7%     8000  100  107  800  107  80  Rs.8560 100

 

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5. EXPONENTS & RADICALS SOLUTIONS Here 212, 28 have same bases.  12 > 8 (Powers of the given numbers) It is obvious that 212 > 28. 2) Given 510, 315 To know that the first number is greater than the second number or vice vice versa, we have to convert both the numbers in such a way that either bases or powers are same so, that the two numbers can be compared. 510 = (5)5×2 = (52)5 = (25)5 315 = (3)5×3 = (33)5 = (27)5 Here 255 and 275 have same powers.

FORMATIVE WORKSHEET 1.

1) 44 Base = 4, Index = 4 2) (–2)4 Base = –2, Index = 4 2

2.

9  9  3)   Base = , Index = 2 25  25  4) (–1)2009 Base = –1, Index = 2009 1) Let us express 192 in exponential form using prime factorization method. 3 192 2 64 2 32 2 16 2 8 2 4 2 2 1

 275  255  315  510 1

5.

32   32 1

6



1   n  a  n  a  

1



1

2)

3

a 6 y 3   a 6 y 3  3 6

a   3  y 

3 729 3 243 3 81 3 27 3 9 3 3 1

1) 27 – 72 = (2 × 2 × 2 × 2 × 2 × 2 × 2) – (7 × 7) = 128 – 49 = 79 2) (2.5)3 – (1.5)3 Formula: a3 – b3 = (a – b)(a2 + ab + b2) (2.5)3 – (1.5)3 = (2.5 – 1.5) ((2.5)2 + (2.5)(1.5) + (1.5)2) = (1.0) (6.25 + 3.75 + 2.25) = 12.25 To know first number is greater than the second number or vice vice versa, we have to convert both the numbers in such a way that either bases or powers are same so, that the two numbers can be compared.

6

1 3

a3

1   n  a  n   3 a   y3   a m  p a mp   n   np  b  b 

1

 a2     y 

 729 = 36

4.



 1 5    32 

1 5  32     32  5

 1  1  6   3 6   26   36  2   6 3  2     2) Let us express 729 in exponential form using prime factorization method.

3.

1 1 5

1

 192 = 31 × 26 6

1)

5

6.

Given,







   

2n  4  2.2n  23 2.2n  3

2n.24  2.2n  23 2.2n.23 2n  24  2  n

2 .2

4



1 23

 a

m+n

 a m  an 

1   -n  a  n  a  

16  2 1 14  2 16    1 16 8 16 16

8th Class Mathematics

266

7.

9n  35   27 

Given,

3  81

 2

4

n



 32   35  33  3   34  2n



5

3

3

4

9

 a

3 3 3 3  316

32 n  5  9  3116

m n



 a mn



 a

n

×a = a

m+n

2



22 n 1 n  2 2 n  2  n 1



22 n 1 n  2 2 n  3 n



2

2n + 1 n2  2n  3 + n2

 0.6  1

  0.1

  y 

1

3

 3   3   1   3      2  2  3 



 1  1   10  1 1  3   27   1         8  8   3 



1 1    10  1  27  1    3 8   1     8  3

   

31 2

2 2

2 2



y  x



y z x × × x y z

   x . y  xy 



z x  1    an  n   y z  a 



a × b = ab





2 10. Given,

n n 1

2 

4



n 1 n 1

2 

2

n2 n

n n 1



2 

 n 1 n 1

 2

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9993  3336  1117 32  94  115 3

 2n 1

 a

m n



0

 1

 a mn





 1 b   a a    b 

9 9 3   93 6 2

12. Given,

n 1

     

1  10  8  27       3  3  8 

1  1 1  1  1 n 1

 a

1   n  a  n  a  

x 1 y  y 1 z  z 1 x

Given,

1

1

 7  5  3  1   a m m-n    x 2 2  y 2 2   n  a       a   x 

1 1  22 4

1

3  2 y    1    y2 

75 2

2

 22 

5

 72 x   5  x2 



m  2 n 1 n   2 n  3 n   a  a m-n  2   n  a 

0

x2 . y



2

2 2 n  3 n

7

9.

2

2

11. Given,

Given,

2

2

x 2 . y3

8.

2

22 n 1 n

 am m-n   n  a   a 

= 32n – 3

 2 n  2    n2 1

2 n 1 n  n  2

=2

32 n 14   32 n 14 17 17 3

2

 am m-n   n  a   a 

 2 n  2   n 1 n 1



 m

 n 1  n2  n 

 9 111   3 111

6

 1117

32  94  115

93  1113  36  1116  1117 4

32   32   115

Exponents & Radiclas Solutions

 ab 

m

 am  bn

2 3





3 

267



 xb  14. Given,  c  x 

367

 36  111 2

8

 a

5

3  3  11

m

 a n  a m n 

36  36  11116 36 6  11116 312  11116  6  1 1 32 8  115 3  115  5 6 3 11



1 3



 a   a  1 4

 81 

16 6



3  4

2 3

 1     27  1 4 12

1 4



2

 2  3 3

 3 

2 3

2 3

2 3

m n



1

1 3

2 3

   



4

2 3

4 3

3

3

   

 3 2 2 3 3  2 3

 23 . 3

6 3 3

 2.3=6

  x

 2 3 . 33

c- a  c+a   b c- a 

c2  a2  bc+ ba

  x

a2  b2  ca+cb

b 2  c 2  ab+ac+c2  a2  bc+ba+a2  b2  ca+cb

 x

× a n  a m +n 

1 1  n-m 1 x 1  x m-n

1 1  n x xm 1 m 1 n x x

 am  m-n  a  n  a  

1 1 xm xn    xm + x n xn + xm xm + x n x n + x m xm xn

y

1  1  a   a   b  b  y y 16. Given ,  1  1 b   b   a  a 

1   1  43 3

 a-b  a+b-c 

×  x



x

1 2  3   1  1 2  23  2 33

a+b-c

   1 b    xm + xn 1  a a   m x + xn  b     

33  9

1 3

  x a-b 

 c-a  c+a-b 

b-c b+c  a b-c

x0  1



 1 9  3  3 

a+b-c

 xa   b  x 

×  x

 x        x   a-b a+b  c a-b   x



mn

1



 a mn

m

 a   a 

1

23

1



 a



4

2 

b-c  b+c-a 

m n

15. Given,

4 9

c+a-b

b 2  c 2  ab+ac

mn

c+a-b

  x c-a 

  x

1

1 12

 x 

 a

 14  3 1 1 4 3  16    3 3     1  3    81 4    1   27   2 3   6     m n

b-c b+c-a

x 

 xc   a  x 

 am m-n   n  a   a  

1   n  a  n  a   12 16 = 3 × 111 × 36 × 115 = 312 +6 × 11116 ×115 = 318 ×11116 × 115 1 4   3  4 16 4  1  3  13. Given, 81  62   27        



b+c-a

1 2  3 3

3

2 4  1 3 3

x

y

 ab  1   ab  1      b   b    x y  ab  1   ab  1       a   a 

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8th Class Mathematics

268 x

 ab  1  ab  1 

bx by x y  ab  1  ab  1 ax ay x





19. Consider, 2x + 3 = 4y – 2.  2x + 3 = (22)y – 2

y

 ab  1  ab  1 bx

by

ax ay  a   b x b y  b 

x

y



a .  b

ax

ay x

 ab  1  ab  1

y

y

 a m  a m   m     b    b

x+ y

a    a m × a n = a m+n   b   17. Given, 2x + 5 = 4x + 2  2x + 5 = (22)x + 2

 2x + 5 = 22(x + 2)

 a

m n



 a mn



 2x + 5 = 22x + 4 Since bases are equal, powers can be equated.  x + 5 = 2x + 4  x – 2x = 4 – 5  –x = – 1  x=1 18. Given, 4x – 4x–1 = 24  4x 

4x  24 4

 a   a  m n

 2x + 3 = 22y – 4

 m-n a m   a  n  a  

 1  4 x 1    24  4

mn

Since, bases are equal powers must be equated.  x + 3 = 2y – 4  x + 3 – 2y + 4 = 0  x – 2y + 7 = 0 –––––––––––– (1) Consider, 3x – 2 = 93y – 2x.  3x – 2 = (32)3y – 2x

 3

x–2

=3

2(3y – 2x)

 a

m n



 a mn



 3x – 2 = 36y – 4x Since bases are equal, powers can be equated.  x – 2 = 6y – 4x  x – 2 – 6y + 4x = 0  5x – 6y – 2 = 0 –––––––––––– (2) Solving equations (1) & (2) Multiply equation (1) with 5, we get 5(x – 2y + 7) = 0 × 5  5x – 10y + 35 = 0 Now, 5x – 10y + 35 = 0 5x – 6y – 2 = 0 – + + –––––––––––––– –4y + 37 = 0 –4y = – 37 y

37 37  4 4

Now, substitute y 

37 in x – 2y + 7 = 0 to get the 4

value of x.

 4 1  4    24  4  x

 4x 

 37   x  2   7  0  4 

3  24 4

37 37  14 23  37   x  2   7  7   2 2 2  4 

4 3 x  4 =8×4  4x = 32  (22)x = 25  4 x  24 

 a   a  m n

2x

 2 =2

5

mn

Since bases are equal, powers can be equated.  2x = 5

5 2 www.betoppers.com  x

23 37 , y 2 4 mn m n 20. Given, a = a .a To find the value of m(n – 2) + n(m – 2): Consider, amn = am.an



x

 amn = am + n

 a

m

× a n  a m+n 

Since base are equal powers must be equated.  mn = m + n ––––––––––––(1) Consider, m(n – 2) + n(m – 2)

Exponents & Radiclas Solutions

269

 mn – 2m + nm – 2n  2mn – 2 (m + n)  2mn – 2mn = 0 ( mn = m + n, from equation –––––(1))

1 x  x -b  1 By multiplying numerator and denominator by x– c, we get Consider,

 m  n  2  n  m  2   0  a 2

21.

2  Given,

xc

 24 23b.4 To find the value of 3b – 5a  a 2

2a



a

.8a  32a 3b

2 .4 a

2 . 2

3 a

5 a

  2 

3b

2 .2

2



22 a.23a  25 a 1  4 23b.22 2



2 2 a  3 a  25 a 1  4 23b.22 2 5a





1 24

 a

m

 a n  a m+ n 

2.25 a 1  4  x  x  2 x  3b 2 2 .2 2 By rearranging the terms, we get



 2

5 2

3b  5a

2



x c+b+c x  c  1  xb

 1 b   a a    b 

(  from (1))

1 1 1  b  c b c x  x  1 x  x  1 x  xa  1 a

xb 1 1  b  c b c x 1 x x  x  1 x  x a  1 (  from (4))

 

3b

21 4 23b  5a 22 2

x  c .x b .x c x a+b  1  xb









     

1 xb  ________(4) x a  x b  1 x  c  1  xb Consider,

2 2 1  4 3b 2 2 .2 2

2.2 2  5a 2 2 2

x c x a .xb  1  x b xb x c



5a

4



 24

1   m n mn n   a   a and a  n  a   

x c xa 1 1   x c x b .x c x c



2 .8  32 1  4 3a 2 .4 2

2a

x c x c x a  x c .xb  x c



x  c  x a  x  b  1

.8a  32a

2  Consider,

a

c

xb  1 1  c c b x 1 x x  xa  1 x c  x b  1 x x c

c

1 x

b





1 x  x -a  1 c

(  Multiplying and dividing the 1st term by xc)

 a

m

 a  a m+ n  n

 am m-n   n  a   a 

 23 = 23b – 5a Since bases are equal, powers can be equated.  3 = 3b – 5a  3b – 5a = 3 22. Given, a + b + c = 0  a + b = – c ___________(1)  b + c = – a ___________(2)  c + a = – b ___________(3)



x b+c  x c 1  c c c c b+c x x x x  xa  1

x a + xc 1  c 0 c a x x x x  x a  1 (  from (2))





x  a  xc 1  c c a 1 x  x x  xa  1



x  a  xc  1 1 1  xc  x  a

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8th Class Mathematics

270 23. Given, 2x = 3y = 12z. To prove that xy = z(x + 2y): Let 2x = 3y = 12z = k.

 n  1  1  n  

1

1

1

3ky,

 2  kx,

1

1 1  n   n  1   n  n  1 1       n  n  n n      

12  k z

1

1

Consider, 12  k z  3  4  k z  3  22  k

2

1 1   y x  2  k and 3      

1  1  k k x   k z  

1 y

2 x

 k k k 1



2

  a  n  b n         a   b

1 z

1 y

1

1



2

x 3  y 3   z _______(1)

Cubing on both the sides, we get 3

1 2 1 x  2y 1     y x z xy z

2  1  3   x3  y3   z  

 xy = z(x + 2y) 1 b

2

Consider, x 3  y 3  z  0

1

1 a

2

26. Given, x 3  y 3  z  0 To prove (x + y2 + z3)3 = 27 xy2z3. 1

1 z

 ky x kz Since bases are equal, powers can be equated.



3

3

1 2 2  1  2  1    x 3    y 3   3x 3 y 3  x 3  y 3    z 3      

1 c

24. Given, x  y  z and xyz = 1 To find the value of a + b +c: 1 a

1 b

1

Let x  y  z  k .  x = ka, y = kb, z = Consider, xyz = 1  ka × kb × kc = 1  ka + b + c = 1  ka + b + c = k0  a + b + c = 0.

kc –––––––––––– (1)

 25. Consider, 1  1  1  n  



1   1    1  1      1  n  

1

 1  n  1 1   1       1  n  

1

   n  1   1      1  n  

1

  n  1   1       n  1 

1

2

 x  y 2  3x 3 y 3   z    z 3 ( from (1))

1 c

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  a  n  b n         a   b

1

( from (1)) ( am × an = am + n) ( k0 = 1)

 

2

1

2

 x  y 2  z 3  3x 3 y 3 z

Cubing on both the sides, we get

x  y

1 1 1



1

 x  y 2  z 3  3x 3 y 3 z  0

2

z

3 3



 1 2    3x 3 y 3 z   

3

3

 n  a 

1   n a 

3

 13   23  3 2 3 3 3 x  y  z  3    x  .  y  .z     2 3 3  (x + y + z ) = 27 x y2 z3 ((am)n = amn) 27. Given, x2 = y + z, y2 = z + x and z2 = x + y. 

To find

1

 1 x

here ‘  ’s means sum.

1

1

1

1

1  x  1 x  1  y  1  z ( we have 3 variables x, y and z in the given problem)

Exponents & Radiclas Solutions

271 Now,

Consider, x2 = y + z

 x

 1 x  1 



a2  b2  c 2

y+z x

 a  b y+z x+ y+ z  1 x  x x

1 1  x + y+z 1 x x

1 z  1  z x + y + z –––––– (3)

=

1

1

1

1  x  1 x  1  y  1  z

x y z + + x+ y+ z x+ y+ z x+ y+ z x+ y+ z 1 x+ y+ z

2  ab + bc + ca  1  6  ab + bc + ca  3



To find the value of

1 1 1 x

28. Given, a + b + c = 0 a 2  b2  c 2 To find the value of (a  b)2  (b  c ) 2  (c  a ) 2 Consider, a + b + c = 0 Squaring on both the sides, we get , (a + b + c)2 = 0  a2 +b2 + c2 +2ab + 2bc + 2ca = 0  a2 +b2 + c2 = – (2ab + 2bc + 2ca)  a2 +b2 + c2 = – 2(ab + bc + ca) ––––––– (1) Consider, (a – b)2 + (b – c)2 + (c – a)2 = a2 + b2 – 2ab + b2 + c2 – 2bc + c2 + a2 – 2ca = 2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca = 2(a2 + b2 + c2) – 2ab – 2bc – 2ca = 2(–2(ab + bc + ca)) – 2ab – 2bc – 2ca (from equation (1)) = – 4ab – 4bc – 4ca – 2ab – 2bc – 2ca = – 6ab – 6bc – 6ca = –6(ab + bc + ca) ––––––––––– (2)

1 1 1    15 a b c

a b c a c b + + + + + b c a c b a

 1 1 1 Consider,  a +b + c       12  15 a b c 

( from (1), (2) & (3))



2

29. Given, (a + b + c) = 12 and

1 y Similarly, we get 1  y  x + y + z –––––– (2),

1

2

 b  c  c  a 

( form (1) & (2))

1 x  –––––– (1) 1 x x+ y + z

Now,



2

a a a b b b c c c + + + + + + + +  180 a b c a b c a b c

a a b b c c  1  + +  1  + +  1  180 b c a c a b 

a a b b c c + + + + +  3  180 b c a c a b



a b c a c b + + + + +  180  3  177 b c a c b a

x 2  yz y 2  zx z 2  xy   k a b c Now, x2 – yz = ak ––––––––––– (1) y2 – zx = bk ––––––––––– (2) z2 – xy = ck ––––––––––– (3) Adding equations (1), (2) & (3), we get x2 – yz + y2 – zx + z2 – xy = ak + bk + ck  x2 + y2 + z2 – xy – yz – zx = k(a + b + c) ––––––––––– (4) Multiplying (1), (2), (3) by x, y, z respectively and then adding up, we get, x(x2 – yz) + y(y2 – zx) + z(z2 – xy) = xak + ybk + zck.  x3 – xyz + y3 – yzx + z3 – zxy = k (ax + by + cz)  x3 + y3 + z3 – 3xyz = k(ax + by + cz)  (x + y + z) (x2 + y2 + z2 – xy – yz – zx) = k(ax + by + cz)  (x + y + z) (k(a + b + c)) = k (ax + by + cz) ( from –––(4))  (x + y + z) (a + b + c) = ax + by + cz.

30. Let

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8th Class Mathematics

272 1

 1 31. Radical form of  x p  

32.

 2  p  3        q    

3 2

 2  q  3          p      1

q 7    p

q    p

1 7

    

3 2



      1    2    p  3         q   

1 7

    

q  is  

1 7

q

3 2

        

1

x p i.e

q p

x.

1 7

1   n  a  n  a  

   n   1   b     a  n  a        b 

1 q    p

1 7

1   n  a  n  a  

   n   1   b   p   a  n  a    7 q      b 

2 2

108

3

27

3

9

3

3 1

3 3

243

3

27

3

9

3

3 1

5

135

3

27

3

9

3

3 1

54

 108 = 22 × 33 ––––––––– (3)

81

Now,

 243 = 35 –––––––––––– (4)

 135 = 33 × 5 –––––––––––– (5)

 60 

4

128 3 108 

2

2

(60)3 5 128. 3 (108) 2

= 2a.3b.5c 4 243.135 To find the values of a, b and c: Now,

2

243.135

(Radical form)

1 7 5

2

3  . 3

 5

3

 3  5  2

=

33. Given

3 5

  2

1 5 4

3

3

60

2

30

3

15



3

3

7 5

1

5

5

5 1



2 2

128 64

2

32

2

16

2

8

2 2

4 2

5

3 4  33  5



2

7 4 6  5 3

3

 7

 128 = 2 –––––––––––– (2)

1

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4

26  33  53  2 5  2 3  32

2

 33  2  53

5 3 4

90  21  20 15

3

5 12 4

5  35  53 5



(from (1), (2),

2  3  5  2   24  36  3 6

7

 60 = 22 × 3 × 5 ––––––––– (1)



(3), (4) & (5))

3 4  33  5 2

1 3 2 3

Exponents & Radiclas Solutions

273

131



2 15  35  53 17 4



3 5

131

5

2 15  3

17 4

131

3

2  531  2 15  3 4  5



 am mn   n  a   a  3 5



 60 

. 128. 3 108

4

243.135

2

131

3

 2 15  3 4  52 

2a  3b  5c Now, comparing the powers, we get

a

 125  3     64 

 3 216   4  256  264  625 1

4

1









x

x x 35. Given, x  x x . to find the value of x.

x Now, x

x

1

 x

 x

1 2

 1 1  x 2   

x

21 2

 2 1  x 2   

x

x

 xx

 x x

x

1     x  x2   

x x 2

1

0

 am mn   n  a   a 

2 1

5 54 9 1   4 4 4

131 3 , b  , c = 2. 15 4 2

34.

5    4

x

( am × an = am + n)

2

1  125  3   1   1  64   256  4    625 

 3   x2   

3

 x

x2

3x 2

3

1  n   a  a n & a  1  

x2

 x x Since bases are equal, powers can be equated. 3

   n  2 1  1   b    53  3  625  4   n   3    1   a   a   4 256        b    

 5      4  

3

1 4

  54     4   1  4 

3

x2 3   x 2

1

1

 x2 

 x

2 4  5    5  4        1  4    4  

3 x 2

 x2 

3

2 3

x

3 2 2



m 3  a  a m 1    n 2  a 

3 2

1

2

3 2 Squaring on both the sides, we get

2

 12   3 2 x      2

5 5     1 4 4

5   4    1 5   4

 x2 

2

( By BODMAS rule)

 x

9 4 www.betoppers.com

8th Class Mathematics

274

CONCEPTIVE WORKSHEET 1. i) ii)

iii)

Base = 3 33 = 27 Base = –2 (–2)6 = 64

5.

i)

4

2 3 16a 2b 3 = (16a b )

ii)

3

3 2 1/3 27a 3b 2 = (27a b )

Index = 3

1

6.

= (a–1 b1/2) 1

Index = 4

ii)

1

16 2     5  625 Base = –1 (–1)100 = 1

Index = 100 ( (–1)n = 1 if n is even)



i)

81 81   16 256

3

2 4 1

 32 4 31  a b c   

bc

abc 1

a 3 b2 . c 2

4



a 1 6



1

b

8.

4

ii)

4.



9.

 8x3   3   27 a 

2

4

4

4

4

= 49 , 27  494 > 274 i.e., 78 > 96

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6

1 1 2 2

y

3 5  4 8



1 5 2

1

x y8

2

3

  2 x 3   3 3  3 .a 

2

 3  2 x  3  3    1     3a   

2

4a 2 x 2  2 xa     9  3  m

12

  72  ,  33 

1

 5 1  x 2 y8   

10. Consider, 25

=7, 3 = 78 , 3 4 × 3

2

b 3 .c  a

x2 y 4

6

8

1 3

3

x4 . y3

x

4

  32   312

2

x 2 y8



1296  81 1215  256 256

1

5

1

x 2 . y 5

1 1  6

.c 3 2 1  3

256

 3   3  1215       256 2 4 2 2 (3.2) – (1.6) = (3.2 + 1.6) (3.2 – 1.6) ( a2 – b2 = (a + b) (a – b) = 4.8 × 1.6 = 7.68 (3.2)2 – (1.6)2 = 7.68 96 can be written as,

4 3

b4 c 1 

2

a3

16  81  811  

2

a3 b3 c 3 4

4

a

2 / 3 4 1/ 3



3. 4

3 1

a 2b

288 = 2 × 144 = 2 × 12 × 12 = 2 × 2 × 2 × 3 × 2 × 2 × 3 = 2 5 × 32  288 = 25 × 32 1296 = 36 × 36 = (22 × 32) × (22 × 32) = (24 × 34) 1296 = (24 × 34) 3 3     2 4

1 3  2

1

 3.a 2 . b 1 

7. 2. i)

(3a1/2 b–3/2)

 3.a 2 . b 2

4

iv)

1

  a 2b  2 . 3  ab3  2

a b . 3 ab 3 2

Index = 6

2 Base  5

1/4

5.5m  5

1 4

  52 

4 m 1 4

5

4 m 1 2

and

1 m 2

m Also 5. 5  51.5

m 2

 51.5

m 2

1 m 2

5

5

2 m 2

 m  14  25 . 5.5m Substituting these values,  m  5. 5 

1

    

m

Exponents & Radiclas Solutions 4 m 1 2

 .5 5  2 m  5 2 

1 m 2

1

    

275

m

 4 m 112 m  2  m   5   

1

 xa

x

11.

15.

1 2

1 1 2   2 2 1 9 16 4       2   4  16;         4  3  16   9  

= (8 ×10–16) –2/3 = 8–2/3 × 104

1     10,000   2500  25  102 4   Similarly, (0.0081) 3/4 = [(0.3)4]3/4 = 0.027 = 27 × 10–3 (0.0256)–1/2 = (256 × 10–4)–1/2 1 100 102  = (256) ×10   100  16 16 16 (0.125)1/3 = [(0.5)3]1/3 = 0.5 = 5 × 10–1 (0.3)2 = 0.09 = 9 × 10–2 Substituting all these values in given sum, we get, –1/2

2

 25 10    27 10  10 2

3

1 1 = a and 4  b 4 3 Given sum reduces to,

 x a   xb   x c  14.  b  .  c  .  a  x  x  x  = (xa – b)a + b . (xb – c) b + c . (xc – a)

c2

. xc

2

 a2





 a2

c2

= x0 = 1

 xb  . c  x 



ac

1 a

 x x  1  b  c  x x  

b+c

 xc  . a x 

c+a

1

1 1 x

bc

x

ba



1



b

1 1 x



b

 x x  1  c  a  x x  

c a

 xc b

1 c

 x xc  1     x a xb  

1 1  c a c a b b a x x x x x x x x  x x  xb x c x b .x c xc xa b

c

a

1 x x  x x  xc xa xa xb a

b

x  

b+c

c b

1 c+a

x x x b+c

a+b

  



1 x  

c+a

 x a+b  xb+c   x c+a 

1 x  

a+b

b+c

 x  x c+a   x a+b 

x b+c

x x

2

c+a

a+b

x

b+c

a b

 x c+a  x a+b 

x

a 2  ab+b2

2

 xb   c  x 

 ab + b 2

 a  b  a 2  ab+b2 

 xa

x

3

 c3

.x

b 3

3

 b3  b3  c3  c3  a3

. xb

b2  cb+b2

. (x b-c )b

3

 xa

x c+a



c+a

 x a+b  x b+c 

 x b  c  x ca  x a  b   1 xab   x b  c  x c a   x b  c  x ca  x a  b 

= (x a-b ) a

a 4  b 4   3 1    4 1   169  169  4225     16 9 144  4   3 a 2  b2 b+c



a b

 xa  16.  b  x 

13. Let 3

a+b





25 16  27  103 240   2.4 5  101  9 100

2

b 2  c2 

a



2

102  5 101  9 102 16 

a 2  b2 

1 x 

–2/3

2

1

4 4 16  16  12   4   3 3 3

12. We know that, (0.000008)

. xb

 xa    b x 

1  62m  3m 3 m 5   5   5  125  

2

b 2

m

1 m

2 0 1 9 3  4   3  8  4    16     

2

. xc

3

2

 xc   a  x 

 cb + a 2

 b  c   b2  cb+b 2 

c 2  ca+a2

. (x c-a )c

.x

2

 ca + a 2

 c  a   c 2  ca+a 2 

 a3

= x0 = 1

c+a

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8th Class Mathematics

276 5

17.

 3  9 3

5/2

3

.3

2

 3x  3  3 x

4

1

=3 ×3 ×3

5 4 2

20. 1/2

13

3

x

3 2

4

3 13 x  2 2

2

x=5

2



3 x 3 

 22 x 1  29 3 x

1

 256 

y

x Consider, 64  2 2

 26 x  2

3

2 1  2x    5 3 2

4 14 x 3 3

14 7  4 2 21. Given, ax = m, ay = n and a2 = (my nx)z

 2 2 (given)

We know that 64 = 26 ; 256 = 28 and 2 2  2

6 x 

 25

x 

 2 x  1  9  3x 2x + 3x = 9 + 1  5x = 10  x = 2 x 19. 64 

 25

4 1 4 1 x 5  x 5 3 3 3 3

1 8x  3

2 x 1

 25

2 1 2 x  2



1  3   4  4 3   

 25

1 1  2x  3 2

23

10 2

22 x 1 

1 2

 32

1 1  2x 2  

13  3 x 2

18.

2x

1 2

 22  3 

13 3 x  2 2

x

2x 

4

 13  4   

= 3x + 1 + 1/2

32

  3

3 2

1 3 3 2 2 x and y 2 12  256 

(1)

 m  a x ; n  a y 

a2 = [(ax)y . (ay)x]z

a2 = [axy . ayx]z a2 = [a2xy]z a2 = a2xyz  2 = 2xyz  xyz = 1 22. Let ax = by = cz = k 1

2

___________

1

1

y so that, a  K x ; b  K ; c  K z Also given b2 = ac Substituting a, b, c values

2

  256 

  28 

y

y

 8 y 

2

2

3

3

2

1 1  1y  K   Kx . Kz    

2

3 3 y 2 16

3  3  Now, 3 x  4 y  3    4    12   16  3 3  0 4 4 Hence, 3x + 4y = 0 

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2

1 1    z

K y  K x

1 1 2     x z y 23. Given that, a = xy p – 1 ; b = xy q – 1 ; c = xy r – 1 Consider, aq – r, br – p, cp – q = (xy p – 1)q – r . (xyq – 1)r – p . (xy r – 1)p – q = xq – r. y(p – 1)(q – r). xr – p.y(q – 1)(r – p). x(p – q).y(r – 1)(p – q) = (x q – r + r – p + p – q)(y (p – 1) (q – r) + (q – 1) (r – p) + (r – 1) (p – q) ) = x0 . y0 =1

Exponents & Radiclas Solutions

277

24. 2x = 3y = 6–z = k Now, 2x = k  2 = k1/x 3y = k  3 = k1/y 6–z = k  6 = k–1/z We know that 2 × 3 = 6 Substituting 2, 3, 6 values in terms of ‘K’, we get 1 y

1 x

K .K  K K

1 1  x y

K

1 z

1 z

 1 1  1     x y z

 3 1 1  1  1  x  



  x   1   3  1  x 

1



  

1

1 1 1  2 2  2 2 2 2 2 a  b  c b  c  a c  a  b2



1 1 1   2ab 2bc 2ca



1  1 1 1  1  c + a +b         2  ab bc ca  2  abc 

2

1

  

1 3

1

3



1  0 , if (a + b + c) = 0  a  b2  c2  2

28. Given that x 

y

a b a b

bc ca ; z bc ca

ab 1    1 x  a b  a    1  x  1  a  b  b   ab 1

 1  x3  1 1  3   1   3    1  x  

 1  x3   1  3  x  

 1 y  b 1 z c  Similarly,    and 1  y c 1 z a   1 x 1 y 1 z       1 x 1 y  1 z 

1 3

a b c . . 1 b c a 29. Given, a = xm + n . yp b = xn + p . ym c = xp + m . yn m–n Now, a , bn – p, cp – m = (xm + n . yp)m – n . (xn + p . ym)n – p . (xp + m . yn)p – m = x(m + n) (m – n) . yp(m – n) . x(n + p) (n – p) . ym(n – p) . x(p + m) (p – m) . yn(p – m) 

1

1

3  x 3  1  x 3  3  1  3  1   3        3  x3 x     x   1



Hence

1 3

1     1    1  1    1  x 3       

3

1  a  b2  c 2  2

= zero (since a + b + c = 0 )

1 1 1 Hence      0 x y z 25. Given x = ya ; y = zb and z = xc Consider, x = ya but y = zb b a  x = (z ) but z = xc  x = zab  x = xabc  abc = 1

26.



Now,

1    x  x 27. Given a + b + c = 0 a + b = –c squaring on both the sides, (a + b)2 = (–c)2 a2 + b2 + 2ab = c2  (a2 + b2 – c2) = –2ab Similarly, (b2 + c2 – a2) = –2bc and (c2 + a2 – b2) = –2ca _____________ (1)

= xm

2

 n2

= xm

2

 n2

. y p (m  n ) . x n  n2  p2  p2 m 2

2

 p2

.y

. y m (n  p ) . x p

2

m 2

. y n (p  m )

p m  n   m n  p   n  p  m 

= x0 . y0 = 1 Hence, am – n . bn – p . cp – m = 1 www.betoppers.com

8th Class Mathematics

278 30. Given ax = bc Multiply with a on both sides ax . a = abc  a(1 + x) = (abc)  a = (abc)1/(x + 1) Similarly b(1 + y) = (abc)  b = (abc)1/(y + 1) c(1 + z) = (abc)  c = (abc) 1/z + 1   abc    abc 

33. Consider,

4

2 2

4

2 3 2

3

2

4

2

1

3

4

1

4

2

 2

1 1 1   x 1 y  1 z 1

 2

Bases are equal powers also equal, i.e.,

1 3

3

 

 2

2

3

1 4

1

 26

1 1 1   1 1 x 1 y 1 z

2

and



5 4

2 3 2

2

1  y 1  z   1  z 1  x   1  x 1  y   1  1  x 1  y 1  z   (1 + x) (1 + y) + (1 + y) (1 + z) + (1 + z) (1 + x) = (1 + x) (1 + y) (1 + z)  (1 + x + y + xy) + (1 + y + z + yz) + (1 + z + x + zx) = 1 + (x + y + z) + (xy + yz + zx) + xyz  3 + 2 (x + y + z) + (xy + yz + zx) = 1 + (x + y + z) + (xy + yz + zx) + xyz  (x + y + z) = –2 + xyz  (x + y + z) – xyz = –2

2

5

1

 26  2

also

3

 

 2

2 3

Hence,

4

3

1

2

6

2

1

6

2 3 2

2 6 2 Hence proved.

34.

3

1

2

2 5 1

1

1 6

6

5

5

1

5

 2

1

6

3

1 3

x  51210  512100  5121000 ---------Consider

2

31.

12 12 2 3  3   1  4   1  4 8  1 4  4 1         x3  x    x x

3 3 3   ..... (it is in G.P.) 10 100 1000 (In G.P. sum to infinity S  

32.

  3

2x

4

but 4

2

1

3

1 2

 

 32  4

 2

2 1  2x   3 2

2



1

3

 25

2

2

1

3

2x

1 2

 25

3

2 1   2x    5 3 2

a = Ist term, r = common ratio) 3 3 10 1  10    1 10 9 3 1 10 1

x  512 3

1  15  15 1  2x  2   2  2 x    2x = 7 2 2  

  29  3

7 x  2

 23 = 23 = 8

1

9

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a , where 1 r

Exponents & Radiclas Solutions



35. Given x 



2  1  3 2 1 1



 x 

3

279

 

2 1



3

1

2 1

3

a ____________

(1)

a

 x3   

x3 

  





 

2 1

 

 

2 1

3

  a  b 

3

x3  2  3  



1

2 1

3

  



2 1

3

2 1



2 1

3



2 1

1

3

4.

 a 2 b   ab 1    3 4   3 2   a .b  a b  3



2 1  

1

3



3

 a 5   b3    5   4  b  a 

x 

1

  2  1  x3 = 2 – 3x



3

1

  x from 1   3  x + 3x = 2

 



2 1

3

5.

1.

1    81 

5

5

5

a15 b15 1  20  5 15 b a a 3 n

n 3

 x  y     x + y   3 n

 x  y  3n x+ y

SUMM ATIVE WORKSHEET 3 4

zx zx

xyz

 b5   a4   5   3  a  b 

  

 a 3  b3  3ab  a  b 



a

3

 

2 1  3 1

y z yz

x  y y z z  x   xy yz zx

a = a0 = 1

3

1

 

2 1  1



3

a

 x  y  z+ y  z  x+ z  x  y

Cubing on both the sides, 1

x y xy

  x  y  1      x + y  

3n

3

 1  4  4      33  27  3  

1       x + y  x  y  

3n

1 2 x

p q   q  p

2.

1

3n

 p 2  p     q q  2x – 1 = 1/2

2x  1  2x 

3.

xy



6.

1 2

= (x2 – y2)–3n

We know that, 6m + 1 = (2 × 3)m + 1 = 2m + 1 . 3m + 1 10m – n = 2m – n . 5m – n 15m + n – 2 = 3m + n – 2 . 5m + n – 2 4m = (22)m = 22m Substituting all these values, we get 2n.6m 1.10m  n.15m+n  2 4m .32m+n .25m 1

3 4

ax  ay xy

2 x 1

3 2

 x

  1  2 2    x  y  

yz

ay  az yz

zx

az ax

a x  y  a y  z  zx a z  x



2n.2m 1.3m 1.2 m n.5m n .3m+n  2 .5m+n  2 22 m .32 m n .25m 1



2 2 m 1 . 32 m n1 . 520  2 22 m . 32 m n . 52 m 2

1 2  2.  3 3 www.betoppers.com

8th Class Mathematics

280 1

7.

1

1

 x b bc  x c  ca  x a  ab Given that,  c  .  a  .  b  x  x  x  1



10.

1

  x b c  bc .  x c a  ca .  x a b  ab bc

 x bc

+

c  a a b + ca ab

n

 1  1  p   . p   q  q  m n Given that  1  1  q   . q   p  p  m

n



n

  pq  1  pq  1  .   m qn  q   m n   pq  1  pq  1    m pn   p 

 pq  1 

m

qm

 pq  1 



n



qn

pm

 pq  1

m



pn

 pq  1

 n

m+n

9.

p m+n  p   m+n    q q Given that, ab = ba Let us consider, L.H.S

a   b

a b



a b



a b



11.

ab

b  a

1 b

a



ab 1 b b

a 

 b

a

= ab 

 LHS = RHS

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

1 n

 x xn  1  p  m  x x  

n

p

m

1 x  

n+ p

p+m

+x +x x n+ p

m+n

  



a a 1 ab   a b  RHS a

1 x  

p+m

m+n

+ x + x n+ p   x p+m 

1 x  

m+n

n+ p

+ x + x p+m   x m+n 

x n+ p

x

n+ p

+ x p+m + x m+n 



x p+m

x

p+m

+ x m+n + x n+ p 

 x n+ p + x p+m + x p+m   1 x m+n   x m+n + xn+ p + x p+m   x n+ p + x p+m + x m+n  1

a

a b

m

1 1  p m p m n n m x x +x x +x x x x + x x + xn x p x n .x p x p xm 1  m n p n x x + x x + x p xm xm xn 

 pq  1   pq  1    .  q   q    m n  pq  1   pq  1    .   p   p  m

m

 x x  1  n  p  x x   1  p  x xp  1  m  n  x x  

abc

m

8.

1



a  b c +b c  a +c  a  b 

x = x0 = 1

1 1  m p n p 1 x + x 1  x + xn m 1  p m 1  x + x p n m n

 x p 1q  p q q 1 p 1 q p    . x . x  q 1p    x   1

 p q  p  q q 1  p 1 x  q  .x q p  p  x  1

2 pq   p+q   p  q  p q  xq p  . x pq    

Exponents & Radiclas Solutions

281

1 2

 p q   x pq  

x

p+q pq

.x

2

Subsitutute 22x + 1 = 2a2 and 2x – 1 

 p  q 2 pq   p+q   . x pq  

a 2a 2  33    4  0 2 2  4a – 33a + 8 = 0  4a2 – 32a – a + 8 = 0  4a (a – 8) – 1 (a – 8) = 0  (a – 8) (4a – 1) = 0

 2 pq   p+q     pq  

 p+q   2 pq  p+q  pq

x = x2

1 4 Case (i) If a = 8 then 2x = 8 x=3

2

12.

 a = 8 (or)

 n  9 n.32  3 2   27n 32 n.32.3n  33 n    33 m . 23 33 m . 23

1 1 x then 2  4 4 Hence x = – 2 (or) 3 16. Given that xp = yq = (xy)pq _____________ (1) Consider (xy)pq = xpq . ypq = (xp)q . ypq = (yq)q . ypq ( xp = yq) Case (ii) If a 

33n  2  33n  3m 3 3 .2 



33 n  2 33n  33 m . 23 33 m.23 1 3

3 m n   2

.2

3



1 3 m  n 

3

1 1  3 3 3 3.2 3 .2



1  1 1 2  3  3. 2  3 

. 23

 xy 

pq

2

 yq

 pq ___________

2

y q  y q  pq  q2 + pq = q  q (p + q) = q  (p + q) = 1 1

1 8 1    3  8  9  27 13.

2x  1  4x  1 81  x

2

17. Given, x  2  2 3  2 3



1

3  (x – 2)  2  2

2

3



____________

(1)

Cubing on both the sides

= 64

2 x 1 . 22 x  2  64 2 3 3 x  2(x + 1) + (2x – 2) – (3 – 3x) = 64 = 26x = 26  x=1 14. Given, 5x . 2x + 1 = 200 5x . 2 x + 1 = 5 2 × 2 3 We get, x = 2 15. 22x + 1 – 33.2x – 1 + 4 = 0 ______________ (1) Let 2x = a so that 22x + 1 

= (2x)2 . 2 = 2a2 and 2x – 1 

(2)

from (1) and (2)

Put (m – n) = 1, we get



a in (1), we get 2



1

(x – 2)3  2 3  2

2

3



3

x3 – 8 – 6x (x – 2) 3

3

     3.2

 2

1

3

 2

2

3

1

3

.2

2

3

2

1

3

2

2

3



 x3 – 8 – 6x2 + 12x = 2 + 22 + 3.2 (x – 2)





 2 13  2 2 3   x  2  from 1      3 2  x – 8 – 6x + 12x = 6 + 6x – 12  x3 – 6x2 + 6x – 2 = 0

2x a  2 2

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8th Class Mathematics

282 x+y+4=4 x + y = 0 ______________ (2) Solving (1) & (2) we get x = – 1 and y = 1

18. Given 2x = 4 y = 8 z  x = 2y = 3z

x x (or) z  2 3 Also given, 5x = p(y + z) y

3.

x x  5x  p    2 3

So that 3  k ; 5  k But 75 = (52 × 3) 2 y

1 z

i.e., K  K  K 1 z

K K





2 1     y x

 x   a  a 2  b3  3

xy

and 75  k

1 z

1 2 1    z  y x



4.

5.

2

2

 n2  2n 1 2.

 n2  2 Given 22x + 1 . 23y + 1 = 8  22x + 1 + 3y + 1 = 23  2x + 3y + 2 = 3 2x + 3y = 1 _____________ (1) and 2x + 2 . 2y + 2 = 16 2x + 2 + y + 2 = 2 4

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1 3





1 3

  

3



a  a 2  b3



1 3

x

a 2 b a

a

Now substitute a = 2b and

n 1

n 1

a b

3

1  a b Also given    a b b

n

 a n  a2

  a 

2

[ (a + b)3 = a3 + b3 + 3ab (a + b)]  x3 = 2a + 3 (–b3)1/3 x  x3 = 2a – 3bx  x3 + 3bx – 2a = 0 Given, 9x – 10.3x + 9 = 0 Put 3x = a, the equation reduces to a2 – 10a + 9 = 0  (a – 1) (a – 9) = 0  a = 1 or 9 If a = 1, then 3x = 1 so that x = 0 If a = 9, then 3x = 9 so that x = 2 Hence x = 0 (or) 2 Given a = 2b



HOTS WORKSHEET an  a2 . 2

1 3

 

3 a  a 2  b3

1

20. Given a = 2x ; b = 4y and c = 8z Also given that ayz . bzx . cxy = 46 Substitute a, b, c values, we get 2xyz × 4xyz × 8xyz = 46  2xyz . 22xyz . 23xyz = 212  26xyz = 212 Since bases are equal, powers are also equal.  6xyz = 12 Hence xyz = 2

1.





1 y

1 2x + y  z xy

z  2x + y 

 

 a  a2  b3

 x 3  a  a 2  b3  a  a 2  b3 

1 x



1 3

Cubing on both the sides we get

 5x  5x  p    6   p=6 x 19. Let 3 = 5y = 75z = K 1 x



Given that x  a  a2  b3

6.

22 = (2b)2 – 1  (2b) = 4 b=2 x2 – 3y2 = 2 __________ (1) 3x2 – y2 = 11 __________ (2) (2) – (1) = 2x2 + 2y2 = 9 2(x2 + y2) = 9 x2  y 2 

9 2

a  2 , we get b



1 3

Exponents & Radiclas Solutions 7.

n Given, 2m  2m

n2

2

 (2m ) n

 2mn

283

2



1  1   n

2



2  m n  mn 2

1

  a  2  2 1     a   b  





2

1

 2 ab

 1  x  1    n 10. Given that a+b+c=0 a = –(b + c) Multiply with a on both sides we get a2 = –a (b + c)  a2 = –ab –ac Subtract bc on both sides, then a2 – bc = –ab –bc – ca Hence a2 – bc = –(ab + bc + ca)

1 1 1 1    b+c a+b a+b c+ a

  b 2  2 1     a   a    1 a + b  2 ab  2 ab

11.

 a 2  b2  2  a a2    1  a  b

a c c b  b+c c+a a2 – c2 = c2 – b2 a2 + b2 = 2c2 12. Given that, (4.2)x = (0.42)y = 102 Consider, (4.2)x = 102



a 2  b2 1 a + b

 (a – b) = 1 ______________ (1) Also given (a + b) = 5 _________ (2) Solving (1) & (2) a = 3, b=2 9.

x

n

 m   n 2  n2 1

a b

 1  1    n

1

2  mn  1  n 2

Given

x

x

1  1   x n 

 1  x n  1    n

2 mn   n2 1 m

8.

x

1  1   n 

Given that x 1

x

x

x .x n

1 1 x n



x .n x

 x. n x

1     x.x n   

 1 1  x n   

1  1   n 



x

x

x

1  1   x

 xx  x n  Since bases are equal, powers are also equal.

a+b  b  c c+a  a b   b + c  a + b   a + b  c + a 

x

 42  2    10  10 

42 x  102 10 x Cross multiply, 42x = 102 . 10x  42x = 102 + x  2 x    x 

 42  10

2  1   x 

__________

 42  10 Similarly, (0.42)y = 102

(1)

y

 42  2  2   10 10   y 42 = 102y + 2

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8th Class Mathematics

284  42  10

2 y 2 y

  8b 

 2 y2   

__________

42  10 y  From (1) and (2)  2  1 x  

10 

 10

(2)

b

3 2

3 16

 3   3  Now, 3a  4b  3    4    0  12   16  Hence, 3a + 4b = 0 15. x . x = y + z

 2  2  y 

2  2  1     2   x  y 

y+ z x Adding 1 on both the sides, we get x=

1 1  2    1 x y

1 1 1    x y 2

x 1 

y+z 1 x

13. Given ax – 1 = bc

x 1

y+ z+ x x

ax   bc  a ax = (abc)  a = (abc)1/x Similarly b = (abc)1/y ( by = ca) and c = (abc)1/z Multiplying both the sides we 

 abc    abc 

1 1 1   x y z

1 x  x 1 x + y + z

1 y Similarly, y  1  x + y + z get,

1 z  z 1 x + y + z

.

1 1 1      1 x y z



yz + zx+ xy 1 xyz



1 1 1   x 1 y 1 z 1

=

x y z + + x+ y+ z x+ y+ z x+ y+ z



 x + y + z 1  x + y + z

 xy + yz + zx = xyz

1 a 2 2 14. Given, 64  256b

IIT JEE WORKSHEET

Consider 64a  2 2 a

  26   2

 26 a  2

3

3 2

3  6a  2 a 

1

2

3 1 2 2 and 12 256b

1.

 216  

1 2 3 3



1

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 2 

1



 256 

3  4

1



6  3

(256)–b  2 2 (28)–b = 23/2

2  3

 3 4   4 

1



 32 

1 4

 3



1 5

1



1 5 5

2 

4  



1 5

 1



1 1 1  3  1 2 6 4 2

4 4 6 3 2 5 = (62 + 43 + 21) = (36 + 64 + 2) = 102.

Exponents & Radiclas Solutions

285

150

2.

150

10 

146

 10 

150 146 

 10  3.

n

10  146 10 





 n  2 n 1

3n 1

3 3 3 3   2 n  n 1  3n 1 2n n 1 3 3 3 3

 3

 104  10000

2 n 1

3 n 1 3 n 1

 32  9. 1

1

(18)3.5  (27)3.5 × 63.5 = 2x 1

.5

 18  

10.

 63.5  2 x

3.5

 27 

1 34 1 1 3 4   1    3 3 3 3 3 3  5  8  27   = 5  2    3           

1

1

3.5

  32  2  

 3

2  3.5

3 3.5

3 

 23.5 

 37  23.5 

  2  3

1

 2x

 23.5  33.5  2x

 3 3.5

3

1

3.5

 23.5  33.5  2 x

10.5

3

 27  2 x  x  7 .

4.

a   b

x 1

b   a

x 3

5.

6.

1 a



7.

1

=  5  53  4

 3 x 

1 1 a

m n

1  an 1  m  a



  



1

=  54  4

 

1  am  1  n  a  

Given Exp. = x b  c b+c a . x c  a  c+a b  .x  a  b  a+b  c  b  c  b+c   a  b  c 

b

9.





 am + a n   1 am an    a m + an  a m + an   am + a n  .

= x

8.

 n  m

1 3 4



x 1

a a     b b  x – 1 = 3 – x  2x = 4  x = 2. We know that 112 = 121. Putting m = 11 and n = 2, we get : (m – 1)n + 1 = (11 –1)(2 + 1) = 103 = 1000. 1

1 3 4  1    3 1     3     3  3  3   = 5 2     

= 5  2  3

a   b

 x  3 

1 1 3 4   3 3 3 3 = 5  2    3       

2

.x

 c 2  c   a 2  a2  b

c  a  c+a  b  c  a 

.x

a b  a+b   c  a b 



x . x  a  b  c   b c  a   c  a  b  = (x0 × x0) (1 × 1)=1. 3x – y = 27 =33 ...(i)  x – y =3 x+y 3 = 243 = 35  x + y =5 ...(ii) On solving (i) and (ii), we get x = 4. Given Expression n

n

 n  5 5  

5 2 n 1 243 5  32 n 1  3  5  3  3   

9n  3n 1

2 n

3 

 3n 1

 32 n 1 32 n  3n 1

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8th Class Mathematics

6. POLYNOMIALS SOLUTIONS

FORMATIVE WORKSHEET 1.

4 5 x+ 3 2 It is a polynomial since the variable x has only integral powers.

3.

i) 3x3 –

4.

2

5 2 It is not a polynomial since the variable x has a 3 ii) 3x 3  5x  4x +

2 . 3

rational power

iii) 5 y 2  3 y 3  17 It is a polynomial since the variable y has only integral powers. iv) 7x3 – 3x2 + 5 x  11 It is not a polynomial, since the variable x has a rational power ½. 1

5.

6.

i) Monomial ii) Trinomial iii) Monomial iv) Binomial v) Trinomial vi) Binomial i) – 498 = – 498 x0 (Q it is a constant polynomial) Degree = 0. ii) 3 – y2 + y7 – 14y4 Degree = 7 iii) 4x2y3 – 3x3y + 14x2y2 – 8 Degree = 2 + 3 = 5 iv) x3y2z – 3x2y2z2 + 5xy4z – y7 + x3y5 + 17z5 Degree = 3 + 5 = 8 a)

4, a, x, 3x2, 2x3, 9x5, – 5x7, 5x10,

b)

9, x, 4x2, 3x3, 5x6, 7x9, 6x15

i ) P  x   2 x  , x 

v) 5x 2 y 2 – 3xy3 + 19x2y3 It is not a polynomial, since the variable y has a rational power ½. vi) 3x2y2 – 5xy3 + 12y4 It is a polynomial, since both the variables x and y have integral powers.

5 vii) 2x – 3x + – 8 x It is not a polynomial, since the variable x has a negative power. The above expressions in the descending order of powers of ‘x’ are : 4

2.

3

i) – 3u6 + 4u2 + u + 2 ii) – y4 – 2y3 + y2 + 3y + 5

4 5

4 8  4 P    2        0' 5 5 5 4 is not a zero of P  x  . 5 ii ) P  x   2 x  1, x 

1 2

1 1 P    2 1  2  0 2 2 1 is not a zero of P  x  . 2 iii) P  x   3 x  1, x 

iii) 4x7 + 3x5 + 2x2 + 9x + 2 iv) x6 – x5 + x4 – x3 – x2 + x + 1 v) 14x4 + 2x3 – 6x2 + 4x + 2 vi) – 3x4 – ax3 + x3 + 3ax2 + 2ax + 4 6

5

4

3

2

vii) x + x + x – x + x – x – 1 viii) 2x8 – 5x7 – x5 + 4x4 + x2 + 9x

1 3

 1   1  P    3    1  1  1  0  3   3 



1 is a zero of P(x). 3

1 11 x 5

8th Class Mathematics

288 7.

8.

Let f(x) = ax3 + 3x2 – 13 g(x) = 2x3 – 5x + a given that f(2) = g(2) f(2) = a(2)3 + 3(2)2 – 13 f(2) = 8a + 12 – 13 f(2) = 8a – 1 and g(2) = 2(2)3 – 5(2) + a g(2) = 16 – 10 + a g(2) = 6 + a 8a – 1 = 6 + a (Q f(2) = g(2)) 7a = 7  a = 1 Let f(x) = ax3 + bx2 + x – b Given that –2 is a zero of f(x)  f(–2) = 0  f(–2) = a(–2)3 + b(–2)2 + (– 2) – b = 0  – 8a + 4b – 2 – b = 0  – 8a + 3b = 2 _________ (1) And also given that f(2) = 4  f(2) = a(2)3 + b(2)2 + 2 – b = 4  8a + 4b + 2 – b = 4  8a + 3b = 2 ____________ (2) from (1) & (2) we get, 6b = 4

b

ii)

ii)

iii)

4 2  and a = 0 6 3

 The values of a & b are 0, 9. i)

i)

2 3

Rearranging the terms and simplifying them, we get (2x2 + 7x2 – 6x2) + (–5x – 20x + 18x) + (2 + 5)  3x2 – 7x + 7 Ascending order is : 7 – 7x + 3x2 Descending order is : 3x2 – 7x + 7 Rearranging the terms and simplifying them, we get

1  3  4  1   xy 2   2.7   x2 y   5  1.4  x3 y = 4 + 2  2  2 (1 + 1.5)xy + (2.7 – 0.5)x2y + (5 – 1.4)x3y = 4 + 2.5xy2 + 2.2x2y + 3.6x3y Ascending order is : 4 + 2.5xy2 + 2.2x2y + 3.6x3y Descending order is : 3.6x3y + 2.2x2y + 2.5xy2 +4 10. Given that A = 3x3 – x + 4x2 – 1  A = 3x3 + 4x2 – x – 1 B = 5 – 3x2 + 4x3 + 4x  B = 4x3 – 3x2 + 4x + 5 C = 4 – 5x + x2 – 2x3  C = – 2x3 + x2 – 5x + 4 www.betoppers.com

A + B = (3x3 + 4x2 – x – 1) + (4x3 – 3x2 + 4x + 5) = (3x3 + 4x3) + (4x2 – 3x2) + (– x + 4x) – (1 + 5) = (3 + 4) x3 + (4 – 3) x2 + (–1 + 4) x + (– 1 + 5) = 7x3 + x2 + 3x + 4 B + A = (4x3 – 3x2 + 4x + 5) + (3x3 + 4x2 – x – 1) = (4x3 + 3x3) + (– 3x3 + 4x2) + (4x – x) + 5 – 1 = 7x3 + x2 + 3x + 4  (A + B) = (B + A) B + C = (4x3 – 3x2 + 4x + 5) + (– 2x3 + x2 – 5x + 4) 3 3 2 = (4x – 2x ) + (– 3x + x2) + (4x – 5x) + 5 + 4 = 2x3 – 2x2 – x + 9 C + B = (– 2x3 + x2 – 5x + 4) + (4x3 – 3x2 + 4x + 5) 3 3 2 = (– 2x + 4x ) + (x – 3x2) + (– 5x + 4x) + 4 + 5 = (– 2 + 4)x3 + (1 – 3)x2 + (– 5 + 4)x + 4 + 5 = – 2 x3 – 2x2 – x + 9 B + C = C + B. (A + B) + C = (7x3 + x2 + 3x + 4) + (– 2x3 + x2 – 5x + 4) 3 2 = 5x + 2x – 2x + 8 A + (B + C) = (3x3 + 4x2 – x – 1) + (2x3 – 2x2 – x + 9) 3 2 = (3 + 2)x + (4 – 2)x + (1 – 1)x + 9 – 1 = 5x3 + 2x2 – 2x + 8 A + C = x3 + 5x2 – 6x + 3 (A + C) + B = (x3 + 5x2 – 6x + 3) + (4x3 – 3x2 + 4x + 5) 3 2 = 5x + 2x – 2x + 8  (A + B) + C = A + (B + C)  Addition of polynomial satisfies associative property.  Addition of polynomial satisfies commutative property.

11. i)

ii)

iii)

(3x3 – 4x2 + 5) – (2x3 – 2x2 + 3) = 3x3 – 4x2 + 5 – 2x3 + 2x2 – 3 = (3x3 – 2x3) + (– 4x2 + 2x2) 5 – 3 = x3 – 2x2 + 2 (3x3 – 2x2 – 3x – 3) – (x3 – 2x2 + 3x – 4) = 3x3 – 2x2 – 3x – 3 – x3 + 2x2 – 3x + 4 = (3x3 – 2x2) + (– 2x2 + 2x2) + (– 3x – 3x) +4–3 = 2x3 – 6x + 1 (7x5 – 6x4 + 5x3 – 4x2 + 3x – 2) – (2x5 – 2x + x3 – x2 + x – 5) 5 = 7x – 6x4 + 5x3 – 4x2 + 3x – 2 – 2x5 + 2x4 – x3 + x 2 – x + 5 5 4 3 2 = 5x – 4x + 4x – 3x + 2x + 3

Polynomials Solutions

289

12. A = 5x3 – 3x2 + 2x – 5; B = 3x3 – x2 – x + 2; C = x3 – 5x + 4; i) A – B = (5x3 – 3x2 + 2x – 5) – (3x3 – x2 – x + 2) = 5x3 – 3x2 + 2x – 5 – 3x3 + x2 + x – 2 = 2x3 – 2x2 + 3x – 7 B – A = (3x3 –x2 – x + 2) – (5x3 – 3x2 + 2x – 5) = 3x3 – x2 – x + 2 – 5x3 + 3x2 – 2x + 5 = – 2x3 + 2x2 – 3x + 7 A– B ?B -A ii) (A – B) – C = (2x3 – 2x2 + 3x – 7) – (x3 – 5x + 4) = 2x3 – 2x2 + 3x – 7 – x3 + 5x – 4 = x3 – 2x2 + 8x – 11 A – (B – C) (B – C) = (3x3 – x2 – x + 2) – (x3 – 5x + 4) = 3x3 – x2 – x + 2 – x3 + 5x – 4 = 2x3 – x2 + 4x – 2 A – (B – C) = (5x3 – 3x2 + 2x – 5) – (2x3 –x2 + 4x – 2) = 5x3 – 3x2 + 2x – 5 – 2x3 + x2 – 4x + 2 = 3x3 – 2x2 – 2x – 3 iii) Subtraction of polynomials does not obey commutative property. iv) Subtraction of polynomials does not obey associative property. 13. (a) Let ‘y’ be added to 2x 2 + 6x – 5 to get 3x2 – 2x + 6  y + (2x2 + 6x – 5) = 3x2 – 2x + 6  y = 3x2 – 2x + 6 – (2x2 + 6x – 5) = 3x2 – 2x + 6 – 2x2 – 6x + 5 = x2 – 8x + 11  x2 – 8x + 11 must be added to 2x2 + 6x – 5 to get 3x2 – 2x + 6 (b) Let ‘y’ be subtracted from a2 + b2 +c2 – 3abc to get 2a2 – b2 – 3c2 + abc (a2 + b2 +c2 – 3abc) – y = 2a2 – b2 – 3c2 + abc  – y = 2a2 – b2 – 3c2 + abc – (a2 +b2 +c2 – 3abc) = 2a2 – b2 – 3c2 + abc – a2 – b2 – c2 + 3abc – y = a2 – 2b2 – 4c2 + 4abc  y = – a2 + 2b2 + 4c2 – 4abc  – a2 + 2b2 + 4c2 – 4abc must be subtracted from a2 + b2 + c2 – 3abc to get 2a2 – b2 – 3c2 + abc

14. a) (x2 + y2 + z2) × (xy – yz + zx + 3) = x2 (xy – yz + zx + 3) + y2 (xy – yz + zx + 3) + z2 (xy – yz + zx + 3) = x3y – x2yz + zx3 + 3x2 + xy3 – y3z + xy2z + 3y2 + xyz2 – yz3 + z3x + 3z2 = x3y + zx3 + xy3 – y3z – yz3 + z3x – x2yz + xy2z + xyz2 + 3x2 + 3y2 + 3z2 = x3 (y + z) + y3 (x – z) + z 3 (x – y) + xyz (– x + y + z) + 3 (x2 + y2 + z2) b) (x2 + 3x + 1) × (x2 + 3x + 2) = x2 (x2 + 3x + 2) + 3x (x2 + 3x + 2) + 1 (x2 + 3x + 2) = x4 + 3x3 + 2x2 + 3x3 + 9x2 + 6x + x2 + 3x + 2 = x4 + (3x3 + 3x3) + (2x2 + 9x2 + x2) + (6x + 3x) +2 = x4 + 6x3 + 12x2 + 9x + 2. c) (4p2 + 2p – 5) × (9p2 – 4p + 2) = 4p2 (9p2 – 4p + 2) + 2p (9p2 – 4p + 2) – 5 (9p2 – 4p + 2) = 36p4 – 16p3 + 8p2 + 18p3 – 8p2 + 4p – 45p2 + 20p – 10 = 36p4 – 2p3 – 45p2 + 24p – 10 d) (y2 + 2y + 2) × (6y2 + 3) = y2 (6y2 + 3) + 2y (6y2 + 3) + 2 (6y2 + 3) = 6y4 + 3y2 + 12y3 + 6y + 12y2 + 6 = 6y4 + 12y3 + 15y2 + 6y + 6 e) (m3 + 4m2 + 2m + 1) × (4m3 + m2 + 5m + 9) = m3 (4m3 + m2 + 5m + 9) + 4m2 (4m3 + m2 + 5m + 9) + 2m (4m3 + m2 + 5m +9) + 1 (4m3 + m2 + 5m + 9) = 4m6 + m5 + 5m4 + 9m3 + 16m5 + 4m4 + 20m3 + 36m2 + 8m4 + 2m3 + 10m2 + 18m + 4m3 + m2 + 5m + 9 = 4m6 + 17m5 + 17m4 + 35m3 + 47m2 + 23m + 9 15. (2x2 – 3x + 4) × (3x2 – 2x + 1) = 2x2 (3x2 – 2x + 1) – 3x (3x2 – 2x + 1) + 4 (3x2 – 2x + 1) = (6x4 – 4x3 + 2x2) – (9x3 – 6x2 + 3x) + (12x2 – 8x + 4) = 6x4 – 4x3 + 2x2 – 9x3 + 6x2 – 3x + 12x2 – 8x +4 = 6x4 – 13x3 + 20x2 – 11x + 4

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8th Class Mathematics

290 16. p(x) × q(x) = (x2 – 2x + 1) × (x3 – 3x2 + 2x – 1)

20. Given that a2 + b2 + c2 = ab + bc + ca

= x2 (x3 – 3x2 + 2x – 1) – 2x (x3 – 3x2 + 2x – 1) 3

 a2 + b2 + c2 – ab – bc – ca = 0

2

+ 1 (x – 3x + 2x – 1)

But a2 + b2 + c2 – ab – bc – ca = 0 1  [(a – b)2 + (b – c)2 + (c – a)2 ] 21  [(a – b)2 + (b – c)2 + (c – a)2 ] = 0 2  (a – b)2 + (b – c)2 + (c – a)2 = 0

= (x5 – 3x4 + 2x3 – x2) – (2x4 – 6x3 + 4x2 – 2x) + (x3 – 3x2 + 2x – 1) = x5 – 3x4 – 2x4 + 2x3 + 6x3 + x3 – x2 – 4x2 – 3x2 + 2x + 2x – 1 = x – 5x + 9x – 8x + 4x – 1.

If the sum of the squares are zero then each square is 0

 The degree of p(x) × q(x) is ‘5’.

a–b=0?a=b

5

4

3

2

2

1 2  1 1 17. i)  x     x   2  x       x  x x

 x2  2 

a=b=c 21. a + b = c

1 x2

Cubing on both sides, (a + b)3 = (c)3 2

2

Formula used : (a + b) = a + 2ab + b ii)

Similarly, b – c = 0 ? b = c also c = a

2

2

2

2

a3 + b3 + 3ab (a + b) = c3 2

(n – 1) (n + 1) (n + 1) = (n – 1) (n + 1)

a3 + b3 + 3abc = c3

= (n2)2 – (1)2 = n4 – 1

(a + b = c)

Formula used : a2 – b2 = (a + b) (a – b) 2

iii)

1 2  1 1  2x     2x   2  2x      x   x  x

22.

2

1 x2 Formula used : (a – b)2 = a2 – 2ab + b2  4x2  4 

2

18. i)

1  1   50    50   2  2 

2

2 1 1   50   2  50       2 2

 2500  50  ii)

2

1 1  2550 4 4

107 × 93 = (100 + 7) (100 – 7) = (100)2 – (7)2 = 10,000 – 49 = 9,951 Formula used : a2 – b2 = (a + b) (a – b)

iii)

(96)2 = (100 – 4)2 = (100)2 – 2(100) (4) + (4)2 = 10,000 – 800 + 16 = 9216 Formula used : (a – b)2 = a2 – 2ab + b2

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3

23.

3

x3  8 y 3  x    2 y   x  2y  x  2 y

Applying the identity a3 + b3 = (a + b) (a2 – ab + b2) to (x)3 + (2y)3 Where, a = x and b = 2y, we get,

 x  2 y   x2  x  2 y   2 y  R.H .S   x  2y 

x3  8 y 3 3 2 x  2 y  x  2 xy  4 y

2

 

Polynomials Solutions

24.

291 26. 4 (p – q)2 – 12 (p – q) (p + q) + 9 (p + q)2 by p + 5q

x 7  64 x x 7  64 x  x  x  2 x  4  x  x  2x  4



2

4  p  q   12  p  q  p  q   9  p  q   p  5q

x  x 6  64  x  x  2 x  4 2

Arranging x6 – 64 as (x3)2 – (8)2 and applying the identity a2 – b2 = (a + b) (a – b) where a = x3 and b = 8, we get, 3 2

 x    8      2  x  2x  4

x  3

3

2

2



2

 23  x 3  23 

p 

  p 2

2q

2q) 2

2

 2q 2 

p  2q

 p 

2q



2q  p 2  2q 2 

p  2q



 p  2q

x2  2x  4 = (x3 – 23) (x + 2)

 p

2

 2q 2 

28. 8p3 – 36p2q + 54 pq2 – 27q3 by 4p2 – 12pq + 9q2



8 p3  36 p 2 q  54 pq 2  27q 3 4 p 2  12 pq  9q 2

xy  x  y  x 7 y  xy 7   x  y   x 2  xy  y 2   x  y   x 2  xy  y 2 

By rearranging the terms, we have,

Adjusting x6 – y6 as (x3)2 – (y3)2 and applying a2 – b2 = (a + b) (a – b) where a = x3 and b = y3, we get,



6

25.

p  5q

p  2q 2

x 3  23   x  2   x 2  2 x  4   x 6  64  2   x  2 x  4  x2  2 x  4 

p  5q

 2q 2  p 2  2q 2 

2

 p  

3 2

x2  2 x  4



p

  p  5q 2

 p  5q 2

2

 x   2  

3





2 2 p 4  4q 4  p    2q   p  2q p  2q

a + b = (a + b) (a – ab + b ) Applying the identity

x

2

2

= p + 5q 27. Given (p4 – 4q4)  (p –

 2 x  4

3

3 2

 2  p  q   3  p  q    p  5q  1 p  5q    p  5q

 23  x 3  23 

x

(2(p – q))2 – 2.2(p – q) 3(p + q) + (3(p + q))2 [  a2 – 2ab + b2 = (a – b)2, where a = 2(p – q) and b = 3(p + q)]

2

x 3  8  x 3  8  x 6  64   x2  2 x  4  x2  2 x  4 





xy  x 6  y 6 

 x  y   x 2  xy  y 2  xy  x 3  y 3  x 3  y 3 

 x  y   x 2  xy  y 2 

2

6

2 2 xy  x 3    y 3      2  x  y   x  xy  y 2 



xy  x 3  y 3  x 3  y 3 

x

3

 y3 

[since, a3 + b3 = (a + b) (a2 – ab + b2)] = xy (x3 – y3)

8 p 3  27q 3  36 p 2 q  54 pq 2

 2 p

8 p 

3

2

 12 pq   3q 

2

 27q3   18 pq  2 p  3q 

 2 p

2

 12 pq   3q 

2

  2 p 3   3q 3   18 pq  2 p  3q    2 2  2 p   2  2 p  3q   3q  

 2 p  3q   2 p 2   2 p  3q    3q 2   18 pq  2 p  3q   2 p  3q 2 www.betoppers.com

8th Class Mathematics

292 [ a3 – b3 = (a – b) (a2 + ab + b2) and a2 – 2ab + b2 = (a – b)2] 3





2

2

8 p  36 p q  54 pq  27q 4 p 2  12 pq  9q 2

3

 2 p  3q  4 p 2  6 pq  9q 2  18 pq  2  2 p  3q 

30. (i) 6(2a + 3b)2 – 8 (2a + 3b) = (2(2a + 3b) [3(2a + 3b) – 4] Taking out (2a + 3b) common, = (2(2a + 3b) (6a + 9b – 4).  The two factors of the given polynomial are 2(2a + 3b), (6a + 9b – 4) (ii) x(x – y)3 + 3x2 y(x – y)



4 p 2  12 pq  9q 2 2 p  3q

= x(x – y) [(x – y)2 + 3xy] Taking out x(x – y) common,

2 2 2 p   2  2 p  3q   3q   

2 p  3q

2 2 p  3q     2 p  3q

2 p  3q

= x(x – y) (x2 + y2 – 2xy + 3xy) = x(x – y) (x2 + y2 + xy).  The two factors of the given polynomial are x(x – y), (x2 + y2 + xy). 31. (i) x3 – x2 – xy + x + y – 1

3

2 2

29. i) Given 36y z + 48y z

= (x3 – x2) – (xy – y) + (x – 1)

2

By taking out y z common, we get  y2z (36y + 48z) The highest common factor of 36 and 48 is 12. i.e. we can take 12 common from 36 and 48.

= x2 (x – 1) – y(x – 1) + (x – 1) Taking out (x – 1) common, = (x – 1) (x2 – y + 1).

 12y2z (3y + 4z)

 The two factors of the given polynomial are (x – 1), (x2 – y + 1).

 The two factors of 36y3z + 48y2z2 are (12y2z) and (3y + 4z).

(ii) ax – (ax + by)2 + a2x + aby + by

ii) Given 15y2z3 – 20y3z4 + 35y2z2. By taking out y2z2 common, we get.  y2z2 (15z – 20yz2 + 35) The highest common factor of 15, 20 and 35 is 5. i.e., we can take 5 common.  (5y2z2) (3z – 4yz2 + 7)  The two factors of the given polynomial are (5y2z2) and (3z – 4yz2 + 7)

= (ax + by) – (ax + by)2 + (a2x + aby) = (ax + by) – (ax + by)2 + a (ax + by) = (ax + by) [1 – (ax + by) +a] = (ax + by) (1 + a – ax – by).  The two factors of the given are (ax + by), (1 + a – ax – by). 32. (x2 + 2x)2 – 3(x2 + 2x) –y(x2 + 2x) + 3y = [(x2 + 2x)2 – 3(x2 +2x)] – [y(x2 + 2x) – 3y]

iii) Given 14m5n4p2 – 42m7n3p7 – 70m6n4p3

Taking out (x2 + 2x) common

By taking out m5n3p2 common, we get

= (x2 + 2x) (x2 +2x – 3) – y(x2 + 2x – 3)

 m5n3p2 (14n – 42m2p5 – 70mnp)

Taking out (x2 + 2x – 3) common

The highest common factor of 14, 42 and 70 is 14 i.e. we can take 14 common.

= (x2 + 2x – 3) (x2 + 2x – y).

 14m5n3p2 (n – 3m2p5 – 5mnp)  The two factors of the given polynomial are (14m5n3p2), (n – 3m2 p5 – 5mnp) www.betoppers.com

 The two factors of the given are (x2 + 2x – 3), (x2 + 2x – y).

Polynomials Solutions 33. x2 – z2 – 2xy + 2yz = x2 – 2xy – z2 + 2yz Adding and subtracting y2, we get, = x2 – 2xy + y2 – y2 –z2 + 2yz = (x2 – 2xy + y2) – (y2 + z2 – 2yz) = (x – y)2 – (y – z)2 (a)2 – (b)2 = (a +b)(a – b)) = (x – y + y – z) (x – y – y + z) = (x – z) (x – 2y + z) The two factors of the given polynomial are (x – z), (x – 2y + z). 34. a4 + 4(a – 1)2 – 4 (a3 – a2) = a4 + {2(a – 1)}2 – 4a2 (a – 1) = (a2)2 – 2a2.2 (a – 1) + {2(a – 1)}2 Clearly, it is in the form of x2 – 2xy + x, other x = a2 and y = 2(a – 1). = {a2 – 2(a – 1)}2 = (a2 – 2a + 2)2  The two factors of the given are (a2 – 2a + 2), (a2 – 2a + 2). 35. i) Using the product (a + b)2 above with a = x, b = 3 we have x2 + 6x + 9 = (x + 3)2  The two factors are (x + 3), (x + 3) ii) Using the product (a – b)2 above with a = 1, b = 4x we have 1 – 8x + 16x2 = (1 – 4x)2  The two factors are (1 – 4x), (1 – 4x) iii) Using the product (a2 – b2) above with a = 2x, b = 9y we get 4x2 – 81y2 = (2x)2 – (9y)2 = (2x – 9y) (2x + 9y) The two factors are (2x – 9y), (2x + 9y). iv) Using the product (a3 + b3) with a = xy, b = 9 we get x3y3 + 729 = (xy)3 + (9)3 (a3 + b3) = (a + b)(a2 – ab + b2) = (xy + 9) (x2y2 – 9xy +81)  The two factors are (xy + 9), (x2y2 – 9xy +81). v) Using the product (a – b)3 with a = x, b = 2, we have x3 – 6x2 + 12x – 8 = x3 – 3(x2)(2) + 3(x)(2)2 – (2)3

293 It is in the form of a2 – 3a2b + 3ab2 + b3 = (a – b)3 (here a = x, b = 2)  (x – 2)3  The three factors are (x – 2), (x – 2), (x – 2). vi) Using the product a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) with a = x, b = y, c = –z we have x3 + y3 – z2 + 3xyz = (x + y – z) (x2 + y2 + z3 – xy + yz + zx)  The two factors are (x + y – z), (x2 + y2 + z2 – xy + yz + zx). vii) n4 + 4 is in the form of a2 + b2 with a = n2, b = 2 (N2)2 + 22 = a2 + b2 = (a2 + b2 + 2ab) – 2ab = (n4 + 4 + 2n2 (2)) – 2n2 (2) = (n2 + 2)2 – 4n2 = (n2 + 2)2 – (2n)2, Which is in the form a2 – b2 with a = n2 + 2, b = 2n  (n2 + 2) – (2n)2 = (n2 + 2n + 2) (n2 – 2n + 2)  n4 + 4 = (n2 + 2n + 2) (n2 – 2n + 2)  The two factors are (n2 + 2n + 2), (n2 – 2n + 2). 36. (2x + 3y)2 + 2(2x + 3y) (x + y) + (x + y)2 Put 2x + 3y = a ; x + y = b  Given expression = a2 + 2ab + b2 = (a + b)2 = (2x + 3y + x + y)2 = (3x + 4y)2 The two factors of the given polynomial are (3x + 4y), (3x + 4y). 37. a2 b2 + c2 d2 – a2 c2 – b2 d2 = a2 b2 – a2 c2 – b2 d2 + c2 d2 = a2 (b2 – c2) – d2 (b2 – c2) Taking out (b2 – c2) common = (a2 – d2) (b2 – c2) = (a + d) (a – d) (b + c) (b – c) The factors of the given polynomial are (a + d), (a – d), (b + c), (b – c). 38. (i) 9 – a6 + 2a3 b3 – b6 = 9 – (a6 – 2a3 b3 + b6) = 32 – {(a3)2 – 2 × a3 × b3 + (b3)2} = 32 – (a3 – b3)2 (This is in the form of (a – b)2 where a= a3, b = b3 ) = {3 + (a3 – b3)} {3 – (a3 – b3)} www.betoppers.com

8th Class Mathematics

294 ( (a2 – b2) = (a + b)(a – b)) = {3 + (a3 – b3) (3 – a3 + b3) = (a3 – b3 + 3) (– a3 + b3 +3) The two factors are (a3 – b3 + 3), (– a3 + b3 +3). (ii) x16 – y16 + x8 + y8 = {(x8)2 – (y8)2} + (x8 + y8) = (x8 – y8) (x8 + y8) + (x8 + y8) Taking out (x8 – y8) common = (x8 + y8) (x8 – y8 + 1) The two factors are (x8 + y8), (x8 – y8 + 1). (iii) (p + q)2 – (a – b)2 + p + q – a + b = {(p + q)2 – ( a – b)2} + (p + q) – (a – b) = {(p + q) + (a – b)} {(p + q) – (a – b)} + {(p +q) – (a – b)} ( (a2 – b2) = (a + b)(a – b)) = (p + q + a –b) (p + q – a + b) + (p + q – a + b) = (p + q – a + b) (p + q + a – b +1)  The two factors are (p + q – a + b), (p + q + a – b +1). 39. (x + y)3 + z3 Clearly, it is in the form of a3 + b3 where a = (x + y), b = z. We have a3 + b3 = (a + b) (a2 – ab + b2) = {(x + y + z)} {(x + y)2 – (x + y)z + z2} = (x + y + z) (x2 + 2xy + y2 – xz – yz + z2) = (x + y + z) (x2 + 2xy + y2 – xz – yz + z2)  The two factors are (x + y + z), (x2 + 2xy + y2 – xz – yz + z2). 40. x2 + 7x + 12 7x is to be written as the sum of the two terms such that their product is 12x2.  3x + 4x = 7x; (3x) × (4x) = 12x2 = x2 + (3x + 4x) + 12 = (x2 + 3x) + (4x + 12) = x (x + 3) + 4 (x + 3) Taking out (x + 3) common = (x + 4) (x + 3)  The two factors are (x + 4), (x + 3). 41. 7x2 – 8x – 12 = 7x2 – 14x + 6x – 12 –8x is to be written as the sum of the two terms such that their product of them is 7x2 × 12 = 84x2.  – 14x + 6x = – 8x and – 14x × 6x = 84x2 = 7x – 14x + 6x – 12 www.betoppers.com

= 7x (x – 2) + 6 (x – 2) Taking out (x – 2) common = (x – 2) (7x + 6) The two factors are (x – 2), (7x + 6). 42. We have x4 – 2x3 + 2x2 – 2x + 1 = (x4 + 1) – (2x3 + 2x) + 2x2 = ((x2)2 + (1)2) – 2x (x2 + 1) + 2x2 = (x2 + 1)2 – 2x2.1 – 2x (x2 + 1) + 2x2 (?(a2 + b2) = (a + b)2 – 2ab, where a = x2, b = 1 = (x2 + 1)2 – 2x (x2 + 1) Taking out (x2 + 1) common = (x2 + 1) (x2 + 1 – 2x) = (x2 + 1) (x – 1)2 The factors are (x2 + 1), (x – 1), (x – 1). 43. When 1 is substituted for a in the given expression it vanishes. Thus (a – 1) is a factor of the expression. Thus a3 – 2a2 – 5a + 6 By splitting midterms a3 – a2 – a2 + a – 6a + 6 = a2 (a – 1) – a (a – 1) – 6 (a – 1) Taking common (a – 1) = (a – 1) (a2 – a – 6) = (a – 1) (a2 – 3a + 2a – 6) (By splitting the middle term method) = (a – 1) [a ( a – 3) + 2 (a – 3)] Taking (a – 3) common = (a – 1) (a – 3) (a + 2)  The factors are (a – 1), (a – 3), (a + 2). 1 1 44. Given, x 2  2  3   x 2  2  2  1 [Rewriting – x x   3 as (– 2) + (– 1)].

1 . x Thus replacing 2, we get, We have, 2 = 2 × x ×

2

1 2  1  2 1 2 R.H.S   x  2  2. x.   1   x    1 . x x x   2 2 This is in the form of a – b = (a + b) (a – b),

where, a = x –

 x2 

1 and b = 1. x

1  1  1   3   x   1  x   1 x2 x  x  

 The factors of

x2 

1  3 are x2

 1   1   x  x  1 and  x  x  1 .

Polynomials Solutions

295

45. Given dividend = y3 + y2 – 2y + 1 divisor = y – a

 Quotient = y2 + (1 + a) y + [a(1 + a) – 2] = y2 + (1 + a)y + (a2 +a – 2) Remainder = a [a(1 + a) – 2] + 1 = a[a2 + a – 2] + 1 = a3 + a2 – 2a + 1 By division algorithm, we have Dividend = Divisor × Quotient + Remainder  y3 + y2 – 2y +1 = (y – a) [y2 + (1 + a)y + (a2 + a – 2)] + (a3 + a2 – 2a + 1) = (y – a)y2 + (y – a) (1 + a)y + (y – a) (a2 + a – 2) + (a3 + a2 – 2a + 1) = y3 – ay2 + (y2 – ay) (1 + a) + ya2 – a3 + ay – a2 – 2y + 2a + a3 + a2 – 2a + 1 = y3 – ay2 + y2 – ay + ay2 – ay2 + a2y + ay – 2y + 1 = y3 + y2 – 2y + 1  L.H.S = R.H.S. Hence verified. 46. i) Given p(x) = 2x3 – 5x2 + 3x + 7 and divisor = 2x –1 1  Zero of divisor = 2

1  Remainder  P    2 3

2

1 1 1  2     5    3   7 2 2 2

1 1 1  2   5   3   7 8 4 2 

1 5 3 1  5  6  28 30   7  4 4 2 4 4

ii) Divisor = x – 2 ? zero of divisor = 2  Remainder = p(2) = 2(2)3 – 5(2)2 + 3(2) + 7 = 16 – 20 + 6 + 7 = 9 iii) Divisor = x + 2  zero of divisor = – 2 Remainder = p(– 2) = 2 (– 2)3 – 5 (– 2)2 + 3(– 2) + 7 = – 16 – 20 – 6 + 7 = – 35 47. Given, g(a) = a6 – 3a4 + 24a2 – 3 g(–1) = (–1)6 + 3(–1)4 – 24 (–1 )2 – 3 = 1 + 3 – 24 – 3 = – 23 g(–2) = (–2)6 + 3(–2)4 – 24 (–2)2 – 3 = 64 + 48 – 96 – 3 = 13 g(–3) = (–3)6 + 3(–3)4 – 24 (–3)2 – 3 = 729 + 243 – 216 – 3 = 753 48. Divisor = x + 3 x + 3 = 0  x = – 3.  Zero of x + 3 is – 3. Remainder = – 20 [Given]  P (–3) = – 20  k(–3)3 + 9(–3)2 + 4(–3) –8 = –20  –27k + 81 – 12 – 8 = –20  27k = 81 81 k 3 27 49. Let f(x) = (x + 1)7 + (2x + k)3 f(x) is divisible by x + 2.  f(–2) = 0 ? (–2 + 1)7 + (–4 + k)3 = 0  –1 + (k – 4)3 = 0  (k – 4)3 = 1 k–4=1 k=5 50. Let f(x) = kx3 + 3x2 – 3 and g(x) = 2x3 – 5x + k Divisor = x – 4 x–4=0x=4  Zero of x – 4 is 4. f(4) = k(4)3 + 3(4)2 – 3. = 64k + 48 – 3 = 64k + 45 g(4) = 2(4)3 – 5(4) + k www.betoppers.com

8th Class Mathematics

296 = 128 – 20 + k = 108 + k. Given that f(4) = g(4)  64k + 45 = 108 + k  64k – k = 108 – 45  63k = 63 k=1  The value of k = 1

5. 6.

CONCEPTIVE WORKSHEET 1.

Let us recall the definition of polynomials. If f(x) = a0 + a1x + a2 x2 + ..... + anxn Where a0, a1, a2, .... an are real numbers and n is a non negative integer (i.e., ‘n’ should be a positive integer)it is called a polynomial in x, of nth degree. The 2nd term in (iii) i.e. x x  x  x1/ 2  x 3 / 2 here the power of ‘x’ is not a positive integer. Hence it cannot be a polynomial. Also 1st term of (v) is u–1/2; here the power of ‘u’ is not a positive integer. Hence it cannot be a polynomial. II is not a polynomials.

2.

3.

4.

(iii) x5 + 2x4 + x2 – 3x (v) u4 – 3u2 + 6u + 2

(iv) – t4 – 3t3 + 2t + 6

=

=

1 3 x 2

1 ×2×2×2=4 2

When x = 3 ; the value of

1 3 x 2

1 × 3 × 3 × 3 = 13.5 2

When x = – 1.5, the value of

1 3 x 2

1 × –1.5 × – 1.5 × –1.5 2 = – 1.6875 (iv) 2x3 When x = 2, then the value of 2x3 = 2 × (2)3 = 2 × 8 = 16 When x = 3 ; then the value of 2x3 = 2 × (3)3 = 2 × 27 = 54 When x = – 1.5, then the value of 2x3 = 2 × (–1.5)3 = – 6.75 P(x) = 2x + 5 P(x) = 0  2x + 5 = 0 2x = –5 =

7.

x 

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1 3 x 2

When x = 2, the value of

1/ 2

1 11 x 5

b) 9, x, 4x2, 3x3, 5x6, 7x9, 6x15 (i) 3x2 When x = 2, the value of 3x2 = 3 × (2)2 = 12 When x = 3, the value of 3x2 = 3 × (3)2 = 27 When x = – 1.5, the value of 3x2 = 3 × (–1.5)2 = 6.75 (ii) – 1.2 x2 When x = 2, the value of –1.2 x2 = – 1.2 × (2)2 = – 4.8 When x = 3, the value of – 1.2 x2 = –1.2 × (3)2 = – 10.8 When x = – 1.5, the value of – 1.2 x2 = – 1.2 × (–1.5)2 = –2.7 (iii)

The 2nd term in (ii) i.e., x x  x  x1/ 2  x 3 / 2 ; here the power of ‘x’ is not a positive integer. Hence it cannot be a polynomial. Also 1st term of (v) is 2x  2x ; here the power of ‘x’ is not a possible integer. Hence it cannot be a polynomial. i) x2 + 4x2 + 4x3 + x Degree = 3 Ascending order: x + 5x2 + x2 + 4x3 i.e., x + 5x2 + 4x3 ii) t4 – 3t3 – 6t + 2 Degree = 4 Ascending order: 2 – 6t – 3t3 – t4 iii) 7x8 + 8x12 + x5 + 9x2 + 4x + 3 Degree = 12 Ascending order: 3 + 4x + 9x2 + x5 + 7x8 + 8x12 iv) 12x11 + 15x9 + 2x3 + 11x4 + 3x Degree = 11 Ascending order: 3x + 2x3 + 11x4 + 15x9 + 12x11 . We know that, in the standard form of a polynomial, the terms are in the descending order of the powers of the variable. i) –2y4 – y3 + y2 + 3y + 4 (ii) x4 – 4x3 + 5x + 4

a) 4, x, 3x2, 2x3, 9x5, – 5x7, 5x10,

5 2

5 is a zero of the polynomial P(x). 2

Polynomials Solutions 8.

9.

297

(i) (–3x2) + (6x2) – (0.5x2) + (1.5x2) = –3x2 + 6x2 + 0.5 x2 + 1.5 x2 = (–3 + 6 + 0.5 + 1.5) x2 = 5x2 (ii) (–3x) + (–4x) – (–4.5) x + (2.5 x) = (–3 – 4 – 4.5 + 2.5) x = –9x (iii) (3x) + (–4x) – (–3x) + (–7x) = 3x – 4x + 3x – 7x = (3 – 4 + 3 – 7) x = – 5x (iv) (–5x2) + (5.2x2) + (1.5x2) – (0.7x2) = (–5 + 5.2 + 1.5 – 0.7) x2 = (6.7 – 5.7) x2 = (1) x2 = x2 (a)

13. A = 3x2 – 4x + 5; B = x2 – 2x – 1; C = 3x – 4 (i) A × B = 3x2 – 4x + 5 x2 – 2x – 1 Multiply by x2 : 3x4 – 4x3 + 5x2 – 6x3 + 8x2 – 10x

Multiply by –2x :

Multiply by – 1 : – 3x2 + 4x – 5 3x4 – 10x3 + 10x2 – 6x – 5

B ×A= x2 – 2x – 1 3x2 – 4x + 5 Multiply by 3x2 : 3x4 – 6x3 – 3x2

x 2 – 2x + 1

Multiply by –4x :

– 4x3 + 8x2 + 4x

Q(x) = x3 – 3x 2 – 2x – 1

Multiply by + 5 :

5x2 – 10x – 5

P(x) =

3x4 – 10x3 + 10x2 – 6x – 5

P(x) + Q(x) = x 3 – 2x 2 – 4x + 0

 A× B = B ×A (b)

P(x) =

4t3

–

3t2

+2

Q(x) =

t4 – 2t3 + 0t2 + 6

R (x) =

t3 + 4t2 – 4

P(x) + Q(x) + R(x) =

t4 + 3t3 + t2 + 4

10. Let ‘y’ be added to 2x2 + 6x – 5 + 6  y + (2x2 + 6x – 5) = 3x2 – 2x + 6  y = (3x2 – 2x + 6) – (2x2 + 6x – 5) = 3x2 – 2x + 6 – 2x2 – 6x + 5 = (3 – 2)x2 + (–2 – 6)x + 6 + 5 = x2 – 8x + 11  x2 – 8x + 11 must be added to 2x2 + 6x – 5 to get 3x2 – 2x + 6 11. a) Let a2 – 2ab – b2 be more than 2a2 + 2ab – 2b2 by P.  (a2 – 2ab – b2) = 2a2 + 2ab – 2b2 + P

 P = (a2 – 2ab – b2) – (2a2 + 2ab – 2b2)  P = – a2 – 4ab + b2 b) Let 2(a2 + b2) be less than a2 + 2ab + b2 by Q.  (a2 + 2ab + b2) – Q = 2(a2 + b2)  (a2 + 2ab + b2) – Q = 2a2 + 2b2  Q = – 2a2 – 2b2 + a2 + 2ab + b2  Q = – a2 + 2ab – b2 12. We have, (y2 – 5y + 2) × (3y2 + 2y – 5) = 3y4 + 2y3 – 5y2 – 15y3 – 10y2 + 25y + 6y2 + 4y – 10 = 3y4 – 15y3 + 2y3 + 6y2 – 10y2 –5y2 + 4y + 25y – 10 = 3y4 – 13y3 – 9y2 + 29y – 10.

(ii) C × A = 3x – 4 3x2 – 4x + 5 Multiply by 3x2 : 9x3 – 12x2 Multiply by – 4x : – 12x2 + 16x Multiply by 5 : + 15x – 20 9x3 – 24x2 + 31x – 20

A×C= 3x2 – 4x + 5 3x – 4 3 2 Multiply by 3x : 9x – 12x + 15x Multiply by – 4 : – 12x2 + 16x – 20 9x3 – 24x2 + 31x – 20

 C ×A= A× C (iii) B × C = x2 – 2x – 1 3x – 4 Multiply by 3x : – 6x2 – 3x Multiply by – 4 : – 4x2 + 8x + 4 3 3x – 10x2 + 5x + 4 3x3

C×B= 3x – 4 x2 Multiply by

x2

:

3x3

– 2x – 1 – 4x2 – 6x2 + 8x – 3x + 4

Multiply by– 2x : Multiply by– 1 :

3x3 – 10x2 + 5x + 4

 B×C=C×B

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8th Class Mathematics

298 14. a + b = 8; ab = 15 (a + b)3 = a3 + 3a2 b + 3ab2 + b3 = a3 + b3 + 3ab (a + b) 3 3  a + b = (a + b)3 – 3ab (a + b) = (8)3 – 3 (15) (8) = 512 – 360 = 152  Multiplication of polynomials satisfies commutative 15. a + b = – c cubing on both sides. (a + b)3 = (–c)3 a3 + b3 + 3ab (a + b) = – c3 a3 + b3 + 3ab (–c) = – c3 ( a + b = – c)  a3 + b3 – 3abc = – c3 16. x + y + z = 0  x+y=–z cubing on both sides, (x + y)3 = (–z)3 x3 + y3 + 3xy (x + y) = – z3 x3 + y3 + 3xy (–z) = – z3  x3 + y3 – 3xy + z2 = 0  x3 + y3 + z3 = 3 xyz 17.

19.

Divisor

Dividend

x – 2 ) x3 – 5x2 + 11x – 10 (x2 – 3x + 5 x3 – 2x2 – +  Quotient – 3x2 + 11x – 3x2 + 6x + – 5x – 10 5x – 10 0  Remainder

Verification : Dividend = divisor × quotient + Remainder = (x – 2) × (x2 – 3x + 5) + 0 = x3 – 5x2 + 11x – 10 = Dividend 20. 4x2 + 2x – 3 ) 8x4 – 8x3 – 10x2 + 15x + 2 ( 2x2 – 3x + –

8x4

4x3 –

+ – + – 12x3 – 4x2 + 15x – 12x3 – 6x2 + 9x 2x2 + 6x + 2 3 2x2 + x –

1 2

2

1 x 3 x

1 5x + 3 2

1 ? x3 Cubing on both sides. x3 

Verification : (4x2 + 2x – 3) × (2x2 – 3x +

1 ) 2

7 2 = 8x4 – 12x3 + 2x2 + 4x3 – 6x2 + x – 6x2

3

+ 5x +

1  3 x  3 x 

x3 

1 1 1  3  x   x    27 3 x x x

x3 

1  3  3  27 x3

1  27  9  18 x3 18. a + b + c = 5 ; a2 + b2 + c2 = 29 a2 + b2 + c2 + 2 (ab + bc + ca) = (a + b + c)2 29 + 2 (ab + bc + ca) = (5)2 = 25  2 (ab + bc + ca) = 25 – 29 = – 4  x3 

 ab + bc + ca = –

6x2

4 = –2 2

3 7 + 5x + 2 2 4 3 2 = 8x – 8x – 10x + 15x + 2 = Dividend. 21. Given dividend = x4 + x3 – 9x2 – 3x + 5 divisor = x2 + 4x + 2. + 9x –

x2 – 3x + 1 x2 + 4x + 2

x4 + x3 – 9x2 – 3x + 5 x4 + 4x3 + 2x2

– –

–

3x3

– – 11x2 – 3x + 5 – 3x3 – 12x2 – 6x +

+

+

–

x2 + 3x + 5 x2 + 4x + 2 – – – x+3

Here, x2 – 3x + 1 is the Quotient and (–x + 3) is the Remainder. Remainder (–x + 3) is a polynomial of degree less than that of the divisor x2 + 4x + 2. www.betoppers.com

Polynomials Solutions

299

a 3  a 2  2a  8 a2 Rearranging the terms and simplifying by applying the identity

22. Given

3 2 a 3  a 2  2a  8  a  8   a  2a   a2 a2

a  



3

25.

4x 2  17xy  4y 2  x  4y 





a2

a  2  a

2



a2 2

2

  a  a  2

a2

 a  2   a

2

4x  25y  30y  9  2x  5y  3



 2x 

2

2

 2a  2

2

  a 

 2x 

2

  25y2  30y  9  2x  5y  3

2 2   5y   2  5y  3   3    2x  5y  3

(5y)2 – 2× 5y × 3 + (3)2 is of the form a2 – 2ab + b2 = (a – b)2 2

 2x    5y  3   2x  5y  3

2

Applying the identity a2 – b2 = (a + b) (a – b) to numerator when a = 2x and b = 5y – 3, we get, 

 2x  5y  3 2x  5y  3 2x  5y  3 2

24.

 2x  5y  3

3 3 x 6  729 x 6  36  x    3    2 x2  9 x2  9 x 2   3





x

3

x  4y

2

 33  x 3  33 

 x  3 x  3  x  3  x 2  3x  9   x  3  x 2  3x  9   x  3 x  3

= (x2 + 3x + 9) (x2 – 3x + 9)

 x  4y  4x  y  x  4y

  4x  y 

26. x7 + xy6 by x2 + y2



a2 = a – 2a + 4 + a = a2 – a + 4 23.

4x  x  4y   y  x  4y 

 23   a  a  2 

2

2

4x 2  16xy  xy  4y 2 x  4y



x 7  xy6 x 2  y2

x  x 6  y6  x 2  y2

3 3 x   x 2    y 2      2 2 x y

2 2 x  x 2  y2     x 2    x 2  y 2    y 2      x 2  y2

x 7  xy6  x  x 4  x 2 y 2  y4  x 2  y2 27. (i) 3x(a + 2b) – 2y (a + 2b) = (a + 2b) (3x – 2y). [ Taking out (a + 2b)] (ii) 2a (a2 + 1) – 3(a2 + 1) = (a2 + 1) (2a – 3). [ Taking out (a2 + 1)] 28. (i) We have, a2 – b + ab – a = a2 + ab – b – a = (a2 + ab) – (b + a) = a (a + b) – (a + b) = (a + b) (a – 1)  The two factors are (a + b), (a – 1). (ii) xy – ab + bx – ay = xy + bx – ab – ay = x (y + b) – a (b + y) = x (y + b) – a (y + b) = (y + b) (x – a) The two factors are (y + b), (x – a).  2 (iii) 6ab – b + 12ac – 2bc = 6ab + 12ac – b2–2bc = 6a (b + 2c) – b (b + 2c) = (b + 2c) (6a – b)  The two factors are (b + 2c), (6a – b). (iv) a (a + b – c) – bc = a2 + ab – ac – bc = (a2 + ab) – (ac + bc) = a (a + b) – c (a + b) = (a + b) (a – c)  The two factors are (a + b), (a – c). 

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8th Class Mathematics

300 (v) a2x2 + (ax2 + 1) x + a = a2x2 + ax3 + x + a = ax2 (x + a) + (x + a)  The two factors are (x + a), (ax2 + 1). (vi) 3ax – 6ay – 8by + 4bx = 3ax + 4bx – 6ay–8by = (3ax + 4bx) – (6ay + 8by) = x (3a + 4b) – 2y (3a + 4b) = (3a + 4b) (x – 2y) The two factors are (3a + 4b), (x – 2y).  29. We have, (i) a2 + 2a + ab + 2b = (a2 + 2a) + (ab + 2b) [ Grouping the terms] = a (a + 2) + (a + 2) b = (a + 2) (a + b) [ Taking (a + 2 common] The two factors are (a + 2), (a + b).  2 (ii) x – xz + xy – yz = (x2 – xz) + (xy – yz) [ Grouping the terms] = x (x – z) + y (x – z) = (x + y) (x – z) [ Taking (x – z) common] The two factors are (x + y), (x – z).  3 2 30. (i) We have, x – 2x y + 3xy2 – 6y3 = (x3 – 2x2y) + (3xy2 – 6y3) = x2 (x – 2y) + 3y2 (x – 2y) = (x – 2y) (x2 + 3y2)  The two factors are (x – 2y), (x2 + 3y2). (ii) We have, 6ab – b2 + 12ac – 2bc = b (6a – b) + 2c (6a – b) = (6a – b) (b + 2c)  The two factors are (6a – b), (b + 2c). 31. 4x2 – 4xy – z2 – 2yz = (2x)2 – 2. 2x. y + y2 – y2 – z2 – 2yz = (2x – y)2 – (y2 + z2 + 2yz) = (2x – y)2 – (y + z)2 = (2x – y + y + z) (2x – y – y – z) = (2x + z) (2x – 2y – z)  The two factors are (2x + z), (2x – 2y – z). 32. (i) x2 + 2xy + y2 – a2 + 2ab – b2 = (x2 + 2xy + y2) – (a2 – 2ab + b2) = (x + y)2 – (a – b)2 = {(x + y) + (a – b)} {(x + y) – (a – b)} = (x + y + a – b) (x + y – a + b)  The two factors are (x + y + a – b), (x + y – a + b). (ii) 25x2 – 10x + 1 – 36y2 = (5x)2 – 2 × 5x × 1 + 12 – (6y)2 = (5x – 1)2 – (6y)2 = (5x – 1 + 6y) (5x – 1 – 6y)  The two factors are (5x – 1 + 6y), (5x – 1 – 6y). (iii) 1 – 2ab – (a2 + b2) = 1 – (2ab + a2 + b2) = 1 – (a + b)2 www.betoppers.com

= {1 + (a + b)} {1 – (a + b)} = (1 + a + b) (1 – a – b) The two factors are (1 + a + b), (1 – a – b).  33. (x + y)3 + z3 = {(x + y + z)} {(x + y)2 – (x + y)z + z2} = (x + y + z) (x2 + 2xy + y2 – xz – yz + z2)  The two factors are (x + y + z), (x2 + 2xy + y2 – xz – yz + z2). 34. 7 = 4 + 3; 4 × 3 = 12 Therefore 2x2 + 7x + 6 = 2x2 + 4x + 3x + 6 = 2x (x + 2) + 3 (x + 2) = (x + 2) (2x + 3) The two factors are (x + 2), (2x + 3).  35. Let (x2 + x) be a.  The given expression becomes a2 – 8a + 12 – 6a – 2a = –8, 6a × 2a = 12a2  a2 – 6a – 2a + 12 = a (a – 6) – 2 (a – 6) = (a – 6) (a – 2) Substituting (x2 + x) in the place of ‘a’  (x2 + x – 6) (x2 + x – 2) = (x2 + 3x – 2x – 6) (x2 + 2x – x – 2) = [x (x + 3) – 2 (x + 3)] [x (x + 2) –1 (x + 2)] (x + 3) take out is first expression, (x + 2) in second expression = (x + 3) (x – 2) (x – 1) (x + 2)  The four factors are (x + 3), (x – 2), (x – 1), (x + 2). 36.

1 4 x Squaring on both sides x

2

1   x   = (4)2 = 16 x 

x2 +

1 + 2 = 16 x2

1 = 16 – 2 = 14 x2 Again squaring on both sides x2 +

2

 2 1   x  2  = (14)2 x  

x4 +

1 + 2 = 196 x4

x4 +

1 = 196 – 2 = 194 x4

Polynomials Solutions 37. (i) In order to factorize x2 – 23x + 132, Clearly, –12 –11 = – 23 and – 12 × –11 = 132. We now split the middle term – 23x of x2 – 23x + 132 as – 12x – 11x  x2 – 23x + 132 = x2 – 12x – 11x + 132 = (x2 – 12x) – (11x – 132) = x (x – 12) – 11 (x – 12) = (x – 12) (x – 11) (ii) In order to factorize x2 – 21x + 108, Clearly, –21 = –12 – 9 and – 12 × – 9 = 108 So, we split the middle term – 21x as – 12x – 9x  x2 – 21x + 108 = x2 – 12x – 9x + 108 = (x2 – 12x) – (9x – 108) = x (x + 12) – 9 (x – 12) = (x – 12) (x – 9) (iii) In order to factorize x2 + 5x – 36, Clearly, 9 + (–4) = 5 and 9 × – 4 = –36. So, we write the middle term 5x of x2 + 5x – 36 as 9x – 4x.  x2 + 5x – 36 = x2 + 9x – 4x – 36 = (x2 + 9x) – (4x + 36) = x (x + 9) – 4 (x + 9) = (x + 9) (x – 4) 38. x4 – x2 – 12 = (x2)2 – 4x2 + 3x2 – 12 = x2 (x2 – 4) + 3 (x2 – 4) = (x2 – 4) (x2 + 3) = (x2 – 22) (x2 + 3) = (x + 2) (x – 2) (x2 + 3) 39. x–a

x3 + ax2 + a2x + a3 x4 – 0x3 – a4 x4 – ax3 – + ax3 – 0x2 – a4 ax3 – a2x2 – + a2x2 – 0x – a4 a2x2 – a3x – + a3x – a4 a3x – a4 – + 0

301 41. Let f(x) = 2x3 – x2 + 3x – 2 When it is divided by 2x – 1, the  2x – 1 = 0 then x = 1/2  Remainder is f(1/2).

1 remainder is f   2 3

2

1  1  1  2       3   2  2  2  2



1 1 3 4   4 4 2

1 2 42. Let f(x) = kx2 – 3x – 2 f(x) is divisible by x – 2.  x–2=0  x=2  f(2) = 0  k(2)2 – 3(2) – 2 = 0  4k – 8 = 0  k=2 43. Given, p(x) = x4 + 15x3 + 6x2 – 12x + 3, Divisor = x + 2  x + 2 = 0 then x = –2.  Zero of divisor = – 2 Remainder on division by x + 2 = p (–2)  p (–2) = (–2)4 + 15 (–2)3 + 6(–2)2 –12 (–2) + 3 = 16 – 120 + 24 + 24 + 3 = – 53 

44.

1 3 x + 2 4

2x 2 + 3x – 6

x 3 + 3x2 – 2x + 4 3x

3x  2 3x  2

 Quotient = x3 + ax2 + a2x + a3 40. Let P(x) = x3 + 3x2 + 3x + 1 x +  = 0  x = –  Remainder = P (–  ) = (–  )3 + 3(–  )2 + 3(–  ) + 1 = –3 + 32 – 3 + 1  The remainder is –  3 + 3  2 – 3  + 1



x3 + – 3x 2 – + +x+4 

– 

9x 4



+

18 4

5x 34  4 4

5x 34  should be added to x3 + 3x2 – 2x + 4 4 4 so that it may be exactly divisible by 2x2 + 3x – 6. 

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8th Class Mathematics

302 3.

The degree of a polynomial in ‘x’ is the exponent of the term containing the highest power of ‘x’. It is always a non-negative integer. The degrees of the given polynomials are: (i) 2 (ii) 1 (iii) 3 (iv) 7 (v) 4

4.

a)

–7x14, 2x11,

b)

2x16, 14x12,

x+ 3

45. x 2 – 5x + 3

x 3 – 2x 2 + 7x – 9 x 3 – 5x 2 + 3x

–

+

–

3x2

+ 4x – 9 3x2 – 15x + 9 – + – 19x – 18

1 10 x , 5x9, 5 x7, –3x5, 6x3, 4x2, 8x.

For the division to be exact, remainder must be zero. i) The remainder (19x – 18) should be subtracted from dividend x3 – 2x2 + 7x – 9 ii) The additive inverse of (19x – 18) = – (19x – 18) = – 19x + 18 should be added to the dividend x3 – 7x – 9.

7x6, 8x5, 4x4, x3, 5.

P(x) = 3x P(x) = 0

SUMM ATIVE WORKSHEET (a) The above expressions in descending order of powers are : (i) –2y4 – y3 + 3y + 4 (ii) x4 – 4x3 + 5x + 4 (iii) x5 – 2x4 + x2 – 3 x

7.

(iv) – t4 – 3t3 + 2t + 6 (v) u4 – 3u2 + 6u + 2 (b) (i) Monomial

(ii) Binomial

(iii) Trinomial (iv) Binomial 2.

(v) Trinomial

Let us recall the definition of polynomials. If f(x) = a0 + a1x + a2x2 + ------ + an xn, where a0, a1, a2, ---- an are real numbers and n is a non-negative integer (i.e. ‘n’ should be positive integer), then f(x) is called a polynomial in x, of nth degree.

8.

(IX) is not a polynomial. 1

Since

x  x 2 , 1/2 is not a integer..

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x

0 0 3

Let f(x) = 5x – 4x2 + 3 value of f(x) at x = –1  f(–1) = 5(–1) –4(–1)2 + 3 = –5 – 4 + 3 = –9 + 3 = –6 Rearranging the terms, we have 12m2 – 9m + 5m – 4m2 – 7m + 10 12m2 – 4m2 + 5m – 9m – 7m + 10 = (12 – 4) m2 + (5 – 9 – 7) m + 10 = 8m2 + (– 4 – 7) m + 10 = 8m2 + (–11) m + 10 = 8m2 – 11m + 10 We first add 2y2 + 3yz, – y2 – yz – z2 and yz + 2z2. 2y2 + 3yz – y2 – yz – z2 + yz + 2z2 y2 + 3yz + z2

(1)

We then add (3y2 – z2)+ (–y2 + yz + z2)

(i), (ii), (iv), (vii), (viii), (x) are polynomials and (iii), (v), (vi) and (ix) are not polynomials. 1 1 The II term in (iii) is  u , here the power of ‘u’ u is not a positive integer. Hence, it is not a polynomial. Also the 1 term in (v) is x–3. Here the power of ‘x’ is not a positive integer. Hence, it is not a polynomial. In (vi), the II term is x–3. ‘x’ is raised to a negative power, so it is not a polynomial.

1 2 x ,x 5

 3x = 0 

6. 1.

4 9 x , 9x7, 3

3y2 – z2 2 – y + yz + z 2 2y2 + yz

(2)

Now we subtract sum (2) from the sum (1): y2 + 3yz + z2 2y2 + yz – – – y2 + yz + z2

9.

(30ab + 12b + 14a) – (24ab – 10b – 18a) = 30ab + 12b + 14a – 24ab + 10b + 18a = 30ab – 24ab + 12b + 10b + 14a + 18a = 6ab + 22b + 32a

Polynomials Solutions

303

Alternatively, we write the expressions one below the other with the like terms appearing exactly below like terms as :

(iv) (x2 + x + 1) (x2 – x + 1) x2 + x + 1 x2 – x + 1

30ab + 12b + 14a 24ab – 10b – 18a – + + 6ab + 22b + 32a

Multiply by x2 :

10. a) Let p(x), be added to 3x2 – 2x + 6 to get 2x2 + 6x – 5.  (3x2 – 2x + 6) + p(x) = 2x2 + 6x – 5  p(x) = (2x2 + 6x – 5) – (3x2 –2x + 6) = 2x2 + 6x – 5 – 3x2 + 2x – 6 = –x2 + 8x – 11.  –x2 + 8x – 11 must be added to 3x2 – 2x + 6 to get 2x2 + 6x – 5. b) Let q (x) be subtracted from a2 + b2 + c2 – 3abc to get 2a2 – b2 –3c2 + abc.  (a2 + b2 + c2 – 3abc) – q(x) = 2a2 – b2 –3c2 + abc  –q(x) = (2a2 – b2 – 3c2 + abc) – (a2 + b2 + c2 – 3abc) = 2a2 – b2 – 3c2 + abc – a2 – b2 – c2 – 3abc) = a2 – 2b2 – 4c2 + 4abc.  q(x) = – a2 + 2b2 + 4c2 – 4abc.  –a2 + 2b2 + 4c2 – 4abc should be subtracted from a2 + b2 + c2 – 3 abc to get 2a2 – b2 – 3c2 + abc. 11. 2x2 – 3x + 4 3x2 – 2x + 1 6x4 – 9x3 + 12x2 – 4x3 + 6x2 – 8x + 2x2 – 3x + 4 6x4 – 13x3 + 20x2 – 11x + 4

12. (i) (3x2 – 5x + 4) × 2x = (3x2 × 2x) + (– 5x × 2x) + (4 × 2x) = 6x3 – 10x2 + 8x (ii) (6x2 – 4x + 3) × 4x3 = (6x2 × 4x3) + (– 4x × 4x3) + (3 × 4x3) = 24 x5 – 16x4 + 12x3 (iii) (3x2 – 5x + 6) × (4x – 3) 3x2 – 5x + 6 4x – 3 Multiply by 4x : 12x3 – 20x2 + 24 x Multiply by –3 :

– 9x2 + 15x – 18

12x3 – 29x2 + 39x – 18

x4 + x3 + x2

Multiply by –x :

– x3 – x2 – x

Multiply by + 1 :

+ x2 + x + 1 x4 + x2 + 1

(i) (2a + b)3 = (2a)3 + 3 (2a)2 (b) + 3 (2a) (b)2 + (b)3 = 8 a3 + 12 a2 b + 6 ab2 + b3 (ii) (3a + 4b)3 = (3a)3 + 3 (3a)2 (4b) + 3 (3a) (4b)2 + (4b)3 = 27 a3 + 3 (9a2) (4b) + 3 (3a) (16 b2) + 64 b3 = 27 a3 + 108 a2 b + 144 ab2 + 64 b3 (iii) (2a – b)3 = (2a)3 + 3 (2a)2 . (–b) + 3 (2a) (–b)2 + (–b)3 3 = 8a + 3 (4a2) (–b) + 3 (2a) (b2) + (–b3) 3 2 2 3 = 8a – 12 a b + 6ab – b (iv) (a2 – 4)3 = (a2)3 + 3 (a2)2 (–4) + 3 (a2) (–4)2 + (–4)3 6 = a + 3 (a4) (– 4) + 3 (a2) (16) + (– 64) = a6 – 12 a4 + 48a2 – 64 (v) (101)3 = (100 + 1)3 = (100)3 + 3 (100)2 (1) + 3(100) (1)2 + (1)3 = 1000000 + 3 (10000) + 3 (100) + 1 = 1000000 + 30000 + 300 + 1 = 10,30,301 (vi) (1002)3 = (1000 + 2)3 = (1000)3 + 3 (1000)2 (2) + 3 (1000) (2)2 + (2)3 = 1000000000 + 3(1000000) (2) + 3 (1000) (4) + 8 = 1000000000 + 6000000 + 12000 +8 = 1006012008. (vii) 983 = (100 – 2)3 = (100)3 + 3 (100)2 (–2) + 3 (100) (–2)2 + (–2)3 = 1000000 + 3 (10000) (–2) + 3 (100) (4) + (– 8) = 1000000 – 60000 + 1200 – 8 = 941192 14. a – b = 2; ab = 15 a3 – b3 = (a – b)3 + 3ab (a – b) = (2)3 + 3 (15) (2) = 8 + 90 = 98 13.

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8th Class Mathematics

304 15.

1 5 x cubing on both sides, x

Verification : (3x2 – 2x + 5) (x2 – 2x +

8x 11  3 3 3x2(x2 – 2x + 1/3) – 2x(x2 – 2x + 1/3)+5 

3

1  3 x  5 x  

x3 



1 1 1  3  x   x    25 3 x x x

(x2 – 2 x + 1/3) 

5 8x 11   3 3 3 = 3x4 – 8x3 + 10x2 – 8x – 2 = Dividend 2

19.

–

5x2

– + + – – 6x3 + 5x2 – 8x – 6x3 + 4x2 – 10x + – + x2 + 2x – 2 5 x2 – 2 x + – +3 – 3 8 11 x 3 3

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 x  8  x  8 

x 8 = (x – 8)

20.

x 4  74x 2  1131 x 2  13 Split midterms of x4 – 74x2 + 131 into – 87x2 + 13x2 





x 4  87x 2  13x 2  1131 x 2  13 x 2  x 2  87   13  x 2  87  x 2  13

x

2

 87  x 2  13

x 2  13 = (x2 – 87) 21. 64x2 – 81z2 by 8x – 9z

Verification : = (2x – 3) (x2 – 2x + 1) + 8 = 2x (x2 – 2x + 1) – 3 (x2 – 2x + 1) + 8 = 2x3 – 4x2 + 2x – 3x2 + 6x – 3 + 8 = 2x3 – 7x2 + 8x + 5 = Dividend

2x3

x 2  16x  64  x   2  x  8    8   x8 x8 

2x – 3 ) 2x3 – 7x2 + 8x + 5 (x2 – 2x + 1 2x3 – 3x2 – + – 4x2 + 8x – 4x2 + 6x + – 2x + 5 2x – 3 – + 8

18. 3x2 – 2x + 5 ) 3x4 – 8x3 + 10x2 – 8x – 2 (x2 – 2x +

2x 3

+ 5x2 – 10x +

1  x  3  125  15  140 x 16. (i) (3x – 4y + z) (3x – 4y – z) = (3x – 4y)2 – (z)2 = 9x2 – 24 xy + 16 y2 – z2 ( a2 – b2 = (a + b) (a – b)) (ii) (5a + 3b + 2c) (5a – 3b – 2c) = [5a + (3b + 2c)] [5a – (3b + 2c)] = (5a)2 – (3b + 2c)2 = 25a2 – (9b2 + 12 bc + 4c2) 2 ( a – b2 = (a + b) (a – b)) = 25 a2 – 9b2 – 4c2 – 12 bc 3

3x4

8x 11  3 3

= 3x4 – 6x3 + x2 – 2x3 + 4x2 –

1 x  3  3  5   125 x 3

17.

1 ) 3

2

1 3

2

64x 2  81z 2  8x    9z    8x  9z 8x  9z Applying the identity, a2 – b2 = (a + b) (a – b) Where, a = 8x, b = 9z, we get, 64x 2  81z 2 8x  9z 

 8x  9z  8x  9z  8x  9z

= 8x + 9z

2

Polynomials Solutions

305

22. x9 – y9 by x3 – y3 3

3 3 x 9  y9  x    y   3  x  y3 x 3  y3

3

a = x3. b = y3. ( a3 – b3 = (a – b)(a2 – ab + b2)) 

x

3

2 2  y3    x 3    x 3 y3    y3     3 3 x y

= x6 + x3y3 + y6 23. 27a3 – 125b3 by 3a – 5b 3

27a 3  125b3  3a    5b    3a  5b 3a  5b Applying the identity a3 – b3 = (a – b) (a2 + ab + b2) where, a = 3a, b = 5b, we get,

3

27a   1253 3a  5b 2   3a  5b    5b     3a  5b = (3a)2 + 15ab + (5b)2 = 9a2 + 15ab + 25b2 24. We have, (i) ax + bx + ay + by = (ax + bx) + (ay + by) [ Grouping the terms] = (a + b)x + (a + b)y = (a + b) (x + y) [ Taking (a + b) common]  The two factors are (a + b), (x + y). (ii) ax2 + by2 + bx2 + ay2 = ax2 + bx2 + ay2 + by2 [ Rearranging the terms] = (a + b) x2 + (a + b) y2 = (a + b) (x2 + y2) [ Taking (a + b) common]  The two factors are (a + b), (x2 + y2). (iii) a2 + bc + ab + ac = (a2 + ab) + (ac + bc) [ Re-grouping the terms] = a (a + b) + (a + b) c = (a + b) (a + c) [ Taking (a + b) common] The two factors are (a + b), (a + c).  (iv) ax – ay + bx – by = a (x – y) + b(x – y) = (a + b) (x – y) [ Taking (x – y) common]  The two factors are (a + b), (x – y).

 3a  5b   3a 

2

25.(i) We have, a3x + a2 (x – y) – a (y + z) – z = a3x + a2x – a2y – ay – az – z = (a3x + a2x) – (a2y + ay – az – z) = a2x (a + 1) – ay (a + 1) – z (a + 1) = (a + 1) (a2x – ay – z)  The two factors are (a + 1), (a2x – ay – z). (ii) (x2 + 3x)2 – 5 (x2 + 3x) – y (x2 + 3x) + 5y = (x2 + 3x) {(x2 + 3x) –5} – y {(x2 + 3x) – 5 = (x2 + 3x – 5) (x2 + 3x – y)  The two factors are (x2 + 3x – 5), (x2 + 3x – y). 26. (x + y) (1 – z) – (y + z) (1 – x) = x – xz + y – yz – y + xy – z + xz = x – yz + xy – z = x (1 + y) – z (1 + y) = (x – z) (1 + y)  The two factors are (x – z), (1 + y). 27. (x + y) (a + bz) – (y + z) (a + bx) = ax + bxz + ay + byz – (ay + bxy + az + bxz) = ax + bxz + ay + byz – ay – bxy – az – bxz = ax – az – bxy + byz = a(x – z) – by (x – z) = (x – z) (a – by)  The two factors are (x – z), (a – by). 28. (i) 9x2 – 4y2 = (3x)2 – (2y)2 = (3x + 2y) (3x – 2y)  The two factors are (3x + 2y), (3x – 2y). (ii) 36x2 – 12x + 1 – 25y2 = ((6x)2 – 2 × 6x × 1 + 12) – (5y)2 = (6x – 1)2 – (5y)2 = {(6x – 1) – 5y} {(6x – 1) + 5y} = (6x – 1 – 5y) (6x – 1 + 5y) = (6x – 5y – 1) (6x + 5y – 1)  The two factors are (6x – 5y – 1), (6x+5y–1). (iii) a2 – 1 + 2x – x2 = a2 – (1 – 2 × 1 × x + (x)2) = a2 – (1 – x)2 = {a – (1 – x)} {a + (1 – x)} = (a – 1 + x) (a + 1 – x) The two factors are (a + x – 1), (a – x + 1).  6 29. x – 1 = (x3)2 – (1)2 = (x3 + 1) (x3 – 1) = (x3 + 13) (x3 – 13) = (x + 1) (x2 – x + 1) (x – 1) (x2 + x +1) (x  The four factors are (x + 1), (x2 – x + 1), – 1), (x2 + x +1). 30. We have, (i) 4x2 – 4xy + y2 – 9z2 = (4x2 – 4xy + y2) – 9z2 = {(2x)2 – 2 × 2x × y + y2} – (3z)2 (a2 – 2ab + b2 = (a – b)2) = (2x – y)2 – (3z)2 = (2x – y + 3z) (2x – y – 3z)  The two factors are (2x – y + 3z),(2x – y–3z). www.betoppers.com

8th Class Mathematics

306 (ii) 16 – x2 – 2xy – y2 = 16 – (x2 + 2xy + y2) = 42 – (x + y)2 = {4 + (x + y)} {4 – (x + y)} = (4 + x + y) (4 – x – y)  The two factors are (4 + x + y), (4 – x – y). (iii) x4 – (x – z)4 = (x2)2 – {(x – z)2}2 = {x2 + (x – z)2} {x2 – (x – z)2} = (x2 + x2 – 2xz + z2) [x + (x – z) {x – (x – z)}] 2 = (2x – 2xz + z2) (x + x – z) (x – x + z) = (2x2 – 2xz + z2) (2x – z) z  The two factors are (2x2 – 2xz + z2),(2x – z) z. 31. 8x3 – 60x2 + 150x – 125 = (2x)3 + 3 (2x)2 (–5) + 3(2x) (–5)2 + (–5)3 ( a3 + 3a2 + 3b2 a + b3 = (a + b)3 = (2x – 5)3  The three factors are (2x – 5),(2x – 5),(2x–5). 32. Given that a3 + b3 + c3 = 3abc  a3 + b3 + c3 – 3abc = 0 (a + b + c) (a2 + b2 + c2 – ab – bc – ca) = 0 (a + b + c)

1 [(a – b)2 + (b – c)2 + (c – a)2] = 0 2

1 (a + b + c) [ (a – b)2 + (b – c)2 + (c – a)2]=0 2 If the product of two factors is zero, atleast one of the them is equal to 0.  a + b + c = 0 or (a – b)2 + (b – c)2 + (c – a)2 = 0 (b – c)2 = 0  a – b = 0  a = b (c – a)2 = 0  c – a = 0  c = a  a + b + c = 0 (or) a = b = c 33. Arranging the terms in descending powers of x, the given expression = 6x2 + (7y + 11z)x + (2y2 + 7yz + 3z2) = 6x2 + (7y + 11z)x + (2y2 + 6yz + yz + 3z2) = 6x2 + (7y + 11z) x + {2y (y + 3z) + z(y + 3z) = 6x2 + (7y + 11z) x + (y + 3z) (2y + z) It is clear that 6(y + 3z) (2y + z) can be broken up into 3(y + 3z) and 2(2y + z), the sum of which is 7y + 11z, that is the coefficient of the middle term x. Thus the given expression = 6x2 + 3(y + 3z) x + 2 (2y + z) x + (y + 3z) (2y+z) = 3x (2x + y + 3z) + (2y + z) (2x + y + 3z) = (2x + y + 3z) (3x + 2y + z)  The two factors are (2x + y + 3z),(3x + 2y + z).

34.

a4 

1  119 a4 2

 1 or (a ) +  2  + 2 = 121 = 1112 a  2 2

2

1     a 2  2   112 a  

1  11 a2 adding (–2) on both sides:  a2 

 a2 

2

1 2    a     3 a 

a

1 3 a 3

1 3    a     3 a 

 a3 

1 1   3  a    27 3 a a 

 a3 

1  3  3  27 a3



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1  2  11  2  9 a2

1  27  9  36 a3 35. x + 1 = 0  x = –1  putting x = – 1 in f(x) = x8 + kx3 – 2x + 1 We get, f (–1) = (–1)8 + k (–1)3 – 2 (–1) + 1  1–k+2+1=0 –k + 4 = 0 –k=–4 k=4 36. Let 5x + y = a and x + 2y = b Thus the given expression = a2 + ab – 20b2 ab can be written as 5ab – 4ab. = a2 + 5ab – 4ab – 20b2 = a(a + 5b) – 4b (a + 5b) = (a + 5b) (a – 4b) = [(5x + y) + 5 (x + 2y)] [(5x + y) – 4 (x + 2y)] = (5x + y + 5x + 10y) (5x + y – 4x – 8y) = (10x + 11y) (x – 7y)  a3 

Polynomials Solutions

307

37. Given p(x) = kx3 + 4x2 + 3x – 4 is divided by x–3 – x – 300  x=3  remainder is p(3).  p(3) = k(3)3 + 4(3)2 + 3(3) – 4 = 27k + 36 + 9 – 4.  p(3) = 27k + 41––––––– (1) Again, q(x) = x3 – 4x + k is divided by x – 3, the remainder is q(3).  q(3) = (3)3 – 4(3) + k = 27 – 12 + k = 15 + k _________ (2)  q(3) = q(3). From (1) and (2), we have 27k + 41 = 15 + k  26k = – 26  k = –1. 38. Let f(x) = x15 + ax13 + 5 f(x) is divisible by x + 1 x+1=0 x = – 1.  f(–1) = 0  (–1)15 + (–1)13 + 5  –1 – a + 5 = 0 a=4 39. Let the other factor be x2 + ax + b. As we know that Dividend = (Divisor × Quotient) + Remainder ..............(1) Applying (1) we have x4 – 5x2 – bx – 5 = (x2 + x + 1) )(x2 + ax + b) ...............(2) x4 – 5x2 – bx – 5 = x4 + ax3 + bx2 + x3 + ax2 + bx + x2 + ax + b x4 – 5x2 – bx – 5 = x4 + x3 (1 + a) + x2 (a + b +1) + x(a + b) + b Equating the coefficient of x3 on both sides  1 + a = 0  a = –1 Equating constant term  b=–5  The other factor is x2 – x – 5. 40. Let f(x) = 67x3 – 5x2 + 2x – 4  2x + 1 = 0

x

1 2

 1  Remainder is f    2  3

2

 1   1   1   6   5   2   4 2 2      2 

 1  5  6    1  4  8  4 3 5 8  5  5 4 4 4 = –2 – 5 = – 7 

HOTS WORKSHEET 1.

A = 2x + 3; B = 3x – 5 ; C = x2 + 3x – 1 i) A × (B × C) (B × C) = x2

3x – 5 + 3x – 1

Multiply by x2 : 3x3 – 5x2 + 9x2 – 15x

Multiply by 3x : Multiply by – 1 : 3x3

+

4x2

– 3x + 5 – 18x + 5

 A × (B × C) = (2x + 3) × (3x3 + 4x2 – 18x + 5) = 2x (3x3 + 4x2 – 18x + 5) + 3(3x3 + 4x2 – 18x + 5) 4 = 6x + 17x3 – 24x2 – 44x + 15 ii) (A × B) × C A × B = (2x + 3) × (3x – 5) = 6x2 – 10x + 9x – 15 = 6x2 – x – 15 (A × B) × C = 6x2 – x – 15 x2 + 3x – 1 Multiply by x2 : Multiply by 3x :

6x4 – x3 – 15x2 18 x3 – 3x2 – 45x

Multiply by – 1 : – 6x2 + x + 15 64x4 + 17x3 – 24x2 – 44x + 15

2.

 A × (B × C) = (A × B) × C The set of polynomials obeys associative property under multiplication. Given expression (i) (2a + b – c) Second expression ( a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)) = ((2a)2 + b2 + c2 – 2ab + bc + 2 ca) = 4a2 + b2 + c2 – 2ab + bc + 2ca  Result = 8a3 + b3 – c3 + 6abc (ii) Given expression = (3a – 2b + 4)  Second expression = (3a)2 + (–2b)2 + (4)2 – (3a) (–2b) – (–2b) (4) – (4) (3a) = 9a2 + 4b2 + 16 + 6ab + 8b – 12 a  Result = 27a3 – 8b3 + 64 + 72 ab (iii) Given expression = 4x2 + y2 + 25 – 2xy – 5y – 10x = (2x)2 + (y)2 + (5)2 – (2x) (y) – y(5) – 5 (2x)  Second expression = 2x + y + 5 Result = 8x3 + y3 + 125 – 30 xy

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8th Class Mathematics

308 3.

4.

(i) Expression with which to multiply = (2x)2 – (2x) (1) + (1)2 = 4x2 – 2x + 1( a3 + b3 = (a + b) (a2 – ab + b2))  Product = (2x)3 + (1)3 = 8x3 +1 (ii) Expression = (3x)2 + (3x) (2y) + (2y)2 = 9x2 + 6xy + 4y2  Product = (3x)3 – (2y)3 ( a3 + b3 = (a + b) (a2 – ab + b2)) = 27x3 – 8y3 2 (iii) 4x – 6x + 9 = (2x)2 – (2x) (3) + (3)2 This is to be multiplied by 2x + 3  Product = (2x)3 + (3)3 = 8x3 + 27 (iv) 16x2 + 20x + 25 = (4x)2 + (4x) (5) + (5)2 This is to be multiplied by 4x – 5  Product = (4x)3 – (5)3 = 64x3 – 125 Given a2 + b2 + c2 = bc + ac + ab

1 ((a – b)2 + (b – c)2 + (c – a)2} = 0 2  (a – b)2 + (b – c)2 + (c – a)2 = 0 So, (a – b)2 = (b – c)2 = (c – a)2 = 0  a–b=b–a=c–a=0  a = b; b = a; c = a  a = b = c ((4) is correct). If we take a = b – 1, c = b + 1 then a2 + b2 + c2 = (b – 1)2 + b2 + (b + 1)2 = 3b2 + 2 and ab + bc + ca + 3 = (b – 1)b + b(b +1) + (b – 1) (b + 1) + 3 = b2 – b + b 2 + b + b 2 – 1 + 3 = 3b2 + 2  a2 + b2 + c2 – [ab + bc + ca + 3] = 3b2 + 2 – (3b2 + 2) = 0

7.

Consider x3 (y – z) + y3 (z – x) + z3 (x – y) + (y – z) (z – x) (x – y) (x + y + z) = = x3y –x3z + y3z – y3x + z3x – z3y + (yz – xy – z2 + xz) × (x2 – y2 + xz – yz) 3 = x y – x3z + y3z – y3x + z3x – yz3 + x2yz – y3z + xyz2 – y2z2 + xy2z – x2z2 + y2z2 –xz3 + yz3 + x3z – xy2z – x3y + xy3 – x2yz + x2z2 – xyz2 = 0 2

1 1    5x  x   4 5x  x   4, x  0    

8.

Let 5x 

2

1 1     5x    4  5x    4 x x   2 = y + 4y + 4 = y2 + 2 × 2 × y + (2)2 ((a + b)2 = a2 + 2ab + b2) Let a = y, b = 2 = (y + 2)2 ––––––– (1) Substituting the value of ‘y’ in (1) we have,



5.

6.

2

1 1 2    5x  x   4  5x  x   4   y  2      1     5x   2  x  

9.

2 1 1 3  3  2  0  x + y + z3 = 3xyz  x3 + y3 + z3 – xyz = 2xyz

  2 2

3

  2  11  3 

2 1 1 3

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We have, 2 2p3  8q 3  27r 3  18 2pqr

 2

 6  3  18  3  2  2   2 1  3  3 2  2  2 3   2 1  2 2  3  2

Given 2 2p3  8q 3  27r 3  18 2pqr

10.

3 2 3

12  6

2

1 1      The two factors are 5x  x  2  ,  5x  x  2 .     2 2 Given, ax – (ax + by) + a x + aby + by = ax + by – (ax + by)2 + a(ax + by) ( rearranging the terms) = (ax + by) [1 – (ax + by) +a] = (ax + by) (1 – ax – by + a).  The factors of the given polynomial are (ax + by + a).

x+y+z= 3

1 y x







2p



3

3

3

  2q    3r   3





2p  2p  3r 

The presence of four terms are such that the first three terms are cubes of some expressions and the remaining term is –3 times the product of the three expressions.Suggest that the given polynomial is in the form a 3 + b 3 + c 3 – 3abc, where, a  2p, b  2q,c  3r.

Polynomials Solutions

309

 2 2p3  8q 3  27r 3  18 2pqr



2p  2q  3r  

 











2p



2

2

  2q    3r 



2p  2q    2q  3r    3r 

2p  2q  3r

 2p

2



2



2p  



 4q 2  9r 2  2 2pq  6qr  3 2pr



 The factors of the given polynomial are





2p  2q  3r and

 2p



2

 4q 2  9r 2  2 2pq  6qr  3 2pr .

13. Given, 12(p2 + 7p)2 – 8(p2 + 7p) (2p – 1) – 15 (2p – 1)2 Let (p2 + 7p) be ‘x’ and (2p – 1) be ‘y’. Now, the given polynomial reduces to 12x2 – 8xy – 15y2. Here, (12) × (–15) = – 180 = – 18 × 10 and – 18 + 10 = – 8.  12x2 – 8xy – 15y2 = 12x2 – 18xy + 10xy – 15y2 = 6x (2x – 3y) + 5y(2x – 3y) = (2x – 3y)(6x + 5y). Substituting the values of x and y, we get, [2(p2 + 7p) –3(2p –1)] [6(p2 + 7p) + 5(2p – 1)] = (2p2 + 14p – 6p + 3) (6p2 + 42p + 10p – 5) = (2p2 + 8p + 3) (6p2 + 52p – 5)  The two factors are (2p2 + 8p + 3), (6p2 + 52p – 5). 8a 3  27b3  125  90ab 2a  3b  5

11. Given, 8p3 – 27q3 – 36p2q + 54pq2. We have, 8p3 – 27q3 – 36p2q + 54pq2 = (2p)3 – (3q)3 – 3(2p)2 (3q) + 3(2p) (3q)2 The first two terms are cubes and the next two terms are multiples of 3, and two terms are negative. This suggests that the given polynomial is in the form a3 – b3 – 3a2b + 3ab2, where, a = 2p and b = – 3q.  8p3 – 27q3 – 36p2q + 54pq2 = (2p – 3q)3  The factors of the given polynomial are (2p – 3q), (2p – 3q) and (2p – 3q).

14.

12. Given, 2 2a 3  16 2b 3  c  12abc

= 4a2 + 9b2 + 25 – 6ab + 15b + 10a 15. (64x3 – 125)  (16x2 + 20x + 25)





2a



3

 2

3

 2  .b

3

 2a 

2a  2 2b  c  

 

 c 3  3. 2.2 2abc

3

 



   2 2b 

2

 c2  16.

 2 2b    2 2b   c   c  2b  c   2a  8b  c 2

2

2



 x

2

 The factors of the given polynomial are

2

P  x  x 



2a  

4ab  2 2bc  2ac

 2a  2 2b  c  and  2a  8b  c  4ab  2

 4x  5 16x 2  20x  25 

16x 2  20x  25 = (4x – 5)

2a

2a  2

2bc  2ac



3

64x 3  125  4x    5    2 16x  20x  25 16x 2  20x  25 

2

2a

3

 2a  3b  5  4a 2  9b 2  25  6ab  15b  10a    2a  3b  5 

where, a  2a,b  2 2b , and c = c.  The given polynomial



3

  3b    5   3  2a  3b  5  2a  3b  5 a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca) Here a = 2a, b = 3b, c = –5

This is in the form of a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)



3



3  2002 2

x  2002  x

3 and P(5) = 2002 2

3 2

4004  3 4001  2 2

4001 3 4004    2002 2 2 2  P(8) = 2002

 P  x 

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8th Class Mathematics

310 17. x100 = (x2 – 3x + 2) Q (x) + lx + m where Q(x) is quotient, lx + m is the remainder x=1 1 = 0 + l + m ...........(1) x = 2  2100 = 0 + 2l + m ...........(2) From (1) & (2) l = 2100 – 1, m = 2 – 2100  remainder = lx + m = (2100 –1) x + (–2100 +2) 18. Let f(x) = x3 + 2x2 – 5ax – 7 Let g(n) = x3 + ax2 – 12x + 6 is divided by x + 1, i.e., x + 1 = 0 x = –1, x + 2, i.e., x – 2 = 0 x=2  f(–1), g(2) are remainders of above polynomials.

21. Let f(x) = 6x4 – x3 + 3x2 + 4x – 2 and g(x) = 3x2 – 2x + 4 2x2 + x – 1 3x2 – 2x + 4 )

6x4 – 4x3 + 8x2 + – 3x3 – 5x2 + 4x 3x3 – 2x2 + 4x – + – – 3x2 + 0 – 2 3x2 + 2x – 4 + – + – 2x + 2 6x4 –x3 + 3x2 + 4x – 2 2x – 2

Given that f(–1) = –1 + 2 + 5a – 7 L1 = 5a – 6 –––––– (1) and g(x) = x3 + ax2 – 12x + 6 g(2) = 23 + a(2)2 – 12(2) + 6 = 4a – 10 –––––(2)  L2 = 4a – 10 Given that 2L1 + L2 = 6  2(5a – 6) + (4a – 10) = 6  10a – 12 + 4a – 10 = 6  14a – 22 = 6  14a = 28  a=2 19. Clearly by well known division

6x4 –x3 + 3x2 + 6x – 4

6x4 – x3 + 3x2 + 6x – 4 = (3x2 – 2x +4) (2x2 + x – 1). 22.

9l + 12 l2 – 2l + 3

9l3 – 6l2 + 3l – k 9l3 – 18l2 + 27l + – 12l2 – 24l – k 12l2 – 24l + 36 – + – – k – 36

–

x91) x2002 – 2001 ( x91 × 22 x2002 – – 2001

For 9l3 – 6l2 + 3l – k to be exactly divisible by l2 – 2l + 3, the remainder should be zero.  –k – 36 = 0  k + 36 = 0  k = – 36.

 Quotient = x91 × 21 Remainder = – 2001 20. By synthetic division

IIT JEE WORKSHEET

x3 – 7x + 5) x5 – 2x4 – 4x3 + 19x2 – 31x + 12 + k (x2 – 2x + 3 x5 + 0x4 – 7x3 + 5x2 – – + – – 2x4 + 3x3 + 14x2 – 31x – 2x4 + 0x3 + 14x2 – 10x + – – + 3x3 – 21x + 12 + k 3x3 – 21x + 15 – + – k–3 3

6x4 –x3 + 3x2 + 4x – 2

5

1.

Given 4x4 – (a – 1) x3 + ax2 – 6x – 1 is divisible by 2x – 1. Dividend = 4x4 – (a – 1) x3 + ax2 – 6x – 1 Divisor = 2x – 1  zero of divisor 

4

3

2

If x – 7x + 5 is a factor of x – 2x – 4x + 19x – 31x + 12 + k  Remainder k – 3 = 0  k = 3

1 and 2

remainder = 0. Let f(x) = 4x4 – (a – 1) x3 + ax2 – 6x – 1

1 The remainder  f    0  2 4

3

2

1 1 1 1  4     a  1    a    6    1  0  2  2  2 2

 4 www.betoppers.com

1 1 1 1   a  1    a    6   1  0 16 2 8  4

Polynomials Solutions 

1  a  1 a    3 1 0 4 8 4



2a   a  1 1 a  1 15  4   8 4 8 4

15  8  a  1  30 4  a = 30 – 1 = 29.  The required values of ‘a’ is 29. Given, (x + 1) is a factor of x4 + (p – 3) x3 – (3p – 5) x2 + (2p – 9) x + 6 Let f(x) = x4 + (p – 3) x3 – (3p – 5) x2 + (2p – 9) x + 6. zero of divisor of x + 1 is x + 1 = 0  x = –1 x + 1 is a factor of f(x)  f (–1) = 0 (2p  (–1)4 + (p – 3) (–1)3 – (3p – 5) (–1)2 + – 9) (–1) + 6 = 0  1 – p + 3 – 3p + 5 – 2p + 9 + 6 = 0

311 and p (–1) = a (–1)3 + b (–1)2 + (–1) –6 = 6  –a+b–7=6 = – a + b = 13 –––––– (2) Solving (1) and (2), we have a+b=5 – a + b = 13

2.

5.

24 4 6  The required value of ‘p’ is 4. Given, p(x) = ax3 + 3x2 – 13 q (x) = 2x3 – 5x + a (x + 2) is the divisor zero of divisor is x = – 2.  Remainders = p (–2) and q (–2). Given, p(x) and q(x) leave the same remainder when divided by x + 2.  p(–2) = q(–2)  a(–2)3 + 3(–2)2 – 13 = 2 (–2)3 – 5 (–2) + a  –8a + 12 – 13 = – 16 + 10 +a  – 1 – 10 + 16 = a + 8a

 –6p = –24  p 

3.

5 . 9  The required value of ‘a’ is 5/9. Given, p(x) = ax3 + bx2 + x – 6 has (x – 1) as a factor x – 1 = 0 x = 1.  p(1) = 0. and p(x) leaves a remainder 6, when divided by (x + 1) divisor x + 1 is x = – 1  p (–1) = 6. We have, p(1) = 0  a(1)3 + b(1)2 + 1 – 6 = 0 a+b–5=0  a + b = 5 –––––––(1)

– 8a + 4b = 10 8a + 4b = – 2 8b = 8  b = 1

Substituting the value of ‘b’ in (1), we get, –8a + 4b = 10  –8a + 4 (+1) = 10  –8a = 10 – 4 = 6.

 5 = 9a  a 

4.

 b = 9. Substituting the value of ‘b’ in (1), we get, a + b=5 a=5–b=5–9=–4  a=–4 Hence the required values are a = – 4, b = 9. Given p(x) = ax3 + bx2 + 4x – 2 has (x + 2) as a factor zero of x + 2 is x = –2  p (– 2) = 0 and p(x) leaves a remainder 4, when divided by x –2 x–2=0  x=2  p(2) = 4. We have, p (–2) = 0  a (–2)3 + b (–2)2 +4( –2) –2 = 0  –8a + 4b – 8 – 2 = 0  –8a + 4b = 10 –––––––– (1) and p (2) = 4  a(2)3 + b(2)2 + 4 (2) – 2 = 4  8a + 4b + 8 – 2 = 4  8a + 4b = 4 – 6  8a + 4b = – 2 ––––– (2) Solving (1) and (2), we get, 2b = 18

 a 1

6 3  . 8 4 Hence the required values are a = –3/4 and b =1 a

3 3 x  x 2  4x  2 . 4 Given x4 + 4x3 + 6px2 + 4qx + r is exaclty divisible by x3 + 3x2 + 9x + 3  remainder = 0. Let the other factor be (x + k). By division algorithm, we have, x4 + 4x3 + 6px2 + 4qx + r = (x3 + 3x2 + 9x + 3) (x + k) x(x3 + 3x2 + 9x + 3) + k(x5 + 3x2 + 9x + 3)  The polynomial is

6.

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8th Class Mathematics

312

7.

8.

9.

= x4 + (3 +k) x3 + (9 + 3k) x2 + (3 + 9k) x + 3k. Comparing the coefficients of like terms, we have, 3 + k = 4  k = 1. 9 + 3k = 6p  9 + 3 (1) =6p  12 = 6p  p = 2 3 + 9k = 4q  3 + 9 (1) = 4q  12 = 4q  q = 3. 3k = r  3 (1) = r  r = 3.  p = 2, q =3 and r = 3 Thus, r(p + q) = 3 (2 + 3) =15.  The required value of r(p + q) is 15. Given, p(x) = x94 – 7x19 + 5x4 – x + k (dividend) divisor = (x – 1)  zero of divisor = 1 remainder is – 1  p(1) = – 1  194 – 7(1)19 + 5(1)4 – 1 + k = – 1  1–7+5–1+k=–1  –2+k=–1  k=1  p(x) = x94 – 7x19 + 5x4 – x + 1. Remainder when p(x) is divided by x + 1 is p(–1).  p(–1) = (–1)94 – 7(–1)19 + 5(–1)4 –(–1) + 1 = 1 + 7 + 5 + 1 + 1 = 15  The required remainder = 15 Given f(x) = x3 + ax + b is divisble by (x – 1)2 = x2 – 2x + 1. Let the other factor be (x + k) By division algorithm, we have x3 + ax + b = (x2 – 2x + 1) (x + k) = x3 – 2x2 + x + kx2 – 2kx + k = x3 + (k – 2) x2 + (1 – 2k) x + k Comparing the coefficients of like terms, we have k–2=0  k=2 1 – 2k = a  a = 1 – 2(2) = 1 – 4 = – 3 and k = b  b = 2.  f(x) = x3 – 3x + 2.  Remainder when f(x) is divided by x + 2 = f(–2)  f(–2) = (–2)3 – 3 (–2) + 2 =–8+6+2=0 Let f(x) = x11; Divisor = 1 + x + x2 + ...... + x10. By division algorithm, we have, x11 = (1 + x + x2 + ..... + x10) (x + k) x(1 + x + x2 ...... + x10) + k(1+ x + x2 ---- x10) = x + x2 + .... + x11 + k(1 + x + x2 + ... + x10) = x11 + x10 (1 + k) + x9 (1 + k) + .... +x(1 + k) + k Comparing coefficients of like terms, we have, 1+ k = 0  k = –1  remainder = – 1

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10. Given, (x – b)5 + (b – a)5 Let (x – b) be ‘p’ and (b – a) be ‘q’ Then p + q = x – b + b – a = x – a. Concept : pn + qn is divisible by p + q when n is odd. Here, n = 5 is odd.  (x – b)5 + (b – a)5 is divisble by x – a. 11. Given x3 – (a + 4) x2 + (4a + c) x + d is divisible by x–a remainder divisor of x – a = 0 x = a.  f(a) = 0  a3 – (a + 4) a2 + (4a + c) a + d = 0  a3 – a3 – 4a2 + 4a2 + ac + d = 0.

 ac + d = 0  a 

d . c

d . c Hence the correct option is (d). 12. x15 + ax11 + 3 is divisible by x + 1 (Given) divisor of x + 1 = 0 is x = –1  f (–1) = 0  (–1)15 + a (–1)11 + 3 = 0  –1–a+3=0  a=2  The correct option is (a).  The required value of a 

 

7. LINEAR EQUATIONS & INEQUATIONS SOLUTIONS

FORMATIVE WORKSHEET 1.

2.

Substituting the values of x and y in the equation, x+y=5 2 + 4 = 6 and 6  5 The values x = 2 and y = 4 do not satisfy the equation  (2, 4) is not a solution set. Given : 6x + 3 = 8x + 10.  6x – 8x = 10 – 3  – 2x = 7 7 x =  2 Verification : Substituting the value of ‘x’ in the given equation, we have 6x + 3 = 8x + 10

3.

4.

or 7x 

4 1 4 (iii) 4x = 15 – 3x 4x + 3x = 15 7x = 15 x 

15 7

15 15 (transposing to R H S) 4 4

21 4

21 or x  4   7  (dividing both sides by –7) or x 

5.

3 7 4  7

3 or x   = solution 4 Let the number of five-rupee coins that Bansi has be x. Then the number of two-rupee coins he has is 3 times x or 3x. The amount Bansi has : (i) From 5-rupee coins, Rs 5 × x = Rs 5x (ii) From 2-rupee coins, Rs 2 × 3x = Rs 6x Hence, the total money he has = Rs 11x But this is given to be Rs 77; therefore, 11x = 77 77 7 11 Thus, number of five-rupee coins = x = 7 and number of two-rupee coins = 3x = 21 Let the number of Rs 50 notes and Rs 20 notes be 3x and 5x, respectively. But she has 25 notes in all. Therefore, the number of Rs 10 notes = 25 – (3x + 5x) = 25 – 8x The amount she has From Rs 50 notes : 3x × 50 = Rs 150x From Rs 20 notes : 5x × 20 = Rs 100x From Rs 10 notes : (25 – 8x) × 10 = Rs (250 – 80x) Hence the total money she has = 150x + 100x + (250 – 80x) = Rs (170x + 250) But she has Rs 590. Therefore, 170x + 250 = 590

or x 

x 

x 

15  7x  9 4

or 7x  9 

 7  7  6     3  8     10  2  2  – 21 + 3 = – 28 + 10  – 18 = – 18  L.H.S. = R.H.S. Hence, verified. (i) On cross-multiplying, we get 6x = 50 50 25  6 3 (ii) 9 – 7x = 5 – 3x – 7x + 3x = 5 – 9 – 4x = – 4

We have

6.

8th Class Mathematics

314

170x = 590 – 250 = 340

1 1 1 i.e., {x – (ab + bc + ca)      0  bc ca ab  We see that the product of two terms is zero, Hence, at least one of them should be zero.

340 2 170 The number of Rs 50 notes she has = 3x The number of Rs 20 notes she has = 5x = 5 × 2 = 10 The number of Rs 10 notes she has = 25 – 8x = 25 – (8 × 2) = 25 – 16 = 9 Multiply both sides of the equation by 2. x

7.

7  3  We get 2   5x    2   x  14  2  2  7  3     2  x    2  14  2  2   or 10x + 7 = 3x – 28 or 10x – 3x + 7 = – 28 (by transposing 3x to LHS) or 7x + 7 = – 28 or 7x = – 28 – 7 or 7x = – 35

 2  5x   2 

35 7 or x = – 5 The equation may be written as

or 8.

x

 x  a  b  c   x  b c  a    1    1  bc ca      x  c a  b    1  0 ab  

1 1 1 Since      0  bc ca ab  as a, b, c are not zeros.  x – (ab + bc + ca) = 0  x = (ab + bc + ca)

9.

x 2  9 5  5  x2 9 Cross-multiplying, we get 9(x2 – 9) = – 5(5 + x2)  9x2 – 81 = – 25 – 5x2  9x2 + 5x2 = – 25 + 81  14x2 = 56 Given

56 = 4 = (?2)2 14  x = 2 (Taking the positive value) 10. The given equation is  x2 

 x  2  2x  3  2x 2  6  2

x 5 (x + 2) (2x – 3) = x(2x – 3) + 2 (2x – 3) = 2x2 – 3x + 4x – 6 = 2x2 + x – 6



2x 2  x  6  2x 2  6 2 x 5

x 2  x 5 1 Cross multiplying, 1 × x = 2(x – 5)  x = 2x – 10  x – 2x =– 10  – x = – 10  x = 10.  x   ab  ac  bc    x   bc  ba  ca   i.e.,   11. Let the present age of the son be x years.    bc ca     Then, the present age of the father = (7x) years  x   ca  cb  ab   Son’s age 2 years ago = (x – 2) years  0 ab Father’s age two years ago = (7x – 2) years   2 years ago, Father was 13 times as old as Taking {x – (ab + bc + ca) common from all the his son terms;  7x – 2 = 13 (x – 2)  7x – 2 = 13x – 26  x  ab  ac  bc   x  bc  ba  ca  i.e.,    bc ca      x  ca  cb  ab   0 ab  

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

Linear Equations & Inequations

 26 – 2 = 13x – 7x  24 = 6x  6x = 24  x = 4  Son’s present age = 4 years Father’s present age = (7 × 4) years = 28 years

315

34x 14  34  34 34 [Dividing both sides by 34]  x = 14 Hence, the two parts are 14, and 34 – 14 = 20 xb xb 15. Suppose I answered x questions correctly. Then,  12. We have, a b a b Number of wrong answers and unattempted ques (x + b) × (a + b) = (x – b) × (a – b) tions = (90 – x) [By cross-multiplication] Since for every correct answer 2 marks are  x(a + b) + b (a + b) = x (a – b) – b (a – b) awarded and for every wrong answer 1 mark is  ax + bx + ba + b2 = ax – bx – ba + b2 deducted, 2 2  ax + bx – ax + bx = – ba + b – ba – b Total score is given as 60  2bx = – 2 ba  2x – (90 – x) = 60  2x – 90 + x = 60 2bx 2ab    x = –a  3x – 90 = 60 2b 2b  3x = 60 + 90 Hence, x = – a is the solution of the given  3x = 150 equation. 3x 150 13. Let the denominator of the fraction be x. Then,    x = 50 3 3 Numerator of the fraction = x – 4  Number of correct answers = 50. x  4 _______  Fraction  (i) 16. The given equations are : x x + y – 5 = 0 ______ (i) If 1 is added to both its numerator and denominay – 2x = 2x ______ (ii) 1 from (i), we get tor, the fraction becomes x+y–5=0?x+y=5 2  x = 5 – y ______ (iii) x 3 1   Substituting x = 5 – y in (ii), we get : x 1 2 y – 2(5 – y) = 2(5 – y) (by transposition) [By cross-multiplication]  y – 10 + 2y = 10 – 2y  2(x – 3) = x + 1  3y – 10 = 10 – 2y  2x – 6 = x + 1  3y + 2y = 10 + 10 x=7  5y = 20 Putting x = 7 in (i), we get on dividing both sides by 5, we get y=4 74 3  Fraction  Substituting y = 4 in (III), we get 7 7 x=5–y 3 x=5–4 Hence, the given fraction is . 7 Hence, x = 1 and y = 4 is the solution of the given 14. Let one part be x. Then, other part is (34 – x). It is equations. given that 17. The given equations are ___________ 3x – y = 5 (i) th th 4 2     ___________ 5x – 2y = 4 (ii)   of one part =   of other part 7 5 Multiplying (i) throughout by 2, we get  20x = 14 (34 – x) 6x – 2y = 10 ___________ (iii)  34x = 14 × 34 Subtracting (iii) from (ii), we get 

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316

(ii) – (iii) = (5x – 2y) – (6x + 2y)= 4 – 10 5x – 2y – 6x + 2y = 4 – 10 –x=–6 x=6 Substituting x = 6 in (ii), we get 5 × 6 – 2y = 4  30 – 2y = 4  30 – 4 = 2y  2y = 26  y = 13  Solution = (x = 6, y = 13) 18. The given equations are x y x y   5;   4 3 2 7 14 L.C.M. of 3, 2 = 6 L.C.M. of 7, 14 = 14



x x 5 1    4 6 12 3

x 1   x 1 12 12 x = 1 in (iii), we get y: 

1 1 y  1  2 2 1 3  2 2

4p  9q 5q  p pq Dividing by q both the numerat or and denominator, we get:

20. Given

2x  y 4 14  2x + y = 56 __________(ii) (i) – (ii)  2y = – 26 26 = – 13 2 Substituting y = – 13 in (ii), we get

y =

69 2

 69   Solution set =  , 13     2 19. Given

x 1 x 5    4 3 6 12

x  y  1

2x  3y 5  ; 6 1  2x + 3y = 30 _________(i)

2x – 13 = 56  2x =



x y 5   _____(i) 4 3 12



4p / q  9

p / q

p 1 q

2

2

p  4   9  0 q 2

p 3     q 2

2



2

p  4   9  0 q 2

p 3     q 2

x x  y  1  y  1  ____(iii) 2 2 Substituting (iii) in (i), we get:



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5

p p p p  4   9   4   9  5 q q q q

x  y  1_____(ii) 2

x 1 x  5   1    4 3  2  12



p 3  q 2

2

p 3  q 2

Linear Equations & Inequations

21. x + y = 12

_____

x 2 y2   18 ____(ii) (i), y x

x 3  y3  18 xy (or) x3 + y3 = 18xy (or) (x + y) (x2 – yx + y2) = 18xy [ (a + b)3 = (a + b) (a2 – ab + c2)] (or) 12 (x2 – xy + y2) = 18xy [by (i)] (or) 2x2 + 2y2 = 5xy, [ ( a2 + b2 = (a + b)2 – 2ab)] (or) 2{(x + y)2 – 2xy} = 5xy [by (i)] (or) 2{(144 – 2xy)} = 5xy 288 – 4xy = 5xy ? xy = 32 _______ (iii) Now (x – y)2 = (x + y)2 – 4xy = 144 – 128 [by (i) & (ii)] (or) (x – y)2 = 16 = (  4)2 Thus x – y =  4 Now, let us take x + y = 12, x – y = + 4; By solving the above, we get: x = 8, y = 4; when x – y = – 4; x + y = 12, By solving the above, we get: x = 4, y = 8  x = 8, y = 4; and x = 4, y = 8 22. 3x + 6y = 4xy _______(i), 9x + 8y = 7xy _______ (ii) (i) × 3  9x + 18y = 12xy _______ (iii) (iii) – (ii)  10y = 5xy  5y (2 – x) = 0  y= 0, x = 2 From (i), when y = 0 ? x = 0 when x = 2  3(2) + 6y = 4(2)y  6 = 8y – 6y  2y = 6 y=3  x = 2, y = 3 23. 4x + 5y = 82 _____ (i) 3x + 2z = 54 _____ (ii) 5y + 4z = 110 _____ (iii) By [(i) × 3] – [(ii) × 4] (i) × 3 =12x + 15y = 246 (ii) × 4 = 12x + 8z = 216 15y – 8z = 30 ______ (iv) (iii) × 2 = 10y + 8z = 220 Adding (iii) and (iv); (12x + 8z) + (10y + 8z) = 216 – 220 

317

Add 25y = 250  y = 10 From (i), x = 8, From (iii), z = 15,  5x + 2y + z = 5(8) + 2(1)) + 15 = 75 24. Let the numerator be ‘x’ and denominator be ‘y’

x y Given that sum of numerator and denominator is 8. i.e., x + y = 8  y = 8 – x________ (i) ‘3’ is added to both numerator and denominator Then the fraction is

and is equal to

3 . 4

x3 3  ________ (ii) y3 4 by cross multiplication, we get  4x + 12 = 3y + 9  4x – 3 (8 – x) + 3 = 0  7x – 21 = 0 7x = 21 21 x=3 7  y = 8 – 3  y = 5 ( y = 8 – x) y=8–x x

x 3  y 5 25. Let unit’s place be ‘y’ & ten’s place be ‘x’.  The number is (10x + y).  The reversed number is (10y + x). (i) 7 times the number = 4 times its reverse.  7(10x + y) = 4(10y + x)  66x – 33y = 0  2x –y = 0 (ii) Difference of the digits is 3. y–x=3  – x + y = 3 ________ (ii) Solving the equations (i) & (ii) 2x – y = 0 x+y=3 On adding: x = 3 From (ii), we have, y = 3 + x y=3+3=6  The number is 10 × 3 + 6 = 36 Thus fraction

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318

26. Let the ten’s and the unit’s digits in the first number be x and y, respectively. So, the first number may be written as 10x + y in the expanded form (for example, 56 = 10(5) + 6) When the digits are reversed, x becomes the unit’s digit and y becomes the ten’s digit. This number, in the expanded notation is 10y + x (for example, when 56 is reversed, we get 65 = 10 (6) + 5). According to the given condition. (10x + y) + (10y + x) = 66 11 (x + y) = 66 x + y = 6 ________ (i) We are also given that the digits differ by 2. Therefore, x – y = 2 _____ (ii) y – x = 2 _____(iii) If x – y = 2, then solving (i) and (ii) by elimination, we get x = 4 and y = 2. In this case, we get the number 42. If y – x = 2, then solving (i) and (iii) by elimination, we get x = 2 and y = 4. In this case, we get the number 24. Thus, there are two such numbers: 42 and 24. 27. Let x be the digit in the units place, y be the digit in hundreds place Then, since the digit in the tens place is 0, the number will be represented by 100y + x And if the digits are reversed, the number so formed will be represented by 100x + y  100x + y – (100y + x) = 495  99x – 99y = 495 x – y = 5 _________ (i) Since the sum of the other digits is 11, we have: We have x + y = 11_________ (ii) From (1) and (2), we get x = 8, y = 3. Hence, the number is 308. 28. Let the usual speed of plane = x km/hr.  Increased speed, y = (x + 250) km/hour. Distance = 1500 km. According to the question, (Scheduled time) – (time in increasing the speed) = 30 minutes

1500 1500 1   x y 2

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 Dinstance   Time  Speed   

1500 1500 1   x x  250 2



1500x  375000  1500x 1  x  x  250  2

 x (x + 250) = 750000  x2 + 250x – 750000 = 0 By factorization, we get:  x2 + 1000x – 750x – 750000 = 0  x (x + 1000) – 750 (x + 1000) = 0  (x – 750) (x + 1000) = 0  (x – 750) (x + 1000) = 0  x = 750 or –1000 But speed can never be negative  Usual speed = 750 km/hr. 29. – x > – 5 – 2 [transposing 2 from LHS to RHS] –x>–7 Multiplying both sides of the inequation by (– 1), we get – x (– 1) > – 7 (– 1) [inequality changes as we multiply the inequation by a negative number] x 14 x

14 x>7 2

Thus, solution set = {8, 9, 10,………..} 32. (i) b – 1 > 9, b is a prime number less than 20. Given b – 1 > 9  b > 9 + 1  b > 10  Solution set = {11, 13, 17, 19} (ii) 2c – 3 < 11, c is a +ve even number. Given 2c – 3 < 11  2c < 11 + 3 14 c – 8, x is a negative integer. Given 2x > – 8 8 x>–4 2  Solution set = {– 3, – 2, –1} (iv) State the smallest integer n for which 4n > 17. Given 4n > 17 (n is the smallest integer)  4n > 17

319 2.

x

108 = 12 9  The number is 12 x

3.

17  n > 4.25 4  Solution set = {5, 6, 7….} ( n > 4.25) 4. 33. x > 5  x = 5 + l where l > 0 y < 6  y = 6 – m where m > 0 x – y = (5 + l) – (6 – m) =5+l–6+m = –1 + (l + m) = –1 + some positive integer x–y>–1  Least value of x – y is – 1. 34. Given that – x + 2  4, x  N . x+24 x4–2 x2  Solution set = {x : x  2 , x  N }. 35. Given: – 3x + 2  – 4 over R. Given, – 3x + 2  – 4  – 3x  – 4 – 2  – 3x  – 6  3x  6 ( Multiplying by ‘–1’ sign, changes inequality sign) n

x

6  x2 3

CONCEPTIVE WORKSHEET 1.

Equation x + y = 8 Solution 1+7=8 (1, 7) 2+6=8 (2, 6) 3+5=8 (3, 5) 4+4=8 (4, 4) 5+3=8 (5, 3) 6+2=8 (6, 2) 7+1=8 (7, 1) Solution set = {(1, 7) (2, 6) (3, 5) (4, 4) (5, 3) (6, 2) (7, 1)} The solution set is a finite set.

Let the number be x Nine times a number = 9x  9x = 108 (given)

5.

5 to the RHS, we get 2

Transposing

x 3 5 8 x     or = –4 3 2 2 2 3 Multiply both sides by 3, x = – 4 × 3 or x = – 12 Given : 3 +



x2 = 10. 6

18  x  2 = 10 6

16  x = 10 6  x + 16 = 60  x = 60 – 16 = 44 Hence, the value of ‘x’ is 44. Verification : Substituting the value of ‘x’ in the given equation, we have: 

3+

x2 = 10 6

3

44  2 6

42 = 10 6  3 + 7 = 10  10 = 10  L.H.S. = R.H.S Hence, verified. 3

4x  5 2  2x  7  3   2 6 3 Multiplying by 12 on both sides, we get: 2(4x + 5) – 8 (2x + 7) = 18  8x + 10 – 16x – 56 = 18  – 8x = 18 + 46  – 8x = 64

Given

x

64 =–8 8

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320

6.

Given

4a 3 (1 + x) – (a – x) 3 4 

3a 1  x  4  a  x  4 3

4a 1  x   4  a  x  3a 1  x   3  a  x   3 4 Multiplying by 12 on both sides, we get:  16a + 16ax + 16a + 16x = 9a – 9ax + 9a – 9x  32a – 18a + 16ax + 9ax + 16x + 9x = 0  14a + 25ax + 25x = 0  25x (a + 1) = – 14a 

x  7.

14a 25  a  1

Let x litres of first solution be mixed with (10 – x) litres of the second solution. Then, quantity of acid in x litres of first solution = 50% of x litres

 50  x   x  litres = litres 2  100  Quantity of acid in 10 litres of new mixture = 40% of 10 litres  40    10  litres = 4 litres  100 

8.

x 10  x   4 2 4  2x + (10 – x) = 16  2x + 10 – x = 16  x + 10 = 16  x=6 Hence, 6 litres of first solution should be mixed with 4 litres of the second solution. Let the distance covered on foot be ‘x’ km. Then, the distance covered on bicycle = (25 – x) km. Time taken to cover x km at 3.5 kmph  x  =  hours  3.5  Time taken to cover (25 – x) km at 9 kmph

 25  x  =  hours  9  

x 25  x  4 3.5 9

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2x 25  x 18x  175  7x  4  4 7 9 63 (by cross-multiplication)  11x + 175 = 252  11x = 77  x = 7 (dividing by 11 on both sides)  Distance covered on foot = 7 km. 9. Let the required number be ‘x’. Given divisor = 8, quotient = 11 and remainder = 6 we know that, Dividend = (Divisor × Quotient) + Remainder Here, Dividend = required number  x = 8 × 11 + 6 = 88 + 6 = 94 Hence, the required number is 94. 10. Let the required numbers be 3x and 8x respectively.Then, 3x + 8x = 165  11x = 165  x = 15 (on dividing both sides by 11)  One number = (3 × 15) = 45 The other number = (8 × 15) = 120 Hence, the required numbers are 45 and 120. 

x x x   7 2 3 4 LCM of denominators 2, 3, 4 on L.H.S. is 12. Multiplying both sides by 12, we get 6x + 4x – 3x = 7 × 12  7x = 7 × 12  7x = 84

11. We have,

7x 84  [Dividing both sides by 7] 7 7  x = 12 

3t  2 2t  3 2   t 4 3 3 The denominators on two sides are 4, 3 and 3. Their LCM is 12. Multiplying both sides of the given equation by 12,

12. We have,

 3t  2   2t  3  2  we get 12    12    12   t   4   3  3 

2   3 (3t – 2) – 4 (2t + 3) = 12   t  3  2 – 12t 3  9t – 6 – 8t – 12 = 8 – 12t  t – 18 = 8 – 12 t  t + 12t = 8 + 18

 9t – 6 – 8t – 12 = 12 ×

Linear Equations & Inequations

321

[Transposing (–12t) to LHS and (–18) to RHS]  13 t = 26

= 150 x paise = Rs

26 13  t=2 t



But, the total amount of the money in the bank is given as Rs. 35.

3x  35 2  2x + 3x = 70 [Multiplying both sides by 2]

x 

5x 70   x = 14 5 5  Number of one rupee coins = 14, Number of 50 paise coins = 3x = 3 × 14 = 42.

2 5 14. Let the number be x. Then,

 5x = 70 

 5x = 2  x 

Quarter of x 

x ; One4

x . 12 Average of third, quarter and one-twelfth of x is twelfth of x 

16. (i) – (ii) =

It is given that the number x is 56 greater than the average of the third, quarter and one-twelfth of x.

1 x x x   x       56 3  3 4 12 

x x x    56 9 12 36

16 4 4

Substituting value of ‘y’ in (i), we get 3x – 4 = 5

9 =3 3 Solution set = {(3, 4)} 17. Find the solution sets of the system of equations 2p + q = 56 ; 2p – q = 11 2p + q = 5 ––––––––– (i) 2p – q = 11 –––––––– (ii) (i) + (ii) = 4p = 16  3x = 9

x x x    56 9 12 36  36x – 4x – 3x – x = 36 × 56 [Multiplying both sides by 36 i.e., theL.C.M. of 9, 12 and 36]  36x – 8x = 36 × 56  28x = 36 × 56 x

28x 36  56   28 28  x = 36 × 2  x = 72 Hence, the number is 72. 15. Let there be x one rupee coins in the bank. Then, Number of 50-paise coins = 3x  Value of x one rupee coins = Rs x Value of 3x fifty-paise coins = 50 × 3x paise

3x – y = 5___ (i) 3x + 3y = 21_ __ (ii) – – – – 4y = – 16 y=

x x x      1 x x x 3 4 12          3 3  3 4 12 

x

3x 

 Total value of all the coins = Rs  x  2   

1 1  0 13. Given 2x  1 3x  1 L.C.M. of the denominators is (2x – 1) (3x – 1)  3x – 1 + 2x – 1 = 0 × (2x – 1) (3x – 1)  5x – 2 = 0

1 One third of x = x; 3

150 3x x = Rs 100 2

 x=

16 =4 4 Substituting in equation (i), we get 2 (4) + q = 5 q=5–8=–3   Solution set = {(4, –3)} P 

18.

x y x y   3;  4 7 15 2 5 15x  7y 3 15x  2y 4  ;  105 1 10 1  15x + 7y = 315 ––––––– (i) 5x – 2y = 40 –––––––– (ii)

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8th Class Mathematics

322 (i) × 1 = 15x + 7y = 315 (ii) × 3 = 15x – 6y = 120 – + – Subtracting

13y = 195

195 = 15 13 Substituting ‘y’ in equation (ii), we get 5x – 2 (15) = 40 5x = 40 + 30 = 70 y=

70 = 14 5  Solution set = {(14, 15)} x=

19. Given: 1 

x 13  ____(i) y 8

2y = 8  y = 4 21. We have x +

y+

4 = 4 ––– (1); y

6 = 6 –– (2) x

From (2), y = 6 –

6 6x  6 = ––––– (3) x x

2x + 3y = 34 Taking ‘y’ common, we get:

Substituting (3) in (1), x +

 x  y  2    3  34 ______(ii)  y  From (i), we get:



x 13 5   1  ____(iii) y 8 8 Now substituting

x 5  in (ii), we get: y 8

 5  y  2    3  34  8 

 5  12   y  34  y = 8  4  Substituting y = 8 in (iii), we get: x =5. x = 5 and y = 8. 20.

By solving 3 & 4 2x = 8  x = 4 2y = 6  y = 3 By solving 3 & 5 2x = 6  x = 3

1 1 7   x y 12 –––––– (1), xy = 12 ––––– (2)

xy 7  xy 12 xy 7   x + y = 7 ––––– (3) 12 12 Now (x – y)2 = (x + y)2 – 4xy = (7)2 – 4 × 12 (x – y)2 = 1 x–y=  1 x – y = 1 ––––– (4); x – y = – 1 –––– (5) www.betoppers.com

4x =4 6x  6

6x 2  6x  4x 4 6x  6

 6x2 – 2x = 24x – 24  6x2 – 26x + 24 = 0,  3x2 – 13x + 12 = 0 By factorization, we get:  3x2 – 4x – 9x + 12 = 0  x (3x – 4) – 3 (3x – 4) = 0 (x – 3) (3x – 4) = 0 x = 3,

From (4), when x =

4 ––––– (4) 3 4 3 , then y = (from (2)) 3 2

From (4), when x = 3, then y = 4 (from (2)) 22. x + y = 9 ––––– (1) x2 + y2 + xy = 61 x2 + y2 + 2xy – xy = 61( 2xy – xy = xy) (x + y)2 – xy = 61(a2 + b2 +2ab = (a + b)2) 92 – xy = 61 xy = 81 – 61 xy = 20 Clearly, x + y = 9, 5 + 4 = 4 + 5 xy = 20, 5 × 4 = 4 × 5  (x, y) = (5, 4) or (4, 5)

Linear Equations & Inequations 23. Let numerator be ‘x’ & denominator be ‘y’. Condition-(i): (N-Numerator, D-denominator) N × 3 and D – 3, Fraction becomes (18/11)

26. (i) Given 3a + 1 < 16  3a < 16 – 1  3a < 15

15  a 4 (x – 3)  y = 50 – 30  y = 20 15 – 5x > 4x – 12   The two numbers are 30, 20  15 + 12 > 4x + 5x 25. Let x be the smaller number then x + 4 is the greater.  27 > 9x One half of the greater is represented by 27  x> 1 1 9 (x + 4), and one-sixth of the lesser by (x). 2 6  x>3 30. x – 3 < 2x – 1 1 1 x  {1, 2, 3, 4, 5, 6, 7, 8, 9} Hence , (x + 4) – x=8 2 6  –3 < 2x – x – 1 multiplying by 6, 3x + 12 – x = 48  – 3 + 1 < 2x – x  –2 10 2x < 15 and 3x > 10 + 9

x

15 and 3x > 19 2

x < 7.5 and x >

19 3

 x < 7.5 and x > 6.3  x = 7 (Integer) 28. Clearly 32 + 42 = 52  x=2

29. Greatest value of x = 5 Greatest value of y = 9 Greatest value of xy = 5 × 9 = 45 30. x + 1 > 5  x > 5 – 1  x>4

 x  {6, 7} 31. P = {x : 5 < 2x –1  11, x  R} Here 5 < 3x – 1  111  5 < 2x – 1 and 2x – 1  111  5 + 1 < 2x and 2x  11 + 1  6 < 2x and 2x  12 

6 12  x and x  2 2

 3 < x and x  6 3 5  x

5 2

16  x=8 5 2  x < 5 and x > 2 x – y = 4 –––––––– (a) Substituting the value of ‘x’ in equation (a) 5  x5 we get 2 8–y=4 27. 3x + 1 < 10 8–4=y  y=4  3x < 10 – 1 Thus :  3x < 9 Speed of man in still water = 8 km/hr.  x3+2  x>5 – x + 7 > 4x – 3 Subtracting ‘–4x’ on both sides  – x – 4x + 7 > 4x – 4x – 3  – 5x + 7 > – 3  – 5x > – 10  5x < 10  x < 2 2 (x + 1)  x + 5  2x + 2  x + 5 Subtracting ‘x’ on both sides  2x – x + 2  x – x + 5  x+2  5  x  3 [Transposing ‘2’ to R.H.S] 3 + 5x > 3x – 3 Subtracting ‘3x’ on both sides  3 + 5x – 3x > 3x – 3x – 3  3 + 2x > – 3  2x > – 3 –3 [Transposing ‘–3’ to R.H.S]  2x > – 6  x>–3 5x + 4 < 2x + 19  5x – 2x < 19 – 4  3x < 15  x – 3 [Multiplying ‘–’ sign inequality is changed] 

3  x > –1.5 2 x = – 2 27. 2x + 10 < 3x + 15  2x – 3x < 15 – 10  –x–5  Solution set = {x : x > – 5, x  R}

 x>

 

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8th Class Mathematics

8. SETS AND RELATIONS SOLUTIONS

FORMATIVE WORKSHEET 1.

2.

3.

4. i) ii) iii) iv) v) 5.

6. 1) 2) 3)

4) 5) 7.

(iii), (iv), (v), (viii), (ix) and (x) are not well defined. Therefore they cannot be considered as sets. Hence, (i), (ii), (iv) and (vii) are sets. i) False ii) True iii) False iv) True v) False vi) True Element of A is x which can be any letter of ‘ADCL’ So x can be A, D, C, L Therefore, A, D, C, L can be written in the place of element x of A.  A = {A, D, C, L} A = {x/x is a square of a natural number, x £ 25} B = {x/x is a month whose name begins with A} C = {x/x is a multiple of 5, x  30} D = {x/x = 1/n, n  6 and nN} F = {x/x is a multiple of 10, x  90} i) A = {6, 7, 8, 9, 10, 11} ii) B = {1, 13} iii) C = {– 3} iv) D = {2, 3, 5, 7} v) G = {A, C, U, M, L, T, O, R}

8. 9.

10.

11.

12.

A = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday} D = {1, 8, 15, 22, 29, ………} C = {I, N, D, A}. (The letter ‘I’ has come twice. But writing the elements of the set, we write one I only) S = {a, e, i, o, u} X = {5, 10, 15, 20, 25, 30, …… 95} 1) A = {1, 4, 9, 16, 25} A = {x2 / x  N and x < 6} 2) B = {1, 3, 5, 7, 9, 11, 13} B = {x/x odd natural numbers and x3 and ‘p’

is prime ,  p  1  p  1 is divisible by ‘3’.Also since p>3 , all primes are odd. Hence ,

 p  1

are even . Therefore

 p  1

 p  1  p  1

multiple of





and

2 ‘8’ and therefore p  1 is multiple of 24.

is

Number Theory Solutions 28.

29.

30.

381

All natural numbers can be divided into groups of ‘6’ as 6k,6k+1,6k+2, 6k+3,6k+4 and 6k+5. 6k, 6k+2,6k+3,6k+4 are composite. Hence for

6.

p1, p2 , p3 ,.... pk . Let us consider the number N which is written as N  p. p2 . p3 .... pk  1 . It is

prime number p  5 has the form 6k+1 or 6k+5. In both cases remainder is ‘1’ when square is divided by 12. ab = 10a+b. 10a+b =3(a+b) or 2b = 7a. Therefore ‘b’ is equal to 7 . Then a = 2. Hence the number is 27.

161 1 9999 1  1 ,  1 160 160 9998 9998

7. and

8.

1 1  . 160 9998

As x n  y n is divisible by (x+y) for ‘n’ being odd. S is divisible by 2007 in this way. Group the terms as below :

2.

3.

4.

5.

2007

 1004

2007

19  9  10  19  5  9  855 . 2

Hence it is divisible by ‘9’. Hence the problem. If A is a square , then it ends in an even number of zeros. B consisting of 300 threes and some zeros. With B ending in 3 . Since B is odd, 2B cannot be a square . It has only one factor 2. n! = 215  36  53  7 2  11 13 . As 13 is the greatest prime in the list n  16 . But as we have 3,

5 15 is present in the product. Hence n! is 16!

 y

under

n

 y n  0 only when ‘n’ is odd. If

any n

circumstances. Hence proved.

n

9.

F(x) = y  y  0 for all ‘n’.

10.

n 2n 11n 2  12 2n1 = 121.11  12.12

= 133.11n  12.11n  12.122 n



= 12 122 n  11n



Each group is divisible by 2007. Hence , S is divisible by 2007. Unit’s digit is ‘3’.Ten’s digit is ‘1’.From 9! onwards all the number factorials will have even the tens digit as zeros. Hence tens digit is the sum of tens digits of numbers which are 1! to 9!. The number of 1’s occurring in the digits from 10 to 19 = 11 and from 20 to 99 = 8. So total of ones is 11+8=19. Similarly , No of 2’s , 3’s , 4’s , .......9’s are equal to 19.So sum of all the digits = 19(1+2+3+4......+9) =

F(x) = x n  y n . To show that it is divisible by x+y only when ‘n’ is odd.

‘n’ is even then F(-y) = 2 y n which is not zero

12007  20062007    2 2007  20052007  .

 .....  1003

not divisible by any of the primes given . Hence there are only two cases that either it is a prime or there exists a new prime which can divide it . Hence the assumption that there are only finite primes is not true. Last digit is 0.

F(-y) =

IIT JEE WORKSHEET 1.

Let there be only finite primes . They be







= 12 144n  11n .

Hence the problem. 11.

No . Since n2  n  1 never ends in 0 or 5.

12.

Let N be the sought for number ; Then N 2  N has three zeros at the end. It is possible only when this difference is divisible by 8 and 125. In the process we get the numbers as 625 and 376.

13.

1  .abc n Then

100c  abc.abc thus , n

999 1000 1    abc . n n n Therefore n(abc) = 999. But 999 = 33.37 So the factors with distinct digits are therefore 27= 027 and 37 = 037 . Therefore n 

considering the power of 2. or n 

999 999   37 abc 27

999 999   27 . abc 37 www.betoppers.com

8th Class Mathematics

382 14.

25

10025  102   1050

6

21.

 10000...00000(50 zeros) Hence when 25 is subtracted the answer is 9999......9975 where 9’s are 48 nines. 15.

11

 1  11  1 .

11

 118  117  116  115  ...  110 

10

9

16.

17.

Hence it is divisible by 100. Let us consider the following cases: If ‘x’ and ‘y’ are positive integers then they are necessarily odd or even Then we have the following combinations for ‘x’ 23. and ‘y’ x is odd and y is odd. x is even and y is even x is odd and y is even 24. x is even and y is odd. That leads to conclusion that if there exists an integer then there lies a contradiction. Clearly x  0 and y  0 . Solving for ‘y’ we get

y

18.

19.

20.

22.

25. 4x 16  4 x4 x  4 .  x  4 divides 16 and

since x  0 and y  0 , x  4 must be in the set { -16,-8,-2,-1,1,2,4,8,16}. So there are nine integer solutions . (-12,3) , (-4,2) , (2,-4) , (2,-4) , (3,-12) , (5,20) , (6,12) , (8,8) , (12,6) and (20,5). The units digits of 171983 and 71983 must be the same. The units digit of 111983 is 1. The final answer is therefore 1. The sides form a Pythagorean triple. So one leg must be even. Since no such triple can contain the number 2, the given primes cannot both be legs. Let the primes be p and q (= p + 50) = 50(2p + 50) = 100(p + 25) = a2. Thus p + 25 must be a square; the smallest such p is 11, so a2 = 100 × 36 and a = 60. If ‘y’ is odd, then xodd + 1 would be divisible by x + 1 and would not be a prime. Thus ‘y’ is even, and must be 2. Also, since ‘z’ must be odd, ‘x’ must be even; thus x = 2. This makes z = 5, so the only answer is (2, 2, 5).

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This is

4  4 5 6  4 5 4 6 66  4 6    6  6     = 46 = 3  35 2  2 5 3 2 3 2

212, so n = 12. 1 down must be 125 or 625. 5 across must be 529 or 576. 3 down must be 243 or 729, of which only 729 can work. 2 down must be 128 or 256 or 512, of which only 512 can work.1 across must now be 157, since 657 is a multiple of 3. The solution is

1

5

7

2

1

2

5

2

9

This is a break-up of the fairly well known fact that 6! 7! = 10! and is easily seen, since 3! 5! = 6! = (3  2) (5) (4) (3) (2) = (2  5) (3  3) (4  2) = 10  9  8. Thus N = 10. [Note: A palindrome is a positive integer that reads backwards the same as it reads forwards. For example: 67276] How many three-digit palindromes have the property that their square roots are two-digit palindromes? [Remainder: A two or three digit number cannot begin with a 0] 1988K = 4  7  71K  K = 7  71 = 497.

 

10. MENSURATION SOLUTIONS Area of surface = 220.5 m2 Height of the board = 24.5 m

FORMATIVE WORKSHEET 1.

Given the base of an isosceles triangle = 12 cm. i)

1 We know that the Area   base  height 2

1 1 Area of PQR  bh   QR  PS 2 2

1  220.5   base  24.5 2

1   4cm  2cm = 4 cm2 2 ii)

2.

i)

4.

1   3cm  2cm = 3 cm2 2 QR = base = 4 cm, PL = height = 5 cm 1 Area of the triangle PQR  bh 2

220.5  2  18m 24.5  Base of the triangular board = 18m. Let AC be the width of the street then, AC = x + a Let us consider the triangle W1 AB which is a right angled triangle. By Pythagoras theorem,  base 

1 1 Area of LMN  bh   MN  LO 2 2

W2 W1

1   4cm  5cm = 10 cm2 2 P

A 2

5cm

M

L

ii)

Q

4cm

R

PR = base = 8cm, QM = height = ? Area = 10 cm2

1 Area of the triangle PQR  bh 2 1  8cm  h  10 2 10 5   2.5 4 2  QM = 2.5 cm Total cost of painting = Rs. 176.40 Rate = 0.80 rupees per square metre  h

3.

 Area of the surface





Total cost Rate per sq.m

176.40 = 220.5 m2 0.80

40

40

24

x 2

32 a

B 2

C

(W1B) = (W1A) + (AB) (40)2 = (24)2 + x2 x2 = (40)2 – (24)2 = 1600 – 576 = 1024 = (32)2  x = 32 m ________ (1) Let us consider the triangle W2BC, which is a right angled triangle, By Pythagoras theorem, (W2B)2 = (BC)2 + (W2C)2 (40)2 = a2 + (32)2  a2 = (40)2 – (32)2 = 1600 – 576 = 576 = (24)2  a = 24m ___________ (2)  The width of the street = AC = x + a = 32 + 24 = 56 m

8th Class Mathematics

384 8.

A

5.

2x

2x

Let the length of the equilateral triangle be ‘a’ cm. Given the area of an equilateral triangle  64 3 sq. cm We know that the area of an equilateral Triangle 

B

3x

C

Let the three sides of a isosceles triangle be a, a, b. Given that their ratio is 2 : 2 : 3. a = 2x a = 2x b = 3x Where x is any constant Given perimeter = 56 cm = a + a + b  2x + 2x + 3x = 56 cm  7x = 56 cm  x = 8 cm  a = 2x = 2 × 8 = 16cm a = 2x = 2 × 8 = 16cm b = 3x = 3 × 8 = 24cm  The equal sides are 16, 16 and the third side is 24cm. 6.

C

h= 4

d = hypotenuse

3 2 3 2 a  a  64 3 4 4

 a2 = 4 × 64 = 256  a = 16 Height of an equilateral triangle. h

9.

3 3 a units   16 cm  8 3 cm 2 2

Given the base of an isosceles triangle = 12 cm. A

x

x

B

C

12 cm

The perimeter of the triangle = 32 cm  Base + 2 × side = 32 cm  12 + 2 × side = 32 cm  2(side) = 20 cm 20  side  cm  10cm 2  The length of equal sides of isosceles triangle = 10 cm. A

A

b= 3

B

10 cm

Given base of a right triangle = b = 3cm Height = h = 4cm  Hypotenuse  d  b 2  h 2



 3

2

  4

2

 9  16

B

10 cm

By Pythagoras theorem,  base  (height)2 = (side)2     2 

 25  5cm

7.

2

2

 Hypotenuse = d = 5cm Perimeter of a right triangle = b + h + d = 3 + 4 + 5 = 12cm

 (height)

Let a = 5, b = 6 and c  61 Now, a2 + b2 = 52 + 62 = 25 + 36 = 61 = c2  The triangle is right – angled triangle with sides equal to 5 and 6 units.

 height =

1 1 Area  ab   5  6 = 15 sq. units 2 2 www.betoppers.com

C

12 cm

2

 12  = (10)     2 = 100 – 36 = 64 cm2 2

64 = 8 cm

1 The area of the triangle   base  height 2 1   12cm  8cm  48cm 2 2

Mensuration

385

10. We can find the area of the shaded region ABDC by subtracting the area of the triangle BDC from the area of the triangle ABC. i.e., Area of the shaded region ABDC = (Area of the triangle ABC) – (Area of the triangle BDC) _________ (1) ABC is an equilateral triangle with side, a = 10cm. A

D

B

10cm

C

We know that the area of an equilateral triangle 3 2 3 2  a   10   3  25 4 4 = 1.732 × 25 = 43.3 sq. cm __________ (2) Given BDC is a right triangle with right angle at ‘D’ and CD = 8cm and BC = 10cm. By Pythagoras theorem, (BD)2 = (BC)2 – (CD)2 = (10)2 – (8)2 = 36 cm2

 BD  36  6cm

Let the height be h cm. 1 27 3 Then,  3 3  h  2 4

27 3 2 9    4.5cm 4 3 3 2 13. Area of the right angled isosceles triangle is 32 sq.cm. h

1  a 2  32 2 2 a = 32 × 2 a2 = 64

a  64  8cms In an isosceles right angled triangle hypotenuse  2 times its side. h  2a units a = 8cms h  8 2 cms Lengths of sides of an isosceles right angled triangle

are 8cms, 8cms 8 2 cms. 14. Let the length of a side be ‘a’.

1 Area of BDC   CD  BD 2 1   8  6 = 24cm2 _________ (3) 2

Area  64 3 sq.cms We know that the area of an equilateral triangle

Substituting (2) and (3) in (1) we get Area of the shaded region ABDC = 43.3 – 24 = 19.3 cm2. 11. Let the bases of the two triangles be x and y and their heights be 3h and 4h respectively. 1  x  3h 4 2   1  y  4h 3 2

3 2 a  64 3 4 a2 = 64 × 4 a2 = 256

a  256 = 16 cm. Perimeter = 3a = 3 × 16 = 48cms. Height =

3 3 a =  16  8 3 cm 2 2

1 2 15. Area of the square    diagonals  x  4 4  16 2     y 3 3 9 1     3.8  3.8  m 2  The ratio of their bases = 16 : 9. 2   12. We know that the area of an equilateral triangle 2 = 7.22 m 3 2 16. Let the diagonals of the squares be 2x and 5x  a 4 respectively. We know that area of the square 2 3  3 3 Area of the triangle  1 2 4    diagonals  2 27 3 2  cm 4 www.betoppers.com





8th Class Mathematics

386

1 2 1 2  Ratio of their areas    2x  :   5x  2 2 2 2 = 4x : 25x = 4 : 25.  Ratio of their areas = 4 : 25 17. The length of the diagonal of a square  12 2 cm If ‘a’ is the side of the square then the length of diagonal is

2a

 2 a  12 2  a = 12 Perimeter of square = 4a = 4 × 12 = 48 cm.



1 d1  2d1  25 2

 d12  25  d1 = 5  d2 = 2 × 5 = 10  Sum of lengths of diagonals = (5 + 10) cm = 15 cm 22. Let d1 , d2 be the diagonals of rhombus. Given that d1 = 24 cm and d2 = 10 cm Area of rhombus

1     24  10  cm2 2   2 = 120 cm

 40  18. Side of first square    cm  10cm  4  ( perimeter of a square = 4a)  32  Side of second square    cm  8cm  4  Area of third square = [(10)2 – (8)2] = (100 – 64) cm2 = 36 cm2 Side of third square  36 cm  6cm ( Area of a square = a2)  the perimeter of third square = (6 × 4) cm = 24 cm.

1  1 OA  d1    24  cm  12cm 2 2   1 1  OB  d 2    10  cm  5cm 2 2   AOB is a right angled triangle (? AOB = 90o) D

C

O

19. Side  2550.25 ( Area of a square = a2)



255025 100

505 10 = 50.5 m. 20. Area of the square room of side 3 metres or 300 cm = (300)2 cm2 Area of the marble slab = (20 × 30) cm2 Number of marbles 

 Area of the Room     Area of the Marble   300  300  Number of marbles     150  20  30   The number of marbles needed = 150 21. Let d1 , d2 be the diagonals of rhombus. Given that d1= d1, d2 = 2d1

1 We know that area of rhombus  d1  d 2 2 www.betoppers.com

1  d1d 2 2

B

A 2

 AB

2

2

= OA + OB = (12)2 + 52 = 169  AB  169  13 cm

 AB = side of rhombus = 13 cm  Perimeter = (13 × 4) cm = 52 cm. 23. We know that perimeter of the rhombus = 4 × side. Given perimeter of the rhombus = 56 m.

56 m  14m 4 Given height of the rhombus = 5 m.  Area = (14 × 5)m2 = 70 m2. 24. Side of square, a = 8m.  Perimeter of the square = 4a = (4 × 8)m = 32m.  Each side of the rhombus 

Diagonal of the square  2a 





2 8 m

= (1.41 × 8)m = 11.28 m. Area of the square = a2 = (8 × 8)m2 = 64 m2.

Mensuration

387

25. Let the breadth of the rectangle be b metres. By Pythagoras theorem, we have (diagonal)2 = (length)2 + (breadth)2 i.e., l2 + b2 = d2 (12)2 + b2 = (15)2

28. Let length =  = 5x metres and breadth = b = 3x metres. Then, area of the park = ( × b) = (5x × 3x) m2 = (15x2) m2  15x2 = 2160

 x2 

b

12m 2

 144 + b = 225  b2 = (225 – 144) = 81

2.5m

 b  81  9  Breadth (b) = 9m.  Area of rectangle = ( × b) = (12 × 9) m2 = 108 m2. 26. Let the length be  cm. We know that the perimeter of rectangle = 2 (  + b)  2 × ( + 8.4) = 36   + 8.4 = 18   = (18 – 8.4) = 9.6.  Length = 9.6cm. Area of rectangle = ( × b) = (9.6 × 8.4) cm2 = 80.64 cm2. Hence, the length of the rectangle is 9.6 cm and its area is 80.64 cm2. 27. 2.5m 25m

38m

Length of the lawn =  = 38 m. Breadth of the lawn = b = 25 m. Area of the lawn = ( × b) = (38 × 25) m2 = 950 m2 Length of the lawn including the path = [38 + (2.5 + 2.5)] m = 43 m. Breadth of the lawn including the path = [25 + (2.5 + 2.5)] m = 30 m Area of the lawn including the path = (43 × 30) m2 = 1290 m2. Area of the path = (area of lawn including path ? area of lawn) = (1290 – 950) m2 = 340 m2. Rate of gravelling = Rs. 4.50 per m2.  Cost of gravelling the path = Rs (340 × 4.50) = Rs. 1530.

2160  144 15

 x  144  12  12  12  Length =  = (5 × 12) m = 60 m. Breadth = b = (3 × 12) m = 36 m  Perimeter = 2 ( + b) = 2(60 + 36) m = 192 m Rate of fencing the park = Rs. 5 per metre.  Cost of fencing the park = Rs (192 × 5) = Rs. 960. 29. From the parallelogram ABCD: D

C 10.8 m M 10.8 m L B

A

We have, diagonal = AC = 42 m perpendicular distance from diagonal to either vertices DL = BM = 10.8 m From the figure, it is clear that diagonal AC divides the parallelogram into two equal triangles.  Area of the field = 2 × Area of  ACD

1  2   42  10.8m 2 = 453.6 m2 2 30. Let ABCD be the given parallelogram in which C D h E

A

F 15 cm

B

AB = 15 cm and AD = 12 cm. Let CE  AD and CF  AB. Then, CE = 7.5 cm. Let CF = h cm.  Area of parallelogram ABCD = (AD × CE) sq.units. = (12 × 7.5) cm2 = 90 cm2 Also, area of parallelogram ABCD = (AB × CF) sq.units. = (15 × h) cm2.  15 × h = 90

 90   h     cm  6cm  15  Hence, the distance between the longer sides = 6 cm. www.betoppers.com

8th Class Mathematics

388 31. Given ABCD is a parallelogram with AB = 7cm and BC = 5cm. D

C

5cm A

7cm

B

Let AC is the diagonal of length 12cm. Consider the triangle ABC. The sum of the two sides AB and BC is equal to the third side AC which is not possible because in a triangle, the sum of two sides is greater than the third side. Hence, no diagonal of length 12cm can exist when the sides are 7cm and 5cm. (ii) In the parallelogram with AB = 7cm and BC = 5cm, diagonal is 8 cm. Then, the area of the parallelogram ABCD = 2 × area of ABC The area of ABC  s  s  a  s  b  s  c 

33. The perimeter of a parallelogram = 32cm Let  be the length and b the breadth of the parallelogram. Given,  : b = 3 : 1   = 3x and b = x, where ‘x’ is any constant. The perimeter of a parallelogram = 2( + b)  2(3x + x) = 32cm  2(4x) = 32cm  8x = 32cm  x = 4cm  The length = 3x = 3 × 4 = 12cm The breadth = 1x = 1 × 4 = 4cm. 34. Let the length and the breadth of a rectangular field be ‘’ m and ‘b’ m respectively Given:  : b = 4 : 3   = 4x and b = 3x. (Where ‘x’ is any constant) The area of a rectangle =  × b sq. m. = 4x × 3x = 12x2 sq. m. But the area of the rectangular field = 8112 m2  12x2 m2 = 8112 m2

8112  676  x  676  26 12  The length = 4x = 4 × 26 = 104 m. The breadth = 3x = 3 × 26 = 78 m The perimeter of the rectangular field = 2( + b) units = 2(104 + 78) m = 2(182) m = 364 m The cost of fencing one metre of field = Rs. 4.00 The cost of fencing 364m of field = Rs. 4.00 × 364 = Rs. 1456.  x2 

abc 8  7  5 20    10cm 2 2 2  Area of  ABC where s 

 10 10  8 10  7 10  5 

 10  2  3  5  10 3 cm 2  the area of parallelogram ABCD = 2 × area of ABC

 b2 

48  16 6  0.50

 b  16  4  = 2b = 2 × 4 = 8m and b = h = 4m  The height of the room = 4m. www.betoppers.com

8cm

12m

 2  10 3 cm 2  20 3 cm 2 32. Let ‘’ be the length, ‘b’ be the breadth and ‘h’ be the height of a room. Given  = 2b and b = h The area of the four walls = 2( + b) × h  The area of the four walls of the room = 2(2b + b) × b = 6b2 m2. The cost of papering one square metre = Rs. 0.50 p The cost of papering 6b2 m2 = 6b2 × Rs. 0.50 p Given the cost of papering the walls = Rs. 48.  6b2 × 0.50 = Rs. 48

35. 6cm

14m

The length of the rectangular region = 14 m = 1400 cm. The breadth of the rectangular region = 12 m = 1200 cm The length of a slab = 8 cm The number of slabs to be placed along the length

1400  175 8 The breadth of a slab = 6 cm The number of slabs to be placed along the breadth of the rectangular region 

1200  200 6 So, the number of slabs required to cover the floor of the hall = 175 × 200 = 35000. of the rectangular region 

Mensuration

389  52 = 32 + CE2  CE2 = 25 – 9

R 2.5 cm

36.

 CE2 = 16  CE  16  4 m  Area of quadrilateral ABCD = Area of rectangle AECD + Area of BEC

S 1.5 cm

Q

P

In this case, d = 5.5 cm, h1 = 2.5 cm, h2 = 1.5 cm.

1   2 1  AE  CE   BE  CE   5  4   3  4  m 2 2   2 2 = (20 + 6)m = 26m 39. Area of the trapezium

1 Area of a quadrilateral  d  h1  h 2  2

D

5.6cm

1   5.5   2.5  1.5  cm 2 2 1   5.5  4cm 2 2 = 11 cm2

A

37.

B

E

1    sum of parallel sides    distance betweem them  2

C

1     18  15   5.6  cm2 = 92.4 cm2 2  Hence, the area of the given trapezium is 92.4 cm2. 40. Let ABCD be a trapezium with AB || CD.

L

D

C

M

D

x

C

B

A

In quadrilateral ABCD, we have AC = 20m Let BL  AC and DM ? AC such that BL = 8.5 m and DM = 11 m  Area of quadrilateral ABCD

1   AC   BL  DM  2

1     20   8.5  11  m2 2  2 = (10 × 19.5)m = 195 m2 38. We have, BE = AB – AE = AB – DC = (8 – 5)m = 3m Using Pythagoras theorem in DEC, we have BC2 = BE2 + CE2 BC2 = BE2 + CE2 D

A

5 cm

C

E 8m

B

7 A

x (x + 6)

E

B

Draw CE  AB Let CD = x cm AB = AE + EB = CD + EB ( AE = CD)  AB = (x + 6) cm The height of the trapezium = 7cm Given the area of the trapezium = 126 sq.cm

1    sum of the parallel sides   height 2 = 126 sq.cm 1    x  x  6   7  126sq.cm 2 126  2 = 18 × 2 = 36 7  2x = 36 – 6 = 30  x = 15cm  CD = 15cm AB = x + 6 = 15 + 6 = 21 cm Hence, the lengths of the parallel sides are 15 cm and 21 cm. www.betoppers.com   2x  6  

8th Class Mathematics

390 41. Let the lengths of parallel sides be (5x) cm and (3x) cm. Then, the area of the trapezium

45. Draw CM perpendicular from C to AB. Clearly, AMCD is a rectangle. D

7 cm

C

1      5x  3x   12.5 cm2 2   = 50x2 cm2

 450  2  50x = 450  x   9  50  2

 96x = 1440

1440 96  x = 15 m Hence, the length of longer parallel side = (5x) = 75m. 43. Area of trapezium x

1    sum of the parallel sides   height 2 Area of trapezium

1     1.5  2.5   6.5 m2 2  = 13 m2 44. Let the altitude of the trapezium be h cm. We have, area of the trapezium = 26 cm2.

1    Sum of the bases  × Altitude = 26 2 1   6.5  Altitude = 26 2 26  2 cm  8cm 6.5 Hence, the altitude of the trapezium is 8cm.  Altitude =

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B

 AM = DC  AM = 7 cm [ DC = 7 cm]  AB – BM = 7  13 – BM = 7  BM = (13 – 7) cm = 6 cm Applying Pythagoras theorem, in right angled triangle BMC, we have BC2 = BM2 + CM2  102 = 62 + CM2  CM2 = 100 – 36 = 64  CM = 8 cm  Height of trapezium ABCD = 8 cm Hence, area of trapezium ABCD

x  9 3 The lengths of parallel sides are 5x = 5 × 3 = 15 cm 3x = 3 × 3 = 9 cm  The lengths of parallel sides are 15 cm and 9 cm. 42. Area of field 1      5x  3x   24  m 2   96x  m2 2 

M 13cm

A

1 1  AB  DC   CM   13  7   8cm 2  80cm2 2 2 46. Here r = 10.5 cm Circumference of the circle = 2r 

10.5 cm

22     2   10.5  cm 7   = 66 cm  22  2 Area of the circle = r2    10.5  10.5  cm  7  2 = 346.5 cm 47. Let the radius of the circle be r cm. Then, circumference = 2r

 2r  22  2 

22 44  r  22  r  22 7 7

7  22  7  r   cm  cm  3.5cm 2  44  7  Radius of the circle = 3.5 cm  cm 2 Area of the circle

 22 7 7  2 = r2      cm  7 2 2 = 38.5 cm2

Mensuration

391 51. Clear, diameter of the circle = Diagonal BD of rectangle ABCD

48. Let the radius of the circle be r cm. Then, area =  r2 cm2.

r 2  55.44 

22 2  r  55.44 7

A

B 8cm

 55.44  7   r2      2.52  7   22 

6cm

D

 252  7   7  6  6  7     100  100   

C

2 2  Diameter  BD  BC  CD

 7  6  42 7 66 7 r  4.2  100  10  10  Radius of the circle = 4.2cm. Circumference of the circle = 2r r

22     2   4.2  cm = 26.4 cm 7   49. Length of wire = Perimeter of the square = (4 × 27.5)cm = 110 cm When this wire is bent into the form of a circle, then clearly the circumference of the circle would be 110 cm. 52. Let r be the radius of this circle. Then, circumference = 2r

 6 2  8 2 cm  10cm Let r be the radius of the circle. Then,

10 cm  5cm 2 Area of the circle r

= r2 = 3.14 × (5)2 cm2 = 78.50 cm2. Area of rectangle ABCD = AB × BC = (8 × 6)cm2 = 48 cm2 Hence, area of the shaded region = Area of the circle – Area of rectangle ABCD = (78.50 – 48)cm2 = 30.50 cm2. C

22    44r    2   r  cm    cm 7    7  44r  110  7   110  r    cm  17.5cm 7  44  Hence, the radius of the circle is 17.5 cm 50. Radius of the wheel = 35 cm. Circumference of the wheel = 2r

A



22     2   r  cm 7  

The centre O of the circle lies an AB and let r cm be the be the radius of the circle. Since radius is perpendicular to the tangent at the point of contact.

1 The area of BOC   13  r sq.cm 2

22     2   35  cm  220 cm = 2.2 m. 7   Distance covered in 1 rotation = 2.2 m. Distance covered in 250 rotations = (2.2 × 250)m = 550 m Distance covered in 5 minutes = 550m.  Distance covered in 1 minute 

B

O c = 15 cm

550 5

Distance covered in 1 hour(60 min)

 550    60  m  6600m = 6.6 km/hr  5 

1 The area of AOC   14  r sq.cm 2 Hence the total area of the ABC = BOC + AOC

1 1     13  r   14  r  sq.cm 2 2  1   r  27 sq.cm 2

_______

(1)

Again if S = semi-perimeter 

13  14  15 cm 2

= 21 cm www.betoppers.com

8th Class Mathematics

392 The area of ABC  s  s  a  s  b  s  c 

R

 21 21  13  21  14  21  15   21 8 7  6 

 R 2    Required area of semi-circle  2   

 7056 = 84 sq.cm _______ (2) 1 56 From (1) and (2):  r  27  84 (or) r  cm 2 9  Circumference of the circle = 2r  2

22 56 1   36 cm 7 9 9

1  Circumference of the circle  36 cm 9 2 53. Area of circle = r We know that,  is an irrational number and r is a rational number. = (Irrational) (square of rational) = Irrational   being irrational. 54. Let the inner and outer radii be r and R metres. The circumference of inner circle = 2r 2r 

352 7

1  22     7  7  cm 2 = 77 cm2 2 7  56. Given: diameter(d) = 40 cm  Radius(R) = 40/2 = 20 cm Distance covered in 1 revolution = 2R 22 880     2   20  cm  cm 7 7   For 1 revolution __________

7    17600    140 880   57. Length of each side of the square

 81  9cm Length of wire = (9 × 4)cm = 36 cm R + 2R = 36  ( + 2)R = 36 R

528 7

 528 7 1  R      12 m  7 22 2   Width of the ring = (R – r) = (12 – 8)m = 4m. 55. We know that perimeter of the semi-circle

880 cm 7

? __________ 176 m or 17600 cm Required number of revolutions

 352 7 1  r     8m  7 22 2  The circumference of outer circle = 2R 2R 

36 cm  36  7   22    cm  7cm  2   36   7  

36  7cm  22   2    7 

Area of the semi-circle

1  1 22   R 2     7  7  cm 2  77cm 2 2 2 7  58.

 2R    R  R  cm2 = R + 2R  2  21m

R

Given: R + 2R = 36  ( + 2)R = 36

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R

Circumference of 1 garden = R Circumference of 4 gardens or fencing = 4R Length of the fence = 4R, where R 

21 m 2

22 21     4    m = 132 m 7 2  Cost of fencing = 132 × 12.5 = Rs. 1650.

Mensuration

393 62.

59. O

O 5

5

r 36o r

3.5 cm

A

B

A

1  Area of the sector    arc  R  2 

22 cm

B

The length of the arc = 22cm. The measure of the arc = 36°

1     3.5  5  cm2 = 8.75 cm2 2  

Then

  AB  m  AB   2r 360

60.  O 7

108o

7

A

 B



108  2  22  Area of the sector    7  7   cm 360   7 = 46.2 cm2 61. The given radius of the circle = 12cm

22  35  220cm 7 The area of a circle = r2  2

22  35  35cm2 7 = 22 × 35 × 5 cm2 = 3850 cm2 

O 90o

7 1  2r 10

7  10 r 2 r = 35 cm  The radius of the circle = 35cm The circumference of a circle = 2r

 2 Area of sector  360  r sq.cm

12

22 36  22 2   r 360 7

12

63. Angle subtended at the centre = 90° Perimeter of segment of a circle



   2r  2r sin 360 2

O 21

90 22 90   2   12  2  12sin 360 7 2 1 2  22  12    24sin 45 4 7 

1    sin 45   2 

132 1  24  7 2

 18.857  12 2

= 18.857 + 16.968



2  1.414

90°

A

21 B

x

Given is the radius of the circle = 21 cm The sector angle = 90° The area of the segment AXB = (The area of sector OAXB) – (The area of OAB)

 1 2rx   1     r     OA  OB  2 360 2    



= 35.825 cm www.betoppers.com

8th Class Mathematics

394 22    1 2  7  21 90  1     21    21  21 360   2   2   

65. Radius of the sector = ‘r’

O r

r

1  44  3  21       21  21   2  4  

A



B

1   33  21  21  21 2

Length of the arc = ‘’

1   21 33  21 2

 Area of the sector

1   21 12 2 = 21 × 6 = 126 cm2 Length of the arc

If the arc is reduced to half, then new arc

r sq.units 2 If the radius is doubled, then new radius = 2r

 Area of the new sector

m  AB     AB    2 r 360



 2

90 2 21  2 21  360 4

2  22  21 = 11 × 3 = 33 cm 74 64. Given the angle subtended at the centre m(AB) = 90°

 1 r  2r    sq.units 2 2 2  Its area is unchanged. 66. Let the cow be tied at the vertex B. The area that the cow grazes in a sector whose arc in EFC and radius is BF.



C 10m

D

A 10m

E 10m B F

O

A 90°

B

The radius of the circle, r = 14cm. The length of the arc AB is given by   AB  



m  AB   2r 360

90 22  2   14 360 7

1   2  22  2  22cm 4  The length of the arc AB = 22cm The area of the circle = r2 

22  14  14 = 616 cm2 7

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Radius of the sector (BF) = 10 m. The angle subtended by sector at vertex B = 360° ? 90° = 270° Area of the sector 



x  r 2 360

270 22   10  10 sq. m 360 7

3 22    10  10 sq. m 4 7  3 

22  5  5 sq. m 7

1650 5 sq. m  235 sq. m 7 7

Mensuration

395

CONCEPTIVE WORKSHEET Given the base of an isosceles triangle = 12 cm. The perimeter of the triangle = 32 cm  base + 2 × side = 32 cm  12 + 2 × side = 32 cm  2 × side = (32 – 12)cm  2(side) = 20 cm

D

cm

10 cm

m

10



 AC    AB 



 25   24 

2

2

2

2

2

2  625  576  49 m  7 m

 The bottom of the ladder must be pulled out 7m from the wall so as to lower the top by one metre. Let a = 13, b = 14 and c = 15. Then,

 Area

2

5.

64  height = 8 cm The area of the triangle 1   base  height 2 1  12 cm  8 cm  48 cm 2 6. 2 The base of the triangular field, b = 540m and the height of the field, h = 235m  the area of the triangular field 1 1   b  h   540  235 2 2 = (270 × 235 m2) = 63450 m2. The cost of levelling 1m2 of field = Rs. 25.  The cost of levelling 63450 m2 of field = Rs. 25 × 63450 = Rs. 1586250.

 length of ladder    length of wall 

1 1 42  a  b  c   13  14  15   21 2 2 2  (s – a) = 8, (s – b) = 7 and (s – c) = 6.

 height =

2.

2



s

2

= 100 – 36 = 64 cm2

C

 The distance of the ladder from the wall

4.

 12   (height)2 = (10)2     2

B

Initially

12 cm 2  base    side      2 

25

20 cm  10 cm 2  the length of the equal sides of isosceles triangle = 10cm. By Pythagoras theorem,

2

D A

25m

 side 

 height 

Given the length of the ladder = 25m. When the bottom of the ladder is pulled away from the wall, AB = 24m.

24m

1.

3.

 s  s  a  s  b  s  c 

 21 8  7  6 = 84 cm2. By Pythagoras theorem, Height of the triangle 

2

13  12 

 Its area

2

cm  169  144  25 cm = 5cm

1   base  height 2

1     12  5  cm 2 = 30 cm2. 2   Let Base = b cm and Height = h cm b + h + 26 = 60  b + h = 34  (b + h)2 = (34)2 Also, b2 + h2 = (26)2  (b + h)2 – (b2 + h2) = (34)2 – (26)2  2bh = (34 + 26) (34 – 26) = 60 × 8 = 480 1  bh = 240  bh  120 2  Area = 120 cm2.

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8th Class Mathematics

396 7.

Let the sides be a metres, a metres and b metres.

10. We have,

C

a

b

A

a

2

 2a  2a  6  3 2

8.



a3

1 1 AB  2cm and AS  AD  2cm 2 2 Using Pythagoras theorem in PAS , we have Also, PS2 = AP2 + AS2



2 cm

10 cm

 Each side of square ABCD  16cm  4cm In  APS, we have

2

= 8 cm2 11. Let ABCD be the given rhombus whose diagonals intersect at O. AB = 13 cm and AC = 24 cm Since the diagonals of a rhombus bisect each other at right angles, AOB is a right triangle, right-angled

B

1  2   Let each side be a cm.

 Area    a  2a  = a2 = 20 cm2.

at O such that

D

2 a 2 Then,    10   a  2

12

3a 2 400  100  a 2  4 3

 3 400  2 100 2    cm  cm  3  3  4

1 AC  12 cm and 2 C

c 12

 a2    a 2    100 4 

 Area of equilateral triangle 

OA 

AB = 13cm.

2





2  Area of the square PQRS  2 2 cm

Then, a2 + (2a2) = (10)2  5a2 = 100  a2 = 20

9.

2

 PS  22  22  4  4  8  2 2 cm Thus, each side of square PQRS is of length 2 2cm .

C

 cm

B

AP 

1   Area    3  3  m 2 = 4.5 m2 2  Let the sides be a cm and 2a cm.

A

P

Area of square ABCD = 16 cm2 2

b 2a

 

Q

A

B

Then, 2a + b  6  3 2 and b = a + a = 2a



C

S

2

a 2 2 3 2 2

R

D

A

3  a2 4

cm

13 cm

m

O

B

Using Pythagoras theorem, in AOB , we have AB2 = OA2 + OB2  132 = 122 + OB2  OB2 = 132 – 122  OB2 = 169 – 144 = 25  OB2 = 52  OB = 5 cm  BD = 2 × OB = 2 × 5cm = 10cm Hence, area of rhombus ABCD

1 1   AC  BD   24  10cm 2 = 120 cm2 2 2 www.betoppers.com

Mensuration

397

12. Let ABCD be a rhombus such that its one diagonal AC = 4 cm. Suppose the diagonals AC and BD intersect at O. D

2m

H D

G C

C

2m A

O

x

B

E A

B

We have, Area of rhombus ABCD = 24 cm2

1   AC  BD  24 2 1   4  BD  24 2

 2 × BD = 24  BD = 12 cm Thus, we have AC = 4cm and BD = 12 cm  OA 

1 1 AC  2cm and OB  BD  6 cm 2 2

F

Area of the square EFGH = (x + 4)2 sq.m.  Area of the path = (Area EFGH) – (Area ABCD) = [(x + 4)2 – x2] m2 = (x2 + 8x + 16 – x2) m2 = (8x + 16) m2.  8x + 16 = 120

 8x = 120 – 16 = 104  x 

 Side of the lawn = 13 m. Area of the lawn = (13 × 13)m2 = 169 m2. 15. Perimeter of the grassy plot = 122 m

 122  Side of the plot    m  30.5m  4  Area of the plot = (30.5 × 30.5)m2  61 61      m2  2 2

Since, the diagonals of a rhombus bisect each other at right angle. OAB is a right triangle, right angled at O. Using Pythagoras theorem in AOB , we have AB2 = OA2 + OB2  AB2 = 22 + 62 = 40

 AB  40cm  2 10cm Hence, perimeter of rhombus ABCD





 4  2 10 cm  8 10cm 13.

1 d1  d 2  150 2 1  10  d 2  150  d2 = 30 cm. 2

14. Let ABCD be the lawn and EFGH be the other boundary of the path. Let each side of the lawn be x metres. Then, area of the lawn ABCD = x2 sq.m. Side of the square EFGH = (x + 2 + 2) m = (x + 4)m.

104  13 8

 3721  2  m  4   3721   Cost of cutting the grass  Rs  4  4    = Rs. 3721 16. Diagonal of the square = 32 cm

1 2  Area of the square    diagonal  2

1     32  32  cm 2 2  2 = 512 cm . Side of the square  Area  512 cm  16 16  2 cm  16 2 cm = (16 × 1.41) cm = 22.56 cm Perimeter of the square = 4 × side = (4 × 22.56) cm = 90.24 cm. www.betoppers.com

8th Class Mathematics

398 17. Length of carpet used

 40  22. Side of first square    cm = 10 cm  4 

Total cost of carpeting  Rate per metre

 810   8100   m    m = 180 m  4.50   45  Width of the carpet = 75 cm  75  3  m   m  100  4 3 2  2 Area of the carpet  180   m  135m 4  Let the breadth of the room be x metres. Length of the room = 18m. Area of the room = (18 × x)m2 = (18 x) m2. Clearly, we have: Area of the room = Area of the carpet  135  15  18x = 135  x   18   2 = 7.5   Hence, breadth of the room = 7.5 m. 18. Given l = 7.5 cm and b = 3.6 cm  Perimeter = 2(l + b) = 2 × (7.5 + 3.6) cm = 22.2 cm Area = (l × b) = (7.5 × 3.6) cm2 = 27 cm2.  80  19. Given: Base = 3.5 m, Height = 80 cm   m  100  Height = 0.8 m.  Area of the parallelogram = (Base × Height) sq.units = (3.5 × 0.8)m2 = 2.8 m2. 20. Other side



2

17   15

2

 289  225

 64 = 8 m.  Area = (15 × 8)m2 = 120 m2. 21. Let length = x and breadth = y. Then, 2(x + y) = 46 (or) x + y = 23 and By Pythagoras theorem, x2 + y2 = (17)2 = 289 Now, (x + y)2 = (23)2  (x2 + y2) + 2xy = 529  289 + 2xy = 529  xy = 120.  Area = xy = 120 cm2.

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 32  Side of second square    cm = 8 cm  4 Area of third square = [(10)2 – (8)2] cm2 = (100 – 64) cm2 = 36 cm2. Side of third square  36 cm = 6 cm.  Required perimeter = (6 × 4) cm = 24 cm. 23. Perimeter = Distance covered in 8 min.

 12000    8  m = 1600 m  60  Let length = 3x metres and breadth = 2x metres. Then, 2 (3x + 2x) = 1600 2 × 5x = 1600 1600  160 10 length = 3x = 3 × 160 = 480 m breadth = 2x = 2 × 160 = 320 m or x = 160.  Length = 480 m and Breadth = 320 m.  Area = (480 × 320) m2 = 153600 m2. 24. Let ABCD be the given ||gm. Let AC = 70 cm. Draw BL  AC and DM  AC . x

D

C L M

A

B

Then DM = BL = 27 cm Area of ||gm ABCD = ar   ABC   ar   ACD   1  1     70  27     70  27   sq.cm  2   2 = 945 + 945 = 1890 sq.cm 25. Area of ||gm= (Base × Height) = (18 × 8)cm2 = 144 cm2. 26. Given d = 8 cm, h1 = h2 = 2 cm. We know that,

1 Area of quadrilateral   d   h1  h2  2 1   8   2  2   16 cm2 2

Mensuration

399

27. We have, Area of the trapezium

1    Sum of the parallel sides  × 2 (Distance between the parallel sides)

Area of the given figure = Area of trapezium ABCD + Area of trapezium CDEF 1  1      21  15  10 cm2    15  24  12 cm2 2 2    

1     10  12   4  cm 2 2 

1  1     36  10 cm2    39  12 cm 2 2  2  2 2 2 = 180 cm + 234 cm = 414 cm 1     22  4  cm 2 = 44 cm2. 31. Let the length of the shorter parallel side be x cm. 2  Then, length of the longer parallel side is 2x cm. 28. Let the distance between the parallel sides be We have, h cm. Then, Area of the trapezium = 180 cm2. 2 Area = 440 cm Height of the trapezium = 12 cm Area of the trapezium = 180 cm2 1 1    30  14   h  440   44  h  440 2 2 1  × (Sum of the parallel sides) × Height = 180 2 440 h  20  22 × h = 440 22 1  × (x + 2x) × 12 = 180 Hence, the distance between the parallel sides is 2 20cm.  6 (x + 2x) = 180 29. We have,  18x = 180 L

D

C

180  10 cm 18 Hence, the lengths of the parallel sides are 10 cm and 20 cm. 32. The circumference of a circle is given by C  2 r The radius of the circle = 12.5 cm  C = 2 × 3.14 × 12.5 cm = 78.5 cm The area of a circle  r 2 =  (12.5)2 = 3.14 × (12.5)2 = 490.625 cm2. perimeter of semicircle = r (2 +  ) = 12.5 (3.14 + 2) = 64.25 cm. 33. The circumference of a circle is given by C  2r . The given circumference C = 28.6 m  2r  28.6 m x

15 m

A

B

20 m

Area of trapezium ABCD = 480 m2



1  AB  CD   AL  480 2



1  20  CD  15  480 2

480  2 15  20 + CD = 32 × 2  20 + CD = 64  CD = 64 – 20 = 44 Hence, the other side of the trapezium is 44m. 30. We have,  20  CD 

24 cm F 12 cm

E

15cm

A

22  d  28.6 (since 2r = d) 7

28.6  7  9.1m 22 34. The given circumference of a circle C = 110cm The circumference of a circle  2r d 

 2r  110cm

C 10 cm

D



21 cm

 2

22  r  110cm 7

B

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8th Class Mathematics

400

r

110  7 35   17.5cm 2  22 2

The area of a circle A  r 2

22 17.5  17.5 7 = 22 × 17.5 × 2.5 cm2 = 962.5 cm2. 35. The given area of the circle = a sq.units.  A

a units  The circumference of a circle  2r  r 2  a sq.units  r 

a  2 a units   A circle of area ‘a’ sq.units has a  C  2

39.

2R1 23 R 23   1 2R2 22 R2 22  R1 

23 R2 22

23R2  R2  5 22  23R2 – 22R2 = 5×22  R2  110  Diameter of inner circle = (2 × 110)m = 220m 40. The diameter of the bigger semicircle = 108 cm  the radius, R = 54 cm Also, R1 – R2 = 5m 

circumference of 2 a units. 36. Let the two radii be r1 and r2. Given the ratio of r1 : r2 = 1 : 4

Area But radius   Let the two areas be A1 and A2 A1 A : 2 =1:4   Squaring on both sides, we get  r1 : r2 

A1 2  1 A1 1 A2  4   A  16 (or) A1 : A2 = 1 : 16 2 

i.e., ratio of their areas is 1 : 16. 37. Area = (13.86 × 10000) m2 = 138600 m2. ( 1 hectare = 10000 m2)

7   R 2  138600  R 2  138600   22    R 2  4410  R  4410  R = 210 m.

22   Circumference  2R   2   210  m 7   = 1320 m  Cost of fencing = Rs. (1320 × 4.40) = Rs. 5808. 38. Let inner radius be r metres. Then, 2r  440

7    r   440   = 70 m. 44    Radius of outer circle = (70 + 14)m = 84 m. www.betoppers.com

The area of the bigger semicircle 

R 2 sq.units 2

22 54  54  7 2 22 32076   27  54cm2  7 7 = 4582.29 cm2 The diameter of the smallest circle d = 36 cm.  the radius r = 18 cm. The area of the smallest circle  r 2 

22  18  18 = 1018.28 cm2. 7 The diameter of the smaller semicircle, d1 = 54cm  the radius r1 = 27 cm 

The area of the smaller semicircle 



r12 2

22 27  27 8019   = 1145.57 cm2 7 2 7

The area of the two smaller semicircles  2 

r12 2

= 2 × 1145.57 cm2 = 2291.14 cm2.  Area of the shaded portion = (area of the bigger semicircle) – [(area of the smallest circle) + (area of the two semicircles)]. = [4582.29 – (1018.29 + 2291.14)] cm2 = 1272.86 cm2 Hence, the area of the shaded portion is 1272.86cm2.

Mensuration

401

1  41. Radius of the required circle    14  cm 2  = 7 cm

 22  2 Area of the circle    7  7  cm = 154 cm2  7  42. Circumference of the circle = 44 cms. 2r  44 22 2   r  44 7 1 7  r  44    7cms 2 22  Area of the circle 22  r 2   7  7  154 sq.cms . 7 43. Radius of the circular field (r) = 77 m. Path of width = 7m. R–r=7 R – 77 = 7 R = 7 + 77 = 84 m. Area of the path =  (R + r) (R – r) 22  84  77  84  77  7 22   161 7 7 = 3542 sq.mts. 44. Given the length of the arc = 26.5 cm and the radius = 63 cm. lr  the area of the sector  sq.units 2 26.5  63  2 1669.5  = 834.75 sq.cm. 2  the area of the sector = 834.75 sq.cm. 45. Angle of sector (  ) = 72°; radius = 14 cm 

  2r 360 72 22   2   14 360 7 88  = 17.6 cm. 5

46. Radius of the sector = 7 cm. Angle (  ) = 240°.   r 2  Area of the sector  360 240 22 308   77  360 7 3 = 102.66 sq.cm 47. Angle of the sector (  ) = 75° Radius = 21 cm  l  2r 360 75 22 55   2   21  = 27.5 cm 360 7 2 Perimeter of the sector = l + 2r = 27.5 + 2 (21) = 27.5 + 42 = 69.5 cm. lr 27.5  21 Area of the sector   2 2 = 288.75 sq.cms. 48. The given perimeter of the sector of a circle = 72cm. Let the angle subtended at the centre be  . The perimeter of the sector of a circle.    2r  2r 360  22   2   14  2  14  72 360 7    88  28  72 360  44   88  44     360 360 88  The angle subtended at the centre.   180 . 49. Find the area of the sector of a circle of radius 16cm, cut off by an arc of length 18.5 cm. Sol: Given is the radius of the circle = 16cm The length of the arc = 18.5 cm Now, the area of the sector 1   length of the arc × radius 2 1  18.5  16cm 2 2 = 18.5 × 8 cm2 = 148 cm2.  the area of the sector is 148 cm2. 50. The angle swept by the minute hand in 60 minutes = 360° 11

 Length of the arc 

12

1

10

2

9

3

8

4 7

5 6

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8th Class Mathematics

402  The angle swept by the minute hand in 10 minutes 0  360   10   60  60   Sector angle = 60°. Now the area swept by the minute hand in 10 minutes = The area of a sector of a circle of radius 21 cm and sector angle 60°. 60 60 22   r 2    21 21 360 360 7 22  21 3   11 21  231cm2 . 6

3.

1 (28 + 21 + 35) = 42, (s – a) = 14, 2 (s – b) = 21 and (s – c) = 7. Then, s 

 Area of the triangle  s  s  a  s  b  s  c 

 42 14  21 7cm2  21 2  7  2  21 7cm2 = (21 × 2 × 7) cm2 = 294 cm2. Longest side = 35 cm. Let the altitude corresponding to longest side be h cm .Then,

SUMMATIVE WORKSHEET 1.

Let the length of each side of an equilateral triangle be ‘a’ cm.

1  2 Area of the triangle    35  h  cm . 2 

Given the area of the triangle  16 3 sq.cm We know that the area of an equilateral triangle

1   35  h  294 2

3 2  a 4

 35 × h = 294 × 2

294  2  16.8cm 35 Hence, the required altitude is 16.8 cm. Here, a = 8 cm. h

3 2 a  16 3 4 a2 = 16 × 4 = 64 

2.

Let a = 28cm, b = 21 cm and c = 35 cm.

 a2 

16 3  4 3

4.

 3  2 Area of the triangle   4  a  sq.units  

 a  64  8cm  the length of each side of the equilateral triangle = 8cm. Let ABC be the isosceles triangle and AD be the altitude. A

x

 3     8  8  cm 2  16 3 cm 2  4  = (16 × 1.73) cm2 = 27.68 cm2. Height of the triangle



 3   3     a  units    8  cm  4 3 cm  2   2 



x

B D C Let AB = AC = x. Then, BC = (32 – 2x) Since, in an isosceles triangle, the altitude bisects the base, so BD = DC = (16 – x). In ΔADC , AC2 = AD2 + DC2  x2 = (8)2 + (16 – x)2  x2 = 64 + 256 + x2 – 32x  32x = 320  x = 10  BC = (32 – 2x) = (32 – 20) cm = 12 cm. 1  Hence, required area    BC  AD  2   1    12 10  cm 2 = 60 cm2. 2  www.betoppers.com



5.



= (4 × 1.73) cm = 6.92 cm. Total cost of cultivating the field = Rs. 4957.20. Rate of cultivation = Rs. 367.20 per hectare.

 4957.20   Area of the field   367.20  hectare   = 13.5 hectares. Let the height of the field be x metres. Then, its base = 3x metres. 1  2  Area of the field   2  x  3x  m  

 3x2  2  m  2 

Mensuration

403

Thus, we have:

9.

3x 2  135000 2

Let ABCD be a rhombus. AC and BD are the diagonals with AC = 18 cm and BD = 12 cm. D

135000  2 x   90000 3  Height = x = 300 m; Base = 3x = (3 × 300)m = 900 m. Let l be the length and b be the width of the rectangular field. It is given that l = 80m We have, Perimeter of the rectangular field = Perimeter of the square field  2(l + b) = 4 (side)  2 (80 + b) = 4 × 60  160 + 2b = 240  2b = 240 – 160  2b = 80  b = 40 m Now, A 1 = Area of the square field = (side)2 = (60)2 m2 = 3600 m2 A 2 = Area of the rectangular field = l × b = 80 × 40m2 = 3200 m2 Clearly, A1 > A2 Hence, the square field has larger area. Area of the square field = 1 hectare = 10000 m2.  Side of the square field

C

2

6.

13    Cost of fencing = Rs  400  2   Rs.2600   8.

Other diagonal  2 

 side 

2

Now one side of a rhombus 

 Other diagonal

 2

 diagonal    2  

146 = 36.5 cm 4

 36.5 

2

 55     2 

 2 1332.25  756.25  2 576 = 48 cm Now, area 

2

1 (product of diagonals) 2

1   48  55 = 1320 sq.cm 2

2

O

A

B

Since the diagonals of a rhombus bisect each other,  AO = OC = 9cm and BO = OD = 6cm Now in AOB , AO = 9cm, OB = 6cm By Pythagoras theorem, (AB)2 = (AO)2 + (OB)2 = (9)2 + (6)2 = 81 + 36 = 117

 AB  117  3 13cm  Each side of the rhombus  3 13cm The perimeter of a rhombus = 4 × side

 43 3  12 13cm 10. Let ABCD be a rhombus. A

D

cm

 Area  10000m  100m Perimeter of the field = (4 × side) = (4 × 100) m = 400 m. Rate of fencing = Rs. 6.50 per metre.

18

66

7.

12

O

B

C

Perimeter of the rhombus = 260 cm  4 × side = 260 cm

260  65cm 4 AC and BD are diagonals with AC = 66cm. Since diagonals of a rhombus bisect each other, AO = OC = 33cm. In AOB , AO = 33 cm, AB = 65 cm. By Pythagoras theorem, (OB) 2 = (AB)2 – (OA)2 = (65)2 – (33)2 = 4225 – 1089 = 3136

 Side of the rhombus 

OB  3136  56cm We have OB = 56cm = OD  BD = OB + OD = 56 + 56 = 112cm. The length of the other diagonal = 112cm Area of rhombus www.betoppers.com

8th Class Mathematics

404



1 (product of the diagonals) 2

1 7392 × 66 × 112 sq.cm  2 2 = 3696 sq.cm. 11. We have, Length = 16m, Breadth = 14m, and Height = 10m  Area of four walls of the room = [2 × (Length + Breadth) × Height)] m2 = [2 × (16 + 14) × 10]m2 = 600 m2. Area of one door = (2 × 1)m2 = 2m2  Area of two doors = (2 × 2)m2 = 4m2 Area of one window = (1.3 × 1.4)m2 = 1.82m2 Area of 4 windows = 4×1.82 = 7.28m2 Thus, the area to be white washed = {600 – (4 + 7.28)}m2 = 588.72 m2.  Cost of whitewashing at the rate of Rs. 5 per sq.metre = Rs. (5 × 588.72) = Rs 2943.6 Area to be painted = (4 + 7.28) = 11.28 m2.  Cost of painting the doors and windows = Rs {(8 × 11.28)} = Rs. 90.24 Hence, total cost of white washing and painting = Rs (2943.6 + 90.24) = Rs 3033.84 12. We have, Diameter of circular ends = 7m 

 r = Radius of circular ends 

7 m = 3.5m 2

Also, l = Length of the rectangle = {20 – (3.5 + 3.5)}m = (20 – 7)m = 13m b = Width of the rectangle = 7m Perimeter of the garden  2l  r  r  2l  2 r

22    2  13  2   7  m 7   = (26 + 44) m = 70 m Area of the garden = Area of rectangular part + Area of semicircular ends 1 1  l  b   r 2    r 2  2 2 2  l  b  r 22    13  7   7 2  m 2 7   = (91 + 154)m2 = 245 m2. www.betoppers.com

13. Let the length and breadth of the rectangular field be 3x metres and 2x metres respectively. Then, Area of the rectangular field = (3x × 2x)m2 = 6x2 m2 It is given that the area of the rectangular field is 3456 m2.  6x2 = 3456 3456  x2  6 2  x  576  x  576  24  Length = (3 × 24)m = 72 m, Breadth = (2 × 24)m = 48 m  Perimeter of the field = 2 × (Length + Breadth) = [2 × (72 + 48)]m = 240 m We have, Rate of fencing = Rs. 3.50 per metre Cost of fencing = Rs. (240 × 3.50) = Rs. 840. 14. The sides of a rectangular field are in the ratio 5 : 3. Let the length of the rectangular field be ‘5x’m and its breadth be ‘3x m. where ‘x’ is any constant.  the area of the rectangular one square metre = Rs. 0.85p.  The cost of turfing 15x2 square metres = Rs. 0.85 × 15x2 But the cost of turfing the field = Rs. 624.75 15x2 × 0.85 = 624.75 624.75  x2   49  x  7 0.85  15  The length of the rectangular field = 5x = 5 × 7 = 35 m and the breadth = 3x = 3 × 7 = 21m. The perimeter of a rectangle = 2(l + b) = 2(35 + 21) = 2(56) = 112m.  Perimeter of rectangular field = 112m. 15. The length of each tile = 20cm The breadth of each tile = 15cm  The perimeter of each tile = 2(l + b) = 2(20 + 15)cm = 2(35)cm = 70cm. The total number of tiles used in the room floor are 50.  The perimeter of 50 tiles = 70 × 50 cm = 3500 cm. 16. The given length of the wooden plank is twice its breadth. i.e., l = 2b ________ (1) The perimeter of the plank = 90cm i.e., 2(l + b) = 90 cm _________ (2) Substituting (1) in (2), we get,  2(2b + b) = 90cm  3b = 45 cm  b = 15 cm  l = 2b  l = 2 × 15cm = 30cm  The length and breadth of the wooden plank are 30cm and 15cm respectively.

Mensuration

405

 Area of ABC  s  s  a  s  b  s  c 

 42  42  30  42  14  42  40 

10 cm D

C

13 c m

A

L

M

B

20 cm

 42  12  28  2 m2 = 168 m2. Hence, area of ||gm ABCD = (2 × 168)m2 = 336m2. 20. We have,

Altitude of the trapezium = 11 cm 

1 × (Sum of the parallel sides ) × Height = 105 2 1  × (x + 6 + x) × 7 = 105 2 1  × (2x + 6) × 7 = 105 2 105  2  2x  6  7  2x + 6 = 30  2x = 24  x = 12. Hence, the lengths of parallel sides are 12 cm and (12 + 6) cm = 18cm. 22. Let ABCD be a trapezium such that AB = 20 cm, CD = 10 cm and AD = BC = 13 cm. 

m 13 c

17. Let original length = x and original breadth = y. Original area = xy. x New length  ; 2 x  3 New breadth = 3y. New area    3 y   xy 2  2 3 1 area increased  xy  xy  xy 2 2 1  1  Increase %   2 xy  xy 100  %  50%   18. Let length = (3x) metres and breadth = (2x) metres. Then, (3x + 5) × 2x = 2600  6x2 + 10x – 2600 = 0  3x2 + 5x – 1300 = 0 By factorization,  (3x + 65) (x – 20) = 0  x = 20.  Breadth = 2x = 40m. 19. Let ABCD be the given ||gm. Area of ||gm ABCD = 2 × (area of ΔABC ) 1  s   30  14  40  m  42m 2

11 m 100

Area of the trapezium = 0.55 m2. Now, 1    sum of the lengths of parallel sides 2 × Altitude = 0.55 1   sum of the lengths of parallel sides 2 11  = 0.55 100  Sum of the lengths of parallel sides 0.55  200  m  10 m . 11 Hence, the sum of the lengths of parallel sides of the trapezium is 10 m. 21. Let the length of the shorter parallel side be x cm. Then, Length of the other(longer) parallel side = (x + 6) cm We have, Height of the trapezium = 7 cm, Area of the trapezium = 105 cm2

Draw CL || AD and CM  AB Now, CL || AD and CD || AB.  ALCD is a parallelogram.  AL = CD = 10 cm and CL = AD = 13 cm In CLB , we have CL = CB = 13 cm  CLB is an isosceles triangle. 1 1  LM  MB  BL  10cm  5cm 2 2  BL  AB  AL   20  10  cm  10cm  Applying Pythagoras theorem in CLM , we have CL2 = CM2 + LM2  132 = CM2 + 52  CM2 = 169 – 25 = 144  CM  144  12

1  BL  CM 2 1  10  12cm2  60cm 2 2 Area of parallelogram ALCD = AL × CM = (10 × 12)cm2 = 120 cm2 Hence, Area of trapezium ABCD = Area of parallelogram ALCD + Area of CLB = (120 + 60)cm2 = 180 cm2.  Area of CLB 

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8th Class Mathematics

406

2.8 cm C A E

7cm

2.8 cm

13cm

D

 22    1.4  2  1.4  cm  7  = (22 × 0.2 + 2.8)cm = (4.4 + 2.8)cm = 7.2 cm ii) Clearly, Path traced by the ant = AB + BC + CD + arc AED = 1.5 cm + 2.8 cm + 1.5 cm  1.4cm 22    1.5  2.8  1.5  1.4  cm 7   = (5.8 + 4.4) cm = 10.2 cm iii)We have, Path traced by the ant = OA + arc ACB + OB   2   1.4  2  cm

22     4  1.4  cm 7   = (4 + 4.4)cm = 8.4 cm Clearly, for (iii) food particle the ant will have to take a longer route. 27. B  7 cm  O  7 cm  c

B

 AM = DC  AM = 7 cm  AB – BM = 7  13 – BM = 7  BM = (13 – 7) cm = 6cm Applying Pythagoras theorem, in right-angled triangle BMC, we have BC2 = BM2 + CM2  102 = 62 + CM2  CM2 = 100 – 36 = 64  CM = 8cm  Height of trapezium ABCD = 8 cm Hence, area of trapezium ABCD 1   AB  DC   CM 2 1   13  7   8cm2  80cm 2 2

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2.8 cm

O

cm

M A

A

i) We have, Diameter of the semicircular path = 2.8 cm r = Radius of the semicircular path = 1.4 cm Path traced by the ant  r  2

C 10

B

1.5 cm 2c m

26.

1      2 x  6   9  cm 2 = (9x + 27) cm2 2  But, area of the trapezium is given as 180 cm2.  9x + 27 = 180  9x = 180 – 27 = 153 153 x  17 9 Thus, the two parallel sides are of lengths 17 cm and (17 + 6) cm = 23 cm. 25. Draw CM perpendicular from C to AB. Clearly, AMCD is a rectangle. D

C

B

m 2c

23. We have, Perimeter of the trapezium = 52 cm Altitude of the trapezium = 8cm  Sum of the parallel sides + Sum of the nonparallel sides = 52cm  Sum of the parallel sides + 2 × 10 = 52  Sum of the parallel sides = (52 – 20) cm = 32 cm  Area of the trapezium 1   (Sum of the parallel sides) × Altitude 2 1   32  8cm2 = 128 cm2 2 24. Let the length of the shorter side be x cm. Then, the length of the other(longer) parallel side is (x + 6) cm.  Area of the trapezium 1      x  x  6   9  cm 2 2 

P

A  7 cm  O  7 cm  D

Area of the shaded region = Area of square ABCD – Area of two semicircles   1 22   14 14   2    72   cm 2 2 7   = (196 – 154)cm2 = 42 cm2 28. Let length = 5x metres and breadth = 3x metres. Then, area of the park = (5x × 3x)m2 = (15x2) m2. 2160 2  144  15x2 = 2160  x  15  x  144  12  12  12

Mensuration

407

 Length = (5 × 12)m = 60 m. Breadth = (3 × 12)m = 36 m.  Perimeter = 2(l + b) = 2(60 + 36)m = 192 m. Rate of fencing the park = Rs. 5 per metre. = Rs (192 × 5)  Cost of fencing the park = Rs. 960. 29. Perimeter of square = 4S = 5024 2024 S=  1256 4 2 area of square = (1256) sq-m. perimeter of circle  2r  5024 5024 5024  7   2 20  22 1256   11 area of circle  r 2 22 1256  7 1256  7    7 11 11 2  1256  1256  7  sq  m 11 ratio of the area of the square to the circle 2  1256  1256  7 = 1256 × 1256 : 11 14  1:  11:14 11 30. Let the side of the square be x. Then, its diagonal  2x .

x 2

7m 20 m

Area of the garden = Area of rectangular part + Area of semicircular ends 1 1  l  b   r 2    r 2  2 2 2  l  b  r 22    13  7   7 2  m2 7   = (91 + 154)m2 = 245 m2. 32. We have, OB = O ' C = 35 m and AB = CD = 14 m  OA = O ' D = (35 + 14)m = 49 m A

D 14 m

35m

B

C

O

Ol

E

H

F x/2

x and 2 2x x  radius of circumcircle  2 2 2 2  x x  1 1  Required ratio   4 : 2   :  1: 2   4 2 31. We have, Diameter of circular ends = 7m. 7  r = Radius of circular ends  cm  3.5m 2 Also, l = Length of the rectangle = {20 – (3.5 + 3.5)}m = (20 – 7)m = 13 m b = Width of the rectangle = 7m Radius of in-circle 

Perimeter of the garden  2l  r  r  2l  2 r 22    2  13  2   7  m 7   = (26 + 44)m = 70m

35m

G 120 m

Now, Area of the shaded region = Area of rectangle ABCD + Area of rectangle EFGH + 2 {Area of the semicircle with radius 49m} – 2 {Area of the semicircle with radius 35m} = (14 × 120) + (14 × 120) 2 2  1 22  1 22 2     49    2     35  m2 2 7  2 7 

22    1680  1680   492  352  m 2 7   22    3360   49  35 49  35  m 2 7  

22    3360   84  14  m 2 7   = {3360 + 44 × 84} m2 = 7056 m2. Hence, the area of the shaded region is 7056 m2.

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8th Class Mathematics

408 33. Given the circumference of a circular path = 264m  2r  264 m 22  2   r  264 7 264  7 r 2  22  r  42m The area of a circular path  r 2

22  42  42 7  The area of the circular path = 5544m2. 34. Let the circumference of the two circles be ‘C1 ’ and ‘C2’ and the area ‘A1’ and ‘A2’. Let ‘r1’ and ‘r2’ be the radii of the two circles. Given C1 : C2 = 2 : 3  2r1 : 2r2  2 : 3

37. Angle of the sector (  ) = 75° Radius = 42 cms.   2r  Length of the arc (l)  360 75 22   2   42 360 7 = 55 cm. 38. Given is the length of radius = 15cm The sector angle = 60° Y



2r1 2   2r2 3 

r1 2  r2 3

Now, A1 : A2  r12 : r22 

A1 r12  A2 r22 2

2

 r  2 4  1     9  r2   3   A1 : A2 = 4 : 9

35.

Side of square paper  784cm = 28 cm.

Radius of each circular plate

O 15 A

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15

X

B

The area of the circle  r 2 = 3.14 × 15 × 15 = 706.50 cm2 The area of the sector OAXB x   The area of the circle 360 60 60   r 2   3.14  15  15 = 117.75 cm2 360 360 OAB is an equilateral triangle.  OA  OB, A  B  60   the area of OAB =

3 2 a 4

3 1.73 2  15   225 4 4 389.25   97.31cm 2 (Approx) 4  The area of the minor segment = (The area of the sector OAXB) – (The area of OAB ) = (117.75 – 97.31)cm2 = 20.44 (Approx) The area of the major segment = (The area of the circle) – (The area of the minor segment) = (706.50 – 20.44)cm2 = 686.06 cm2. 39. Given the radius of the circle = 21 cm 

1     28  cm = 7cm 4   Circumference of each circular plate

 22    2   7  cm = 44cm 7   36. Radius of the sector = 28 cm Length of the arc = 132 cm lr 132  28 Area of the sector   = 1848 sq.cms 2 2

60°

A 21 cm

B



21 cm

C

 The perimeter of the circle  2r 22  2   21cm  132cm 7

Mensuration

409

 The perimeter of the sector = AB + AC + BC  = 43 cm  21 + 21 + BC  BC = 43 – 42 = 1 cm  The length of the arc BC = 1 cm. Let ‘  ’ be the angle subtended by the arc. l  AB  We know that    360 2r where m(AB) =  1 360 30    2.727  2.73 132 11 40. Given the length of the arc = 26.5 cm and the radius = 63 cm. lr  The area of the sector  sq.units 2 26.5  63  2 1669.5  2 = 834.75 sq.cm.

2.

2

 3925    625 c 2

157  25    25  25  c c

3.

2

1250 25 157  252  2

Let AD be the perpendicular from ‘A’ on BC. A

2

1250

HOTS WORKSHEET 1.

Let the three sides of the right-angled triangle be a, b, c with ‘c’ as the hypotenuse. Given a = 3925 cm __________ (1) Difference between the hypotenuse and the other side = c – b = 625 cm.  b = c – 625 cm ________ (2) We know that c2 = a2 + b2 ________ (3) (by Pythagoras theorem) Substituting (1) and (2) in (3), we get,  c2 = (3925)2 + (c – 625)2  c2 = (3925)2 + c2 + (625)2 – 2c × 625  2c × 625 = (3925)2 + (625)2

2

1250 625  24649  625  25274 c  = 12637 cm 1250 2 b = c – 625 (from (2)) = 12637 – 625 = 12012 cm.  The hypotenuse and the other side of the right angled triangle are 12637 cm and 12012 cm respectively. Using the Heron’s triangle we can prove that the area of the two triangles of sides 6.5, 6.5, 5 units and 6.5, 6.5, 12 units is the same. Let us now see how this happens. Consider the rectangle of length 12 units and breadth 5 units. The length of its diagonal will be 12 2  5 2  13 .

30°

D

45°

B

x

D

DAC  90  DCA  90  45  45 DAC  DCA  DC  AD  x say Since ABD : BAD : ADB = 30° : 60° : 90°

5

A

Then AD : BD : AB  1: 3 : 2 If AD = x, then BD  3x

 BD  DC  3x  x  i.e., BC 







3 1 x



3 1 x

 3  1   3  1 x   3  1

but given BC 

 x  1  AD  1unit 1 Area of ABC   BC  AD 2 1 3 1   3  1 1  sq.units 2 2





C 6 .5

C

4.

6.5

O 12

6.5

B

Since diagonals in a rectangle are equal and bisect each other, OA = OB = OC = OD = 6.5. Further, the median in a triangle divides the triangle into two triangles of equal area. OAD = OAB The sides of OAD are 6.5, 6.5 and 5 and OAB are 6.5, 6.5 and 12. The length of each side of ABC = 8 units.  The area of the equilateral triangle ABC. 3 2   side  4 3 3 2   8    64 4 4  16 3 sq.units Given ADE is an equilateral triangle with each side = 4 units. 

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8th Class Mathematics

410 7.

The area of the triangle ADE.

D

3 2   side  4 3 2   4  4 3 sq.units 4  the area of the shaded region, BCED = (the area of ABC ) – (the area of ADE ) 

J

A

 12 3 sq.units Let x be the length of the leg of the right-angled isosceles triangle, originally.

x

G L

E

 16 3  4 3 5.

M

H

K

C

F

B

i) Show that the side of the second inner square The length of each side of the square ABCD = x cm. i) E, F, G, H are the mid-points of AB, BC, CD and DA respectively.  AE = EB = BF = FC = CG = GD

x = DH = HA  cm 2 x

1 2 Its area  x 2 The area of the new triangle 

1  x  4  x  6  2

x In AEH , AE  AH  cm 2 By Pythagoras theorem, (EH)2 = (AE)2 + (AH)2 2

 x  x     2 2 x– 4



x+ 6

6.

1   x 2   x  4  x  6   24 2 2  x2 = x2 + 2x – 24 + 48  x = –12. Area of square = (side)2 = 40 cm × 40 cm = 1600 cm2 It is given that, The area of the rectangle = The area of the square Area of the rectangle = 1600 cm2 Breadth of the rectangle = 25 cm. Area of the rectangle = l × b or 1600 = l × 25 1600 or  l or l = 64 cm 25 So, the length of rectangle is 64 cm. Perimeter of the rectangle = 2(l + b) = 2 (64 + 25)cm = 2 × 89 cm = 178 cm So, the perimeter of the rectangle is 178 cm even though its area is the same as that of the square.

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2

x2 x2  x2 x2    4 4 4 2

x2 x  cm 2 2  Each side of the second inner square.  EH 

x cm 2 ii) J,K,L, M are the mid-points of HE, EF, FG and GH respectively.  EK = KF = FL = LG = GM = MH

EFGH 

 HJ  JE 

x

2 2 In EJK , by Pythagoras theorem, (JK)2 = (EJ)2 + (EK)2 2

 x   x      2 2 2 2



2

x2 x2 2 x2 x2    8 8 8 4

Mensuration

411

1     AC  BD  sq.units 2  1     24 10  cm 2 = 120 cm2. 2  Hence, the area of the given rhombus is 120 cm2.

x2 x  cm 4 2

x  Each side of the third inner square JKLM  cm 2 iii)Let us fence the side of every square. x

The first square 

9.

 x units

11

2

2

2

The second square 

x  21



x units 2

 Other diagonal

2 2 Proceeding in this manner, we get the side of nth square as x

n square  th

 n1

diagonal  2  side     2   146 Now one side of a rhombus  = 36.5 cm 4

Other diagonal  2 

units

x  n1



8.

8



8 6 2



8 8  1 23 8

2 2 2  Each side of the 7th inner square = 1cm. The area of a square = (side)2 sq.units.  The area of the 7th inner square = (1)2 cm2 = 1cm2. Let ABCD be the given rhombus in which AB = 13 cm and diagonal AC = 24 cm. Let AC and BD intersect at O. We know that the diagonals of a rhombus bisect each other at right angles. D

C

O

A

 AO   OB 

13 cm

B

1 AC  12 cm and AOB  90 2 2

 AB    AO 

2



 55     2 

2

576  2  24  48cm S P 25 m

2

13  12 

 25  5cm So, BD = 2 × OB= 10 cm.  Area of Rhombus ABCD

D

E

H

2

C

1.8 m A

F

G R Q 50 m

2 2  the side of the 7th inner square  71

2

10.

2 Hence this is true for nth inner square. iv) Given x = 8 cm From (iii) we have the side of the nth inner square



 36.5 

 2  1332.25  756.25

2



 2

B

40.5 m

 JK 

Area of the field = (50 × 40.5)m2 = 2025 m2. Let ABCD and PQRS be the crossroads. Then, AB = 50m, BC = 1.8 m. Area of the road ABCD = (50 × 1.8)m2 = 90 m2. Also, PQ = 40.5 m, QR = 2.5 m. Area of the road PQRS = (40.5 × 2.5) m2 = 101.25 m2. Clearly, the area EFGH is common to both the roads. Area of rect. EFGH = (1.8 × 2.5)m2 = 4.5 m2.  Total area of both the crossroads = (90 + 101.25 – 4.5)m2 = 186.75 m2. Cost of cementing the roads = Rs. (186.75 × 24) = Rs. 4482 Area of the remaining = (Area of the field) – (Area of the roads) = (2025 – 186.75)m2 = 1838.25 m2. Cost of laying the grass = Rs (1838.25 × 1.60) = Rs 2941.20. Hence, total expenditure = Rs (4482 + 2941.20) = Rs 7423.20 11. Length of the whole plot = 110 m Breadth of the whole plot = 65 m. Area of the whole plot = (110 × 65)m2 = 7150 m2. Length of the plot excluding the path = [110 – (3 + 3)]m = 104 m. Breadth of the plot excluding the path = [65 – (3 + 3)]m = 59 m. Area of the plot excluding the path www.betoppers.com

8th Class Mathematics

412 = (104 × 59)m2 = 6136 m2 Area of the path = (7150 – 6136)m2 = 1014 m2.  80  2 Rate of levelling  Rs   per m .  100 

80    Cost of levelling the path  Rs 1014  100    = Rs. 811.20. 12. Let the length of the rectangle be l units, and the breadth of the rectangle be b units. The area of the rectangle = l × b sq.units. The perimeter = 2(l + b) units Given the length and breadth are increased by 20%. 20 l  The new length l1  l  100 20 b The new breadth b1  b  100  The area of the new rectangle 20  20    l  l  b  b  100  100  2

2

20    1  lb  1    lb  1    100   5 1 2   1  10   lb 1     lb 1  25   25 5   44   11 4    lb 1     lb 1   25 4   100  44  lb  lb 100  The area increase by 44%. The perimeter of the new rectangle  20l   20b    2  l   b  100 100       20  20     2 l 1    b 1    100     100  20    2  l  b  1    100   The perimeter of the rectangle increase by 20%.

13. C

D

ii) A median divides a triangle into two triangles of equal area.  In the figure,

1 AOB  ABC  OB is median  2 1 1   ABCD  AC diagonal  2 2 1  ABCD 4 Similarly ΔBOC, ΔCOD and AOD are equal to

1 ABCD . 4 14. Let x and y be the sides of the rectangle. Then, correct area = xy.  105   96  504 x y  xy Calculated area    100   100  500 4  504  xy   xy  xy Error in measurement   500  500   4  1 4  Error %   500 xy  xy  100 %  5 %   = 0.8 %. 15. Let breadth = x cm. Then, length = (x + 8) cm. area of rectangle = x(x + 8)sqm length is increased by 7 cm New length = (x + 8 + 7) = (x + 15)cm Breadth is decreased by 4 cm Breadth = (x – 4)cm area = (x + 15)(x – 4) sq-cm according to given data  (x + 8) x = (x + 15) (x – 4)  x2 + 8x = x2 + 11x – 60 11x – 8x = 60  x = 20. 16. Let each have base = b and height = h Then, P = b × h, R = b × h. 1 T  b h 2 1 So, P = R, P = 2T and T  R are all correct 2 statements. 17. Draw CE || DA and CF  AB .

O

A

D

C

B

Concepts: i) In a parallelogram the diagonal divides it into two triangles of equal area.

6 cm

A

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9 cm

19 cm

6 cm

h

E

F

8 cm

B

Mensuration

413

Area of CEB= s  s  a  s  b  s  c 

1     10  18   15 cm 2  210cm 2 2  Hence, the area of the given trapezium is 210 cm2. 20. Through C, draw a line parallel to DA, meeting AB in M.

 12  6  2  4 cm2  24cm 2

C 17

cm

12 cm

15 cm

1  Also, area of CEB =   EB × CF  sq.units 2  1    10  h    5  h  cm 2 2   24   5 × h = 24  h    = 4.8 cm.  5  Thus, the distance between parallel sides = 4.8 cm.  Area of trapezium ABCD 1  × (sum of parallel sides) × 2 (distance between them) 1     19  9   4.8 cm2 2   = (28 × 2.4)cm2 = 67.2 cm2. Hence, the area of the given trapezium is 67.2 cm2. 1 18. (sum of parallel sides) × depth = Its area 2 1  (12 + 8) × d = 840 2  d = 84 m. 19. Draw DE  BC

D

C

cm



 225 cm  15 cm  Area of trapezium ABCD 1      AD  BC   DE  sq.units 2 

15

Let CF = h cm. Then, AE = DC = 9cm, CE = DA = 6 cm and EB = (AB – AE) = (19 – 9) cm = 10 cm First we find the area of CEB . Let a = CE = 6 cm, b = EB = 10 cm and c = BC = 8 cm. 1 1 Thus, s   a  b  c    6  10  8  = 12, 2 2 (s – a) = 6, (s – b) = 2 and (s – c) = 4

A

M

B

36 cm

Clearly, AMCD is a parallelogram.  CM = AD and AM = DC  CM = 15 cm and AM = 12 cm  BM = AB – AM = (36 – 12)cm = 24 cm Thus, in triangle CMB, we have CM = CB  CMB is an isosceles triangle,  Altitude of CMB 

1 2  Equalside     base  2 

2

1   Altitude of CMB  15    24  2   Altitude of CMB  225  144  81cm = 9 cm Hence, area of trapezium 1 ABCD    AB  CD   Altitude 2 1      36  12   9 cm 2 2   = (24 × 9) cm2 = 216 cm2. 21. We have, 2

4 cm A1

10 cm

E

B

18 cm

A

D

B1 28 cm

20 cm

A

DE =

DC2  EC 2



17    8

2

4cm

B

EC = (BC – BE) = (BC – AD) = (18 – 10)cm = 8cm From right triangle DEC, we get 2

cm 

 289  64  cm

2

16 cm

4cm 4cm

D1 D

C1

24 cm

C

Area of section AB B1A1 = Area of section CD D1C1 1   24  16   4cm 2 2 www.betoppers.com

8th Class Mathematics

414 = 20 × 4 cm2 = 80 cm2 Area of section AD D1A1 = Area of section BC C1B1 1   28  20   4cm 2 2 = 24 × 4 cm2 = 96 cm2 = 96 cm2 Area of section A1B1C1D1 = A1B1 × B1C1 = 16 × 20 cm2 = 320 cm2 22. We have, AB + BC + CD + AD = 120m  AB + 48 + 17 + 40 = 120  AB + 105 = 120  AB = 120 – 105  AB = 15 m Since AB is perpendicular to the parallel sides AD and BC, AB is the altitude (height) of the trapezium. 40 m

A

Hence, Area of the shaded region = Area of the circle – Area of rectangle ABCD = (78.50 – 48) cm2 = 30.50 cm2. 24. The circumference of the circle  C  2r

C 2 the area of the circle, r

2

2 C  C A  r      4  2  Given the circumference of the circle is increased by 50%. 2

 C1  C 

D

50C   C 2   C1  100    Area of the circle A1  4 4

17 m

 B

48 m

C

Area of the field 1 1   BC  AD   AB   48  40   15m 2 2 2 = 44 × 15 m2 = 660 m2

C2  50  1   4  100 

2

2

2 C 2   50  2  50    1   4   100  100 

Circles, Semi circles & Ring



C 2  1 100  1    4  4 100 



C 2  125  1 4  100 

23. B

A

50C 100

 125  125 A1  A 1   A A  100  100   The area of the circle increases by 125%.

8 cm 6 cm

25.

C C1

D

C

Clearly, diameter of the circle = Diagonal BD of rectangle ABCD  Diameter = BD  BC 2  CD 2  6 2  82 cm = 10 cm Let r be the radius of the circle. Then, 10 r  cm = 5 cm 2 Area of rectangle ABCD = AB × BC = (8 × 6)cm2 = 48 cm2.

Area of the circle

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2

 r 2  3.14   5 cm2 = 78.50 cm2.

A

r

O

r

B

Consider the above semicircle. If AB = 2r is diameter, we can make vertex ‘C’ anywhere on the arc. But to get the triangle with largest area we take it at the centre of the arc as it is the highest point. If ‘O’ is the centre, OC forms the height. If another point is considered at C|, its perpendicular distance is less than that of OC.

1 2 Area   2r  r  r square units. 2

Mensuration

415 27. Let AB be the chord and the tangent drawn, and let AB = a.

26. Case - 1 A

B

r

r

O r

O

B

R D

C

S

A

Metal wasted in case 1: S = 2r 2

S S S     S 2  4 2

2

2



4S 2  S 2 4  S2 4 4

Case - 2 F

r

r

E

G

O S H

S = 2r Now, if side of square ABCD is ‘S’ then metal Metal wasted in Case ‘2’ is: diagonal of square EFGH is ‘S’ In EFGH, 2 (side)2 = (digonal)2  2 side2 = S2 S 2

 side =

S 2 Hence, metal wasted  Side of EFGH  2

r  S        2  2  2



C

2 2 2 2 Area of Ring  R  r    R  r  . From figure, OC = r and OA = R where r and R represent the radius of inner and outer circles. 1 a2 AC  a  R 2  r 2  AC 2  2 4 a 2  Area  4 28. Let original radius = R. 50 R R New radius  100 2 Original area  R 2 and 2

R 2 R New area      4 2 2  Decrease in area  R 

R 2 3R 2  4 4

% area decreased  3R 2  1   2 100  %  75% R  4 

 22  29. Area of the field grazed   14 14  sq. ft 7   = 616 sq.ft Number of days taken to graze the field 616  days = 6 days (approx). 100 30.

C

2

r

2

S S  1 2   S2     S2   4 2 4 2  4 

2 4 2  2  Total metal wasted  S  4   S  4      2 4      2 S    S2    2 4 

A

r

O

r

B

2

Required area 

  AC   AC 2     2  2  2 2

 AB 2  BC 2  2 4 2 2 2   AB BC 2    AB    BC              2  4 4  2  2  2  2  = 81 + 36 = 117 cm2. www.betoppers.com 

8th Class Mathematics

416 31.

22  29  28 29  28 7 22   57  1 7 1254   179.1 sq.mts 7 Cost for levelling the circular path = 179.1 × 1.85 = Rs. 331.34 33. Area of the sector = A Length of the arc = ‘l’ lr 2 2A A r  A  2 l l 2A  Radius of the sector  l 34. Perimeter of the sector = ‘p’ l + 2r = p  l = (p – 2r) radius = r l r  p  2r  r  sq.units  Area of the sector  2 2 35. Radius of the sector = ‘r’ Length of the arc = ‘l’ lr sq.units.  Area of the sector  2 If the radius is doubled, then the new radius = 2r l If the arc is reduced to half, then the new arc  2  Area of the new sector l 1 lr  2r    sq.units 2 2 2  Its area is unchanged. 36. 

A

B

Given, the diameter of the larger circle AB = 28 cm.  The radius OB = OE = R = 14 cm. E

A

O

B

But OE is the diameter of the inner circle. 14 = 7 cm  The radius of the inner circle  2 The area of the inner circle  r 2 22  77 7 = 154 sq.cm. R 2 The area of the semicircle AEB  2 22 14 14 2   cm 7 2 2 = 22 × 14cm = 308 cm2  The area of the shaded portion = (area of the semicircle) – (area of the inner circle) = 308 – 154 = 154 sq.cm.  The area of the shaded region is 154 sq.cm. 32. Area of the circular grass field = 2464 r 2  2464 22 2  r  2464 7 7  r 2  2464  22  r  112  7  7  16  7

 7 4 4 7  r = 7 × 4 = 28 m. Width = 1 R–r=1 R – 28 = 1  R = 1 + 28 = 29 m. Area of the path    R  r  R  r  www.betoppers.com

7m

I 90°

14 m II

III 14 m

21 m

7m

270°

Radius of the 1 sector = 7 mts. Angle (x°) = 90°  Area of the 1 sector 90 22 77   77  360 7 2 = 38.5 sq.mts. Radius of the II sector = 21 mts. Angle (x) = 270° 270  r 2  Area of the II sector  360

Mensuration

417

3 22 2079    21 21  4 7 2 = 1039.5 sq.mts. The measurements of the I sector and III sector and the same.  Area of the III sector = 38.5 sq.mts.  Area of the field that can be grazed = 38.5 + 1039.5 + 38.5 = 1116.5 sq.mts. 37.

I 3.5 m

 Area of the second sector

270 22 21 21    360 7 2 2 = 259.875 sq.mts. Area that can be grazed = 9.625 + 259.875 = 269.5 sq.mts. 

39. 3.5 3.5

90°

21 m

II

3.5

24.5 m 270°

28 m

The strikers are arranged as shown in the figure. Side of the square = 3.5 + 3.5 = 7 cm  Area of the square = 7 × 7 = 49 sq.cm. Area of 1 sector.

24.5 m

Radius of the I sector = 3.5 m Angle (x°) = 90°  Angle of the I sector 90 22 7 7 77      360 7 2 2 8 = 9.625 sq.mts Radius of the second sector = 24.5 m  Angle (x°) = 270°  Area of the II sector 270 22 49 49     360 7 2 2 = 1414.875 sq.mts.  Area of field that can be grazed = 9.625 + 1414.875 = 1424.5 sq.mts. 38.

90 22 7 7 77     sq.cms 360 7 2 2 8 77  4  38.5sq.cms Area of 4 sectors 8  Area of the empty space between the strikers = 49 – 38.5 = 10.5 sq.cms. 40. The radius of the circular disc = 28 cm. The sector angle = 45° 

49 m 2

28 45°

I 3.5 m

90°

The area of the circle

7m 10.5 m

270°

14 m

10 .5

m

II

Radius of the first sector = 3.5 mts. Angle (x°) = 90°  Area of the first sector 90 22 7 7 77      360 7 2 2 8 = 9.625 sq/mts. Radius of the II sector = 10.5 mts. Angle (x°) = 270°

22  28  28 7 = 88 × 28 cm2 = 2464 cm2. Now, the area of the sector x  × the area of the circle 360 45   2464cm 2 = 308 cm2 360  The area of the remaining part of the disc = (The area of the circle) – (The area of the sector) = (2464 – 308) cm2 = 2156 cm2.  r 2 

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8th Class Mathematics

418 2

IIT JEE WORKSHEET 1  A1    15  12  cm 2 = 90 cm2. A = 2A 2 1 2  = 180 cm2.

 Area 

1   20  h  180  h  18cm. 2 2.

1 Area of triangle   Base  Height 2

1 1 1 Ratio of sides = : : = 6 : 4 : 3. 2 3 4

6.

7.

8  160  x  cm  x cm. Then, length   5  100 

6  Perimeter = 52 cm. So, sides are  52   cm, 13  

4.

3  4   52   cm and  52  13  cm . 13    a= 24 cm, b = 16 cm, c = 12 cm.  Length of smallest side = 12 cm. Let the smallest side be x cm. Then, other sides are 13 cm and (17 – x) cm. Let a = 13, b = x and c = (17 – x). So, s = 15.

So,

8. P R

 15  2  15  x  x  2 

A

 x 2  17x  60  0   x  12  x  5  0

2 b l . 3 Then, Area = 216 cm2

x

D

C

Let the side of the triangle be a. Then, www.betoppers.com

8 cm

B

2 l  b  5   2l  2b  5b  3b  2l b 1

A

B a/2

Q

9.

 x = 12 or x = 5.  smallest side = 12 cm or 5 cm.

a

C

Required perimeter = (AB + BC + CP + PQ + QR + RA) = AB + BC + (CP + QR) + (PQ + RA) = AB + BC + AB + BC = 2 (AB + BC) = [2 (8 + 4)] cm = 24 cm.  smallest side = 12cm or 5cm

2

 15  x  x  2   30

5.

8 3 x  x  24  x  24 5 5

 24  5   x   40  3   Length = 64 cm, Breadth = 40 cm. Area = (64 × 40) cm2 = 2560 cm2.

 Area  s  s  a  s  b  s  c 

 30  15  x  x  2    30 

x2 3 square units 3 We have : (l – b) = 23 _________ (1) 2 (l + b) = 206 or (l + b) = 103 _____________ (2) Solving the two equations, we get : l = 63 and b = 40.  Area = (l × b) = (63 × 40) m2 = 2520 m2. Let breadth = x cm.  Area 

1  40  Base  Base  Height 2  Height = 80 cm. 3.

3 2 3 4 2 x2 a   x  4 4 3 3

4 cm

1.

3a 2 4x 2 a a2     x2   x2  a2  4 3 2

2  l × b = 216  l  l  216 3 2 l = 324 l = 18 cm.  

Mensuration

419 2

10. Other side

2

225 81  15   9        ft   ft 4 4  2  2



144 ft = 6ft 4

area of rectangle = l × b

8   6  24sq.units 2 11. Let breadth = x metres. Then, length = (x + 5) metres. Then, x (x + 5) = 750  x2 + 5x – 750 = 0  (x + 30) (x – 25) = 0  x = 25  Length = (x + 5) = 30m. 12. Let breadth = x metres. Then, length = (x + 20)metres.

16. Let the altitude of the triangle be h1 and base of each be b.

1  b  h1  b  h2 where h2 = 100m. 2  h1 = 2h2 = (2 × 100)m = 200m. 17. Let the two parallel sides of the trapezium be a cm and b cm. Then, a – b = 4 ________ (1). Then,

And,

1   a  b  19  475 2

 475  2    a  b     19   a + b = 50 ________ (2). adding (1) and (2), we get 2a = 54  a = 24 by substituting a = 24 in (1), we get b = 23  5300  Perimeter    m = 200m.  length of the parallel sides is equal to  26.50  a = 27, b = 23.  2[(x + 20) + x] = 200  2x + 20 = 100 18. Clearly, the cow will graze a circular field of area  2x = 80  x = 40. 9856 sq.metres and radius equal to the length of Hence, length = x + 20 = 60 m. the rope. 13. Let original length = x metres and original breadth Let the length of the rope be R metres. = y metres. 7   Then, R 2  9856  R 2   9856   = 3136 Original area = (xy)m2. 22    120  6   R = 56. xm   xm ; New length    100  5  length of rope = 56m.  120  6  y  m   y  m. New breadth    100  5  6  2  36  2 6 New area   x  y  m   xy  m 5  5  25  area increased 

5 10  19. Speed = 12 km/hr  12   m / s  m / s 18  3  22   Distance covered   20  2   50  m 7  

36 11xy xy  xy  sq. m 25 25

 11  1  Increase%   25 xy  xy  100  %  44%   14. 2(l + b) = 340 (Given) Area of the boundary = [(l + 2) (b + 2) – lb] = 2 (l + b) + 4 = 344 m2.  Cost of gardening = Rs. (344 × 10) = Rs. 3440. 15. Area of the mat = [(15 – 3) × (12 – 3)]sq.ft = 108 sq.ft.  Cost of the mat = Rs. (108 × 3.50) = Rs. 378.

 Time taken 

44000 m 7

Distance  44000 3    s Speed 10   7

 4400  3 1     min 7 60  



220 3 min  31 min 7 7

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8th Class Mathematics

420 20.

2R = 2(l + b)

( area of sector =

 2R  2  26  18 cm

 88   R  7   14cm  Area of the circle  2  22   22   R 2   14  14  cm 2  616cm 2 7   21.

R 2  24.64

  r 2 ) 360

 49 3     11 0.5  3.5  cm 2  4  2 = (21.217 – 19.25) cm = 1.967 cm2. 26. Let the original radius be R cm. New radius = 2R. Original area  R 2 , 2

New area    2 R   4R 2

 24.64  R   7   7.84  22  2

2 2 2 Increase in area   4R  R   3R

 R  7.84  2.8cm 

22



 Circumference   2  7  2.8  cm  17.60m  

22.

2R  R 2  R  2  2R  4 Hence, diameter = 4. 23. Radius of the ground = 17.5m. Radius of inner circle = (17.5 – 1.4)m = 16.1 m. Area of the garden   × [(17.5)2 – (16.1)2]m2

 22     17.5  16.117.5  16.1  m 2 7   22     33.6  1.4 m2 = 147.84 m2. 7   24.

2 1 2 2

R1 2 2R1 R1 2     R2 3 2R2 R2 3

 Required ratio = 2 : 3. 25. Required area = (Area of an equilateral  of side 7 cm) – (3 × area of sector with   60 & r = 3.5 cm)

 3   22 60   2    7  7    3   3.5  3.5    cm 7 360    4   3.5

3.5 3.5

3.5 3.5

 106   53R  R  cm   New radius    cm  100   60  Original area  R 2 2

 53R  2 Increase in area      R  50  2 2 2  53 2  R   53   50    R    1  2500  50   2

R 2 103  3 2 m 2500 Increase% 

2 1 2 2

R 4 R 4    R 9 R 9



 3R 2  100  %  300% Increase%   2  R  27. Let the original radius be R cm.

3.5

 R 2  309 1    2 100  %  12.36% R  2500  28. Let original radius be R cm. Then, original circumference   2R  cm New radius = (175% of R)cm

7R  175    R  cm  cm 4  100  7R  7 R  cm New circumference   2  cm  4  2  Increase in circumference 3R  7 R    2R  cm  cm 2  2 

1  3R    100  %  75% Increase%   2 2  R   www.betoppers.com

11. PLANE GEOMETRY SOLUTIONS

FORMATIVE WORKSHEET 1.

2.

When the sum of the measures of two angles is equal to 90°, then the angles are said to be complementary. i) 54° Measure of the complementary angle of 54° = 90° – 54° = 36°. ii) 90° Measure of the complementary angle of 90° = 90° – 90°= 0°. iii) 47°22| We know that 1° = 60| Measure of the complementary angle of 47° 22| = 90° – 47°22| = 89° 60|– 47°22| = 42° 38| iv) 65°10|20|| We know that 1° = 60| and 1| = 60|| Measure of the complementary angle of 65°10|20|| = 90° – 65°10|20|| = 89°60| - 65°10|20|| = 89°59| 60|| – 65°10|20|| = 24°49| 40|| v) 87°59|| Measure of the complementary angle of 87° 59|| = 90° – 87° 59|| = 89° 60| – 87° 59|| = 89° 59| 60|| – 87° 59|| = 2° 59| 1|| In 60 minutes, the minute-hand of a clock turns 360°. i) In 5 minutes, the minute-hand of a clock turns

b)

 The two angles are 4x = 4 × 20° - 80° 5x = 5 × 20° = 100° Given, the angles (2a – 10)° and (a – 11)° are complementary.  (2a – 10) + (a – 11) = 90°  2a – 10 + a – 11 = 90°  3a – 21 = 90°  3a = 90 + 21 = 111

111  37 3  The angles are (2a – 10)° = (2 × 37 – 10)° = (74 – 10)° = 64° and (a – 11)° = (37 – 11)° = 26°. Let one angle be x°. Then the other angle = (x + 32)° Given the two angles are supplementary.  x° + (x + 32)° = 180°  2x° + 32° = 180°  2x° = 180° – 32° = 148° a

c)

148  74 2  The two angles are x = 74° and (x + 32)° = 74° + 32° = 106°. As POR and QOR are angles of a linear pair  POR + QOR = 180°  4t + 2t = 180° 6t = 180°  x 

4.

180  t  30 6 Given OP is a ray on line QR. Also POQ = POR ________ (1) t 

5.

360  5 = 6 × 5 = 30° 60 ii) In 21 minutes, the minute-hand of a clock turns

3.

360  21 = 6 × 21 = 126° 60 a) Given two supplementary angles are in the ratio 4 : 5. Let the two angles be 4x and 5x. Where ‘x’ is any constant As the angles are supplementary.  4x + 5x = 180°  9x = 180° ? x 

180  x = 20° 9

But, POQ + POR = 180° (from the above theorem) From (1) and (2), we have  2 POQ = 180°

180  90 2 Hence POQ = POR = 90° Thus proved.  POQ 

________

(2)

8th Class Mathematics

422 6.

7.

As XOZ and YOZ are angles of a linear pair.  XOZ + YOZ = 180° i.e. a + b = 180° (i) Given a = 75° a + b = 180°  75° + b = 180°  b = 180° – 75° = 105°. (ii) Given b = 110° But a + b = 180°  a + 110° = 180°  a = 180° – 110° = 70°. Here t + 10° + t + t + 20° = 180°  3t + 30° = 180°  3t = 180° – 30° = 150°

150  50 3  QOS = t + 20° = 50° + 20° = 70° ROS = t = 50° POR = t + 10° = 50° + 10° = 60°. 8. Given, AB and CD are two coplanar lines and APX = 125°, DQP = 55°. BPQ = APX ( vertically opposite angles)  BPQ = 125° Now, BPQ + DQP = 180° ( if a pair of interior angles on the same side of the transversal are supplementary, then the lines are parallel). Hence, the lines AB and CD are parallel. 9. Given PQ and RS are two coplanar lines and XY is the transversal. Given PEX = 105° Since PEX and ?PEF are linear pair of angles, PEX + PEF = 180° 105° + PEF = 180°  PEF = 180° – 105° = 75°  a = 75° PEF = EFS = 75° i.e., PEF and ?EFS form a pair of alternate angles. Hence PQ||RS 10. Given A || B and C || D and ?1 = 75°. t 

It is required to prove that ?3 = ?1 + angle. Since A || B. ? ?1 = ?2. ( Corresponding angles are equal) Also C || D.

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1 of a right 3

 2 + 3 = 180°  75° + ?3 = 180°  3 = 180° – 75° = 105°  ?3 = 75° +

1 × 90° 3

1  1   90 3 1  1  of a right angle. 3 Hence proved. 11. Let GA and HB be the bisectors of two alternate angles ? PGH and ?SHG respectively.

It is required to prove that GA || HB.

1 

1  PGH 2

_________

(1)

1 and 2  SHG _________ (2) 2 But ?PGH = ?SHG ( alternate angles) 1 1  PGH  SHG 2 2 1 = ?2 But these two are alternate angles. Hence GA || HB. Thus proved. 12. AC is parallel to DF implies AC || FD  CAF = DFB = x Since DE || FB (AB) DFB + ?FDE = 180° (Interior angle on the same side of the transversal)  x + y = 180°

Plane Geometry 13. Given PQ || RS XEQ = 80°, ?YHS = 120°, ?XYZ = a Since PQ ||RS and XY is the transversal, XEQ = ?XGS = 80° Since ?XGS and ?YGS are adjacent angles. XGS + ?YGS = 180° ( they are linear pair)  80° + ?YGS = 180° YHS = 120° In GYH, by exterior angle theorem, Exterior angle is equal to the sum of the two interior opposite angles  YHS = YGS + GYH 120° = 100° + a  a = 120° – 100° = 20° Hence a = 20° 14.

423 Consider a ABC, with angles p, q and r. Extend the side BC to D. Let the exterior angle at ‘C’ be ‘s’. It is required to proved that s = p + q From Interior angle theorem, we have p + q + r = 180° ______ (1) Also r + s = 180° ______ (2) ( linear pair) From (1) and (2), we have p + q + r = r + s  p + q = s or s = p + q Hence the exterior angle of a triangle is equal to the sum of the two interior opposite angles. 16.



Given ABC is a triangle. The interior angles of this triangle are p, q and r. It is required to prove that the sum of the three interior angles is 180°. i.e. p + q + r = 180° Through A, draw a line XY parallel to BC. s and t are the exterior angles on the line XY at point A. Since XY is parallel to BC.  q = s and r = t ( alternate interior angles) Adding p on both the sides, we get p + q + r = s + t + p But p + s + t = 180° (by linear pair axiom)  p + q + r = p + s + t = 180°  p + q + r = 180° Hence the sum of the three interior angles of a triangle is 180°. 15.

Given l||m, CA is the incident ray on the mirror l and AB is the reflected ray from the mirror l. We know that Angle of incidence = angle of reflection Angle of incidence on the mirror l = 1 Angle of reflection from the mirror l = 2  1 = 2 _______ (1) Now the reflected ray AB incidents on the mirror m and reflects along BD. Angle of incidence on the mirror m = 3 Angle of reflection from the mirror m = 4  3 = 4 __________ (2)



Since l||m and AB is the transveral a = b 90 – a = 90 – b  2 = 3 __________ (3) From (1), (2) and (3), we get 1 = 4 Adding 2 on both the sides, we get 1 + 2 = 4 + 2 1 + 2 = 4 + 3 ( 2 = 3)  CAB = ABD CAB and ABD are alternate angles and are equal between two lines CA and BD. Where AB is the transversal.

   CA || BD

Thus proved

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8th Class Mathematics

424 17. We know that the sum of the interior angles in a triangle is equal to 180°  p° + q° + r° = 180° ________ (1). We know that an exterior angle is equal to the sum of the two interior opposite angles.  x° = p° + r° __________ (2) and y° = p° + q° __________ (3) Adding (2) and (3) we get x° + y° = p° + r° + p° + q° Using (1) x° + y° = p° + 180° p° + 180° > 180° ( angle of a triangle cannot be zero or negative)  x° + y° > 180°. 18. Consider the triangles PQR and ABC We have, PQx = ABy PQR = ABC ; PQ = AB ; and QPR = BAC Here, two angles and the included side are equal. Then according to the ASA congruence theorem, we have PQR  ABC 19. In  ABC and ? ADE BC = DE (given) BCA = ADE (given) BAC = EAD (vertically opposite angles)  By ASA congruency axiom.  ABC  ADE. 20. Given in ’s ABC and PQR, AB = PQ, A = 40°, B = 60°, P = 40° and R = 80° In PQR, Q + P + R = 180°  Q = 180° – 120° = 60°  Q = 60° In  ABC and  PQR AB = PQ (given) A = P (given)) B = Q B = Q  by ASA congruence theorem, ABC  PQR 21. Consider the triangles ABC and ACD. We have, AB = AD BC = CD and AC is common side All the three sides of the triangles are equal.  By SSS congruence theorem, we have ABC  ACD. 22. Given in  PQR, PQ  PR and QPS = RPS. ( PS is the bisector of P) It is required to prove that www.betoppers.com

 PQS  PRS and PS  QR In PQR and  PRS, PQ = PR PS is the common side  QPS = RPS  By SAS rule,  PSQ  PRS Since,  PSQ  PRS  PSQ =  PSR But these two angles form a linear pair.  PSQ = PSR = 90° PS  QR

23.

Given : - In  ABC, AB  AC and BAD = CAD R.T.P : - (i) ABD  ACD (ii) AD  BC Proof : - (i) In ABD and ACD AB  AC (given) AD is common. Included BAD = Included  CAD (given)   ABD   ABD   ACD (ii) Since  ABD   ACD ADB = ADC

But these two angles form a linear pair.  ADB = ADC = 90° AB  BC

24. Given : AB  CD . AD and BC are transversals intersecting at ‘O’ and R.T.P :-  OAB  OCD Proof :- AB  CD and AD is the transversal  OAB = ODC (Alternate angles) Now in  OAB and  OCD OAB = ODC (shown above) AOB = COD (vertically opposite angles) AO = OD

Plane Geometry

425

 OAB  OCD (A.S.A. Axiom)

28. Given, In ABC, P and Q are the midpoints of the sides AB And AC. RTP PQ || BC and

1 BC 2 Construction: Extend PQ to R, such that PQ = QR. Join CR. Proof: Now in ’s APQ and QRC, ( Q is the midpoint of AC) AQ = QC AQP = CQR ( vertically opposite angles) and PQ = QR  By SAS congruence axiom,  APQ   QRC PQ 

25. Let ABC be an isosceles triangle and let B = C

Given that one of the base angles is 40° i.e., B = C = 40° Since the sum of the three interior angles of a triangle is 180°,  B + C + A = 180°  40° + 40° + A = 180°  A = 180° – 80° = 100° Hence the vertical angle, A = 100°. 26. Let ABC be an isosceles triangle with AB = AC  B = C

Given the vertical angle, A = 2(B + C) = 2 (2B) ( B = C) = 4B But A + B + C = 180° ( by interior angle theorem)  4B + B + B = 180°  6 B = 180°  B = 30° Also C = 30° ( B + C) a = 2(B + D) = 2 (30° + 30°) = 120° 27. Given ABC is an isosceles triangle in which AB = AC.  B = C ( angles opposite to equal sides are equal) Also given B = 2A  C = 2A In a triangle the sum of angles is 180°  A + B + C = 180°  A + 2A + 2A = 180°  5A = 180°  A 

 AP = CR But AP = PB ( P is the midpoint of AB)  PB = CR Also since  APQ   QRC  PAQ = ACR But these are alternate angles  BP || CR Since BP is equal and parallel to CR, BCRP is a parallelogram. PR || BC ( opposite sides of a parallelogram are parallel)  PQ || BC and PR = BC ( opposite sides of a parallelogram are equal)

1 1 PR  PQ  BC  PR  BC  2 2 Thus the line-segment joining the midpoints of any two sides of a triangle is parallel to the third side and equal to half of it. PQ 

29.

180  36 5 www.betoppers.com

8th Class Mathematics

426 Given that ABC is a right triangle, right angled at C. i.e., C = 90°. AB is the hypotenuse and M is the midpoint of AB. Through M draw MD || BC to meet AC at D.  ADM = DCB = 90° ( corresponding angles are equal) In  AMD and CMD ADM =  CDM = 90° AD = DC (intercept theorem) DM is the common side  By SAS congruence axiom,  AMD   CMD  CM = AM = MB

AM  MB AB  2 2 Hence proved. 31. In ?ABC, ?A = 90°, X, Y, Z are midpoints of AB, BC and CA. CM 

D is the foot of the altitude through A, N is the midpoint of AY. It is required to prove NX = NY = NZ = ND We know that AY = YB = YC In ?ABY, N and X are midpoints of the sides AY and AB.

1 1  NX  BY  AY 2 2

32. Draw a tangent AX at the point ‘A’. Here the point of contact is ‘A’.

It is required to prove that OA  AX. Consider any point ‘B’ on the tangent AX. Join OB. Since AX is a tangent to the circle at the point ‘A’, the point ‘A’ lies on the circle. Other points are outside the circle. Consider any other point ‘C’ between ‘A’ and ‘B’. Join OC… OC is also greater than the radius OA. This is true of any other point on the tangent. Hence OA is the shortest distance. The shortest distance between a point and a line is the ? distance. Hence OA ? AX. But OA is the radius. Hence, “the tangent at any point of a circle is perpendicular to the radius drawn to the point of contact”. 33. Given PO = 17 cm and OQ = 7cm. From the above theorem, “the tangent of a circle is  to the radius at the point of contact”. PQO = 90°  PQO is a right-angled triangle. According to Pythagoras theorem, PO2 = PQ2 + OQ2 ? PQ2 = PO2 – OQ2  PQ2 = (17)2 – (7)2 = 289 – 49 = 240  PQ  240  4 15cm 34. Proof : Given, a circle with centre ‘O’, OX is a radius.

1 1 BY  AY 2 2 In right ? ADY, DN is the median to the hypotenuse AY. Similarly, NX 

1  ND  AY 2 Since N is midpoint of AY, NY 

1 AY 2

 NX = NY = NZ = ND  Hence prove. www.betoppers.com

1 AY 2

A line AB is drawn through ‘X’ such that AB  OX at ‘X’. It is required to prove that AB is a tangent to the circle. Consider any other point ‘Y’ on AB. Since OX  AB, OY is the hypotenuse of ? OXY. Hence OY > OX.  ‘Y’ lies outside the circle (since OY > radius) Similarly, with any other point taken on the line, except point ‘X’, we have a right-angled triangle

Plane Geometry whose hypotenuse will be greater than OX, Which is the radius of the circle.  Hence there is only one point ‘X’ on AB which lies on the circle. Therefore AB is a tangent at point ‘X’. 35. ‘O’ is the centre of the circle. Radius of the circle = OA = OB = 4.8cm Length of the chord AB = 4.8cm

1 4.8 AB  2 2 AC = 2.4cm Considering the right-angled triangle, AOC, By Pythagoras theorem, OC2 = OA2 – AC2 OC2 = (4.8)2 – (2.4)2 OC2 = (7.2) (2.4)  OC2 = 17.28 OC2 = (7.2) (2.4)  OC2 = 17.28  AC 

OC  17.28  4.2 Distance of the chord from the centre OC = 4.2cm OA = OB = AB = 4.8cm   AOB is an equilateral triangle. Area of an equilateral triangle 

3 2 a sq.units 4



3 3 2  4.8   4.8  4.8 4 4

427 37. Given : In two concentric circles with centre ‘O’. AD is the chord of the bigger circle. AD intersects the smaller circle at B and C. R.T.P : AB = CD Construction : Draw OE  AD Proof : AD is the chord of the bigger circle centre ‘O’ and OE  AD.  OE bisects AD  AE = ED ________ 1 BC is the chord of the smaller circle with centre ‘O’ and OE  BC.  OE bisects BC  BE = CE ________ 2 From I and II AE – BE = ED – EC  AB = CD 38.

 3 1.8  4.8 

= 1.732 (1.2) (4.8) = 9.975  10 sq.units 36. Given radius of circle = 13cm and OP = OQ = 12cm

We have to prove that AB = CD In right AOP, OA = 13, OP = 12 AP2 = OA2 – OP2 = 169 – 144 = 25 AP = 5cm AB = 2AP = 10cm Similarly, in COQ CQ = 5cm and CD = 10cm  AB = CD

Given : Chords AB and CD of circle with centre ‘O’ intersects at E and OE bisects BED (i.e.,) R.T.P. : Chord AB = Chord CD Construction : Draw OM ? CD and ON ? AB Proof : In OME and ONE OME = ONE = 90° (Construction) OME = OEN (given) OE is the common side  OME  ONE  OM = ON  Chord AB = Chord CD

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8th Class Mathematics

428

CONCEPTIVE WORKSHEET 1.

2.

When the sum of the measures of two angles is equal to 180°, then the angles are supplementary. i) 50° Measure of the supplementary angle of 50° = 180° – 50° = 130°. ii) 90° Measure of the supplementary angle of 90° = 180° – 90° = 90°. iii) 132°45| Measure of the supplementary angle of 132°45 | = 180° – 132°45| = 179°60| – 132°45| = 47°15| iv) 56°20|48|| Measure of the supplementary angle of 56°20|48|| = 180° – 50°20|48|| = 179°60| – 56°20|48|| = 47°15| iv) 56°20|48|| Measure of the supplementary angle of 56°20|48|| = 56°20|48|| = 179°60| – 56°20|48|| = 123°39|12||. v) 112°48|| Measure of the supplementary angle of 112°48 || = 180° – 112°48|| = 179° 60| – 112° 48|| = 179° 59|60|| – 112° 48|| = 67° 59| 12||. The supplement of x = 180° – x

1 × (180° – x) 3 3x = 180° – x 4x = 180°  x = 45° Given in 24 hours the earth makes 360° i.e., 1440 minutes = 360°

4.

120 = 40° 3 Hence, the value of ‘a’ is 40°.

a =

5.

360 1  1440 4 3 hours 20 minutes = 3 × 60 min + 20 min = 20 minutes In 3 hours 20 minutes the earth makes an angle

Given ON bisects POQ

 PON  NOQ and OM bisects QOR

 QOM  MOR and ON  OM  NOM = 90°

 NOQ  QOM  90  PON  MOR  90

Now POR  PON  NOQ

QOM  MOR   PON  MOR    NOQ  QOM 

= 90° + 90° = 180° Since the adjacent angles formed are supplementary, POR is a straight line. i.e., P, O, R are collinear.

Given x =

3.

Given POQ is a straight line a + 30°, a + 10° and a + 20° are angles of a linear pair.  (a + 30°) + (a + 10°) + (a + 20°) = 180°  3a + 60° = 180°  3a = 180° – 60° = 120°

6.

From the figure,  SOU = 5t° (vertically opposite angles) Since OP and OQ are in the same straight line,   POU +  SOU +  SOQ = 180°  5t° + 5t° + 2t° = 180°

 1min 

1   200  50 4 The earth makes 360° in 1440 minutes 1440 = 4 min. 360  130° = 130 × 4 = 520 minutes = 8 hours 40 minutes Hence, in 3 hours 20 minutes the earth makes an angle 50° and it takes 8 hours 40 minutes to turn through 130°. 1 

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 12t° = 180°  t° =

180 12

 t° = 15°.

7.

Given  POR +  QOS = 85° We know that,  POR +  ROS +  QOS = 180° ( the adjacent angles formed on a straight line are supplementary)

  ROS + 85° = 180°   ROS = 180° – 85° = 95°   ROS = 95°

Plane Geometry 8.

Given  XOZ and  YOZ are linear pair of angles.   XOZ +  YOZ = 180° –––––– (1)  p + q = 180° Given p – q = 80° –––––– (2) Adding (1) and (2), we get

429 P

11.

Y

A

C

T 260 = 130° O Q 2 B  p + q = 180°  130° + q = 180° X Z  q = 180° – 130° = 50° Given OP || XY and OQ || XZ 9. Given that PQ || AB and  1 = 70° Let us extend OQ to meet XY at T XY is the tranversal cutting PQ and AB at L and  OT || XZ M, forming the angles. Since OP || XY,  A =  C  1,  2,  3,  4,  5,  6,  7 and  8. ( corresponding angles) ––––––– (1) It is required to prove that  4 =  6 and  3 = Also OT || XZ,   C =  B 5 ( corresponding angles) ––––––– (2) By corresponding angles axiom,  1 =  5 From (1) and (2), we get Since  1 = 70°   5 = 70° ––––– (I) A = B Since the vertically opposite angles are equal. Hence proved. 1  3 12. Since  1 = 70°   3 = 70° ––––– (II) p 2  1 +  4 = 180° ( linear pair of angles) q   4 = 180° –  1 = 180° – 70° = 110° –––––– (III) 1 r Since  4 and  8 are corresponding angles. 10°   4 =  8 = 110° Given p || q and q || r Since  8 and  6 are vertically opposite angles,  p || r they are equal 10° ––––––– (IV)   8 =  6 = 110° 2 Now from (I) and (II),  5 =  3 = 70° from (III) and (IV),  4 =  6 = 110° Hence each pair of alternate interior angles are 1 3 equal. 10. Let AB, CD and EF be three straight lines which are perpendicular to the line XY. Now  1 = 120° and It is required to prove that AB || CD || EF.  1 +  3 = 180° (linear pairs) E C A  3 = 180° –  1 = 180° – 120° = 60° But  2 =  3 (Alternate angles) 2 3 1   2 = 60°. Y X B D F 13. Given r  p i.e.,  3 = 90° Now  1 =  2 (= 90°) r  q i.e.,  4 = 90° But these are corresponding angles.   3 =  4 (Each 90°)  AB || CD.  p || q Similarly CD || EF. Hence AB || CD || EF. Thus proved.

2p = 260°  p =

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8th Class Mathematics

430 q

r

1 2 t

From the diagram  1 =  2 (Alternate angles) But  1 = 65°   2 = 65° 14.   4 and     180

180 = 36° 5

50°

50°

80°

 = 4.36 = 144°

80°

B

15.   x =  CBD ( BD || CE)  ADB = 180° – 110° – 30° = 40°   BDC = 75° – 40° = 35° In  BDC,  DBC = 180° – 35° – 60° = 85°   x = 85° 16. According to the exterior angle theorem “Exterior angle of a triangle is equal to the sum of the two interior opposite angles.”  p° + 30° = 110°  p° = 110° – 30° = 80°. We know that sum of the interior angles of a triangle is 180°. (Interrior angle theorem)  p° + q° + 30° = 180° Using, p° = 80°, we have, q° = 180° – 30° – 80°  q° = 70°. 17. Let the triangle be ABC and AB, BC the legs of the right angle. A 45°

45°

B

C

Given  BAC =  BCA = 45°  AB = BC (Sides opposite to equal angles are equal) Hence AB : BC = 1 : 1 www.betoppers.com

D

6c

(or)  

A

m

(  ,  are adjacent angles) or   4 = 180°

m

4

3

18. Let two angles of a triangle be ‘x’ and ‘y’ Given that, x + y = 80° ––––– (1) and x – y = 20° ––––– (2) Adding (1) and (2), we get 2x = 100°  x = 50° Substituting the value of ‘x’ in (1), we get 50° + y = 80°  y = 30° Sum of two angles is 80°  third angle = 180° – 80° = 100° ( sum of interior angles is 180°). 19. The two triangles are  ABC and  DEF Given that AB = 6 cm,  BAC = 50°,  ABC = 80°, DE = 6 cm,  EDF = 50°,  DEF = 80°

6c

p

C

E

F

In  ABC and  DEF,, AB = DE  BAC =  EDF = 50°  ABC =  DEF = 80° Here two angles and the included side of one triangle are equal to the corresponding two angles and the included side of the other triangle.  by ASA congruency..  ABC   DEF.. 20. In the adjoining figure, consider the triangles ABC and ADC. C D

A

B

We have,  ACD =  CAB  B =  D and CA is common side Here, two angles and the non-included side are equal. Then according to the AAS congruence theorem, we have  ABC   ADC. 21. In  ABC and  DEF, we have AB = DE ; AC = DF ; BC = EF The three sides of  ABC are equal to the corresponding three sides of  DEF..  by SSS congruence theorem.  ABC   DEF

Plane Geometry 22.

 ABC is a right-angled triangle. In  CDE, the sum of two angles is 90°. Hence the third angle,  D = 90°   CDE is also a right-angled triangle. Consider the right triangles ABC and CDE. AC = CE and BC = CD Hypotenuse and one side of the triangles are equal.  By R.H.S. congruence theorem, we have  ABC   CDE 23. Given PR = QR  SQP =  TPQ  SRP =  TRQ T   SRP +  SRT =  TRQ +  SRT   PRT =  SRQ Now in  TRP and  SQR PR = QR  TPQ =  SQP and  PRT =  SRQ  By ASA rule,  TPR   SQR  TR = SR.

431 26. Let ABC be the isosceles triangle with AB = AC. A

B

C

Since AB = AC, the angles opposite to these sides are also equal. i.e.,  B =  C Given that the length of the equal sides is greater than the length of the unequal side. i.e., AB > BC and AC > Bc Since the angle opposite to the greater side is greater,   B >  A and  C>  A. B +C > 2 A 180 –  A >2  A  180 > 3  A  3  A < 180   A < 60°. 27. Given in  ABC, AD bisects  BAC 24. Given AB || XY and BX and AY are the   BAD =  DAC = x transversals intersecting at ‘O’ and OA  OY . Also AD = DC   DAC =  DCA ( angles opposite to equal sides are equal) X Y and  BDA = 70° We know that the exterior angle is equal to the sum O of the two interior opposite angles. A B  70° = x° + x° It is required to prove that  2x° = 70°  OAB   OXY.. 70 x   35 Since  OAB   OXY. and AY is the transversal, 2   OAB =  OYX (alternate angles)   BAD =  DAC =  DCA = 35° Now in  OAB and  OXY In  ABD. OA = OY  BAD +  ABD +  BDA = 180°  OAB =  OYX and  35° +  ABD + 70° = 180°  AOB =  XOY (vertically opposite angles)   ABD + 105° = 180°  by ASA rule,  OAB   OXY..   ABD = 108° – 105° = 75° 25. Given :- In  ABC, AB  AC and AD  BC D R.T.P : - Since AD  BC ,  ABD and  ACD 28. 4 A are right angled triangles. P In  ABD and  ACD 3 AD is common.

Hypotenuse AB = Hypotenuse AC (given)   ABD   ACD consequently  BAD =  CAD

B

 AD bisects  A

Let AP be the external bisector of the vertical angle A of  ABC.

1

2 C

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8th Class Mathematics

432

Let  B =  1 and  C =  2  PAC =  3 and  PAD =  4 Since AP || BC and AC, AB are the transversal   1 =  4 –––––– (1) ( corresponding angles)

––––– (2) 2 = 3 (alternate angles) Since AP is the bisector of  A,   3 =  4 ––––– (3) From (1), (2) and (3), we have 1 = 4 = 3 = 2   1 =  2  AB = AC ( in atriangle, the sides opposite to equal angles are equal)   ABC is an isosceles triangle.

29. Given that X is the orthocentre of  PQR.PA, QB and RC are the altitudes of  PQR that intersect at X. In  XQR, XA  QR ( PA A  QR) and QC  RX (produced) ( RC  PQ) Two altitudes XA and QC of  XQR meet at P..  P is the orthocentre of  XQR. 30. Given : PQ = 12cm, OQ = 4cm. We know that the tangent of a circle is perpendicular to the radius at the point of contact.

 TPO = 90° ( angle in semicircle)  TP is perpendicular to OP.. Hence TP is a tangent to this circle.  TQO = 90° ( angle in semicircle)  TQ is perpendicular to OQ. Hence TQ is a tangent to this circle. Hence we have two tangents TP and TQ from an external point T. 32. Given : Distance from the external point to the centre of the circle, d = 10 cm Radius of the circle, r = 6cm Length of the tangent of a circle, t  d2  r2 t  10 2  62  100  36  64 = 8

33. Chord AB is at a ditance of 8 cm from the centre.

O A

Q

O

 PQO = 90°   POQ is a right angled triangle :

According to Pythagoras theorem. PO2 = PQ2 + OQ2  PO2 = (12)2 + (4)2 = 144 + 16 = 160

 PO =

160 = 4 10 cm 31. Proof : Consider a circle with centre ‘O’. P T

O Q

8 C

B

 OC = 8cm Radius of circle OA = 10cm We have to find the length of the chord AB.

P

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Let ‘T’ be an external point. It is required to prove that two tangents can be drawn from ‘T’. Join ‘OT’. Taking this as diameter, draw a circle. This circle cuts the circle with centre ‘O’ at only two points ‘P’ and ‘Q’. Join TP, TQ, OP and OQ.

10

Given AP || BC

Consider  AOC, by Pythagoras theorem AC2 = AO2 – OC2 AC  AO 2  OC 2  AC  100  64

 AC  36  AC = 6cm We know that, AC = CB [ OC is the perpendicular bisector of AB]  AB = 6 + 6 = 12cm  length of the chord = 12cm

Plane Geometry

433

SUMMATIVE WORKSHEET 1.

4.

Given ‘p’ and ‘q’ are complementary angles. ––––– (1)   p +  q = 90°

1