CBSE Class 10, Mathematics Term 1 - Book (guide) · proportional and the triangles are similar. iv. (Motivate) If the corresponding sides of two triangles are proportional, their

MATHEMATICS TERM - I

Written as per the syllabus prescribed by the Central Board of Secondary Education.

CBSECLASS X

Printed at: Repro Knowledgecast Ltd., Mumbai

10280_10450_JUP

P.O. No. 13158

No part of this book may be reproduced or transmitted in any form or by any means, C.D. ROM/Audio Video Cassettes or electronic, mechanical including photocopying; recording or by any information storage and retrieval system without permission in writing from the Publisher.

Salient Features

• Extensive coverage of the syllabus for Term - I in an effortless and easy to grasp format.

• In alignment with the latest paper pattern of Central Board of Secondary Education.

• Neat and labelled diagrams.

• ‘Things to Remember’ highlights important facts.

• Variety of additional problems for practice.

• Questions from previous years board papers have been solved.

• Memory Maps at the end of each chapter to facilitate quick revision.

• Sample Test Paper at the end of each chapter designed for student’s Self Assessment.

• Model Question Papers in accordance with the latest paper pattern.

First Edition: March 2016

PREFACE

In the case of good books, the point is not how many of them you can get through, but rather how many can get through to you. “Class X: Mathematics” is a complete and thorough guide critically analyzed and extensively drafted to foster the student’s confidence. The book ensures extensive coverage of the syllabus for Term - I in an effortless and easy to grasp format.

The Topic-wise classified format for each chapter of this book helps the students in easy comprehension.

Each chapter includes the following features:

Theory of each mathematical concept is explained with appropriate references. NCERT exercises and Exemplar questions are covered with solutions. Problems based on NCERT exercises covering all the topics are provided with solutions. Practice problems based on NCERT exercises and MCQs help students to revise the topics thoroughly. One Mark Questions are provided for each chapter. Higher order thinking skills (HOTS) questions have been added for the student to gain insight on the various levels of theory-based questions. Value - Based Questions which emphasize on values have been included. Memory Map has been provided to give a quick overview of the chapter, helping the students in effective learning. Sample Test Paper at the end of each chapter helps to test the range of preparation of the students. ‘Things to Remember’ help the students’ gain knowledge required to understand different concepts.

Two Model Question Papers, designed as per CBSE Paper Pattern, are a unique tool to enable self-assessment for the students. Answers for previous years Board Questions have been included in this book. The journey to create a complete book is strewn with triumphs, failures and near misses. If you think we’ve nearly missed something or want to applaud us for our triumphs, we’d love to hear from you.

Please write to us on : [email protected] A book affects eternity; one can never tell where its influence stops.

Best of luck to all the aspirants! Yours faithfully,

Publisher

UNIT WISE WEIGHTAGE (Term ‐ I)

No. Units Marks

I Number Systems 11

II Algebra 23

III Geometry 17

IV Trigonometry 22

V Statistics 17

Total 90

SYLLABUS Unit I: Number Systems 1. Real Number: Euclid’s division lemma, Fundamental Theorem of Arithmetic - statements after reviewing work done

earlier and after illustrating and motivating through examples, Proofs of irrationality of 2, 3, 5 . Decimal representation of rational numbers in terms of terminating/non-terminating recurring decimals.

Unit II: Algebra 1. Polynomials: Zeroes of a polynomial. Relationship between zeroes and coefficients of quadratic polynomials.

Statement and simple problems on division algorithm for polynomials with real coefficients. 2. Pair of linear equations in two variables: Pair of linear equations in two variables and graphical method of their solution,

consistency/inconsistency. Algebraic conditions for number of solutions, Solution of a pair of linear equations in two variables

algebraically - by substitution, by elimination and by cross multiplication method. Simple situational problems. Simple problems on equations reducible to linear equations.

Unit III: Geometry 1. Triangles: Definitions, examples, counter examples of similar triangles. i. (Prove) If a line is drawn parallel to one side of a triangle to intersect the other two sides in

distinct points, the other two sides are divided in the same ratio. ii. (Motivate) If a line divides two sides of a triangle in the same ratio, the line is parallel to the third

side. iii. (Motivate) If in two triangles, the corresponding angles are equal, their corresponding sides are

proportional and the triangles are similar. iv. (Motivate) If the corresponding sides of two triangles are proportional, their corresponding

angles are equal and the two triangles are similar. v. (Motivate) If one angle of a triangle is equal to one angle of another triangle and the sides

including these angles are proportional, the two triangles are similar. vi. (Motivate) If a perpendicular is drawn from the vertex of the right angle of a right triangle to the

hypotenuse, the triangles on each side of the perpendicular are similar to the whole triangle and to each other.

vii. (Prove) The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

viii. (Prove) In a right triangle, the square on the hypotenuse is equal to the sum of the squares of the other two sides.

ix. (Prove) In a triangle, if the square on one side is equal to sum of the squares of the other two sides, the angles opposite to the first side is a right angle.

Unit IV: Trigonometry 1. Introduction to Trigonometry Trigonometric ratios of an acute angle of a right-angled triangle. Proof of their existence (well defined);

motivate the ratios whichever are defined at 0 and 90. Values (with proofs) of the trigonometric ratios of 30, 45 and 60. Relationships between the ratios.

2. Trigonometric Identities Proof and applications of the identity sin2A + cos2 A = 1. Only simple identities to be given.

Trigonometric ratios of complementary angles. Unit V: Statistics 1. Statistics: Mean, median and mode of grouped data (bimodal situation to be avoided). Cumulative frequency

graph.

QUESTION PAPER

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Contents

NCERT Textbook

Chapter No. Chapter Name Page No.

01 Real Numbers 1

02 Polynomials 24

03 Pair of Linear Equations in Two Variables 62

06 Triangles 136

08 Introduction to Trigonometry 211

14 Statistics 275

Model Question Paper – I 386

Model Question Paper – II 389

Note: * marked questions are not from examination point of view.

1

Chapter 01: Real Numbers

Euclid’s Division Lemma Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r ; 0 r < b where ‘a’ is dividend, ‘b’ is divisor, ‘q’ is quotient and r is the remainder. Note: q and r can also be zero. Examples: Consider the following pair of integers: i. 29, 8 Here, a = 29 and b = 8 By using Euclid’s Division lemma, a = bq + r ; 0 r < b i.e., 29 = 8 3 + 5 ; 0 5 < 8 ii. 77, 7 Here, a = 77 and b = 7 By using Euclid’s Division lemma, a = bq + r ; 0 r < b i.e., 77 = 7 11 + 0 ; 0 0 < 7 iii. 9, 12 Here, a = 9 and b = 12 By using Euclid’s Division lemma, a = bq + r ; 0 r < b i.e., 9 = 12 0 + 9 ; 0 9 < 12 Euclid’s Division Algorithm: Euclid’s division algorithm is a technique to compute the Highest Common Factor (HCF) of two given positive integers. Euclid’s Division Algorithm to find HCF of two positive integers ‘a’ and ‘b’ (a > b): Step I: By Euclid’s division lemma, find whole

numbers ‘q’ and ‘r’ where a = bq + r ; 0 r < b Step II: If r = 0, the HCF is b. If r 0, apply the

division lemma to b and r.

Step III: Continue the process till the remainder is zero. When the remainder is zero the divisor at that stage is the required HCF.

For the above algorithm HCF(a, b) = HCF(b, r) Example: Use Euclid’s division algorithm to find the HCF of 1467 and 453. Solution: Step I:Apply Euclid’s division lemma to 1467 and 453, 1467 = 453 3 + 108 Step II: Since r 0 apply Euclid’s division lemma to 453 and 108, 453 = 108 4 + 21 Step III: Again, r 0 apply Euclid’s division lemma to 108 and 21, 108 = 21 5 + 3 Step IV: Apply Euclid’s division lemma to 21 and 3, 21 = 3 7 + 0 Since, r 0 HCF(1467, 453) = 3 3 = HCF(21, 3) = HCF(108, 21) = HCF(453, 108) = HCF(1467, 453) NCERT Exercise 1.1 1. Use Euclid’s division algorithm to find the

HCF of: i. 135 and 225 ii. 196 and 38220 iii. 867 and 255 Solution: i. Since 225 > 135, we apply the division lemma

to 225 and 135, to get 225 = 135 1 + 90

01. Real Numbers

3453 1467

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09


remainder (r)

divisor (b)….

Things to Remember

Euclid’s division algorithm can be extendedfor all integers except zero i.e., b 0.

38 29

245


remainder (r)

divisor (b)….

2

Class X: Mathematics

2

Since the remainder 90 0, we apply the division lemma to 135 and 90, to get

135 = 90 1 + 45 We consider the new divisor 90 and the new

remainder 45, and apply the division lemma to get 90 = 45 2 + 0 As the remainder is zero, we stop. Since, the divisor at this stage is 45, the HCF

of 135 and 225 is 45. ii. Since 38220 > 196, we apply the division

lemma to 38220 and 196, to get 38220 = 196 195 + 0 As the remainder is zero, we stop. Since the divisor at this stage is 196, the HCF

of 196 and 38220 is 196. iii. Since 867 > 255, we apply the division lemma

to 867 and 255, to get 867 = 255 3 + 102 Since the remainder 102 0, we apply the

division lemma to 255 and 102, to get 255 = 102 2 + 51 We consider the new divisor 102 and the new

remainder 51, and apply the division lemma to get

102 = 51 2 + 0 As the remainder is zero, we stop. Since the divisor at this stage is 51, the HCF

of 867 and 255 is 51. 2. Show that any positive odd integer is of the

form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer. [CBSE 2014]

Solution: Let a be any positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r for

some integer q 0 and r = 0,1,2,3,4,5 because 0 r 6.

a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5

Here, a cannot be 6q or 6q + 2 or 6q + 4, as they are divisible by 2.

6q + 1 6 is divisible by 2 but 1 is not divisible by 2. 6q + 3 6 is divisible by 2 but 3 is not divisible by 2. 6q + 5 6 is divisible by 2 but 5 is not divisible by 2. Since, 6q + 1, 6q + 3, 6q + 5 are not divisible

by 2, they are odd numbers. Therefore, any odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5. 3. An army contingent of 616 members is to

march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Solution: HCF(616, 32) will give the maximum number

of columns in which they can march. Let us use Euclid’s algorithm, to find the

HCF. 616 = 32 19 + 8 32 = 8 4 + 0 the HCF of 616 and 32 is 8. the maximum number of columns in which

they can march is 8. 4. Use Euclid’s division lemma to show that

the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

[Hint: Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]

[CBSE 2015] Solution: Let x be any positive integer and b = 3. Then, by Euclid’s division lemma, x = 3q + r

for some integer q 0 and r = 0,1,2 because 0 r < 3 x = 3q or 3q + 1 or 3q + 2 When x = 3q, (x)2 = (3q)2 = 9q2 = 3(3q2) = 3m, where m is a integer When x = 3q + 1 (x)2 = (3q + 1)2 = 9q2 + 6q + 1 = 3(3q2 + 2q) + 1 = 3m + 1, where m is a integer When x = 3q + 2, (x)2 = (3q + 2)2 = 9q2 + 12q + 4 = 3(3q2 + 4q + 1) + 1 = 3m + 1, where m is a integer the square of any positive integer is either of

the form 3m or 3m + 1 for some integer m. 5. Use Euclid’s division lemma to show that

the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.

Solution: Let a be any positive integer and b = 3. Then, by Euclid’s division lemma, a = 3q + r

for some integer q 0 and r = 0,1,2 because 0 r < 3 a = 3q or 3q + 1 or 3q + 2 When a = 3q, a3 = (3q)3 = 27q3 = 9(3q3) = 9 m, where m is a integer When a = 3q + 1, a3 = (3q + 1)3 = 27q3 + 27q2 + 9q + 1 = 9(3q3 + 3q2 + q) + 1 = 9m + 1, where m is a integer

3


When a = 3q + 2, a3 = (3q + 2)3 = 27q3 + 54q2 + 36q + 8 = 9(3q3 + 6q2 + 4q) + 8 = 9m + 8, where m is a integer the cube of any positive integer is of the form

9m or 9m + 1 or 9m + 8. Problems based on Exercise 1.1 1. Using Euclid’s division algorithm, find the

HCF of 240 and 228. [CBSE 2012] Solution: By Euclid’s division algorithm, 240 = 228 1 + 12 228 = 12 19 + 0 HCF (240, 228) = 12 2. Find the HCF by Euclid’s division

algorithm of the numbers 92690, 7378 and 7161. [CBSE 2013]

Solution: By Euclid’s division algorithm, 92690 = 7378 12 + 4154 7378 = 4154 1+ 3224 4154 = 3224 1 + 930 3224 = 930 3 + 434 930 = 434 2 + 62 434 = 62 7 + 0 HCF (92690, 7378) = 62 7161 = 62 115 + 31 62 = 31 2 + 0 HCF(7161, 62) = 31 HCF (92690, 7378, 7161) = 31 3. Using Euclid’s division algorithm, find

whether the pair of numbers 231, 396 are coprime or not.

Solution: By Euclid’s division algorithm, 396 = 231 1 + 165 231 = 165 1 + 66 165 = 66 2 + 33 66 = 33 2 + 0 HCF(231, 396) = 33 the numbers are not coprime. 4. Two tankers contain 850 litres and

680 litres of petrol. Find the maximum capacity of a container which can measure the petrol of each tanker in exact number of times. [CBSE 2012]

Solution: HCF(850, 680) will give the maximum

capacity of container. 850 = 680 1 + 170 680 = 170 4 + 0 HCF (850, 680) = 170

the maximum capacity of a container which can measure the petrol of each tanker in exact number of times is 170 litres.

5. A sweetseller has 420 kaju barfis and 130 badam barfis. She wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray. What is the maximum number of barfis that can be placed in each stack for this purpose?

Solution: HCF(420, 130) will give the maximum

number of barfis that can be placed in each stack.

By Euclid’s division algorithm, 420 = 130 3 + 30 130 = 30 4 + 10 30 = 10 3 + 0 HCF(420, 130) = 10 the sweetseller can make stacks of 10 for both

kinds of barfi. 6. The length, breadth and height of a room

are 8m 25 cm, 6m 75 cm and 4 m 50 cm respectively. Find the length of the longest rod that can measure the three dimensions of the room exactly. [CBSE 2012]

Solution: Since, 1m = 100 cm 8 m 25 cm = 825 cm 6 m 75 cm = 675 cm 4 m 50 cm = 450 cm HCF(825, 675, 450) will give the length of the

longest rod. 825 = 675 1 + 150 675 = 150 4 + 75 150 = 75 2 + 0 HCF(825, 675) = 75 450 = 75 6 + 0 HCF(450, 75) = 75 HCF (825, 675, 450) = 75 the length of the longest rod is 75 cm. NCERT Exemplar 1. Write whether every positive integer can be

of the form 4q + 2, where q is an integer. Justify your answer.

Solution: No, every positive integer cannot be only of

the form 4q + 2. Justification: Let a be any positive integer. Then by Euclid’s

division lemma, we have a = bq + r, where 0 ≤ r < b

4


4

Putting b = 4, we get a = 4q + r, where 0 ≤ r < 4 Hence, a positive integer can be of the form,

4q, 4q + 1, 4q + 2 and 4q + 3. 2. “The product of two consecutive positive

integers is divisible by 2”. Is this statement true or false? Give reasons.

Solution: True. Justification: Let a, a + 1 be two consecutive positive

integers. By Euclid’s division lemma, we have a = bq + r, where 0 ≤ r < b For b = 2 , we have a = 2q + r, where 0 ≤ r < 2 ...(i) Putting r = 0 in (i), we get a = 2q, which is divisible by 2. a + 1 = 2q + 1, which is not divisible by 2. Putting r = 1 in (i), we get a = 2q + 1, which is not divisible by 2. a + 1 = 2q + 2, which is divisible by 2. Thus for 0 ≤ r < 2, one out of every two

consecutive integers is divisible by 2. The product of two consecutive positive

integers is divisible by 2. 3. “The product of three consecutive positive

integers is divisible by 6”. Is this statement true or false? Justify your answer.

Solution: True. Justification: At least one out of every three consecutive

positive integers is divisible by 2. The product of three consecutive positive

integers is divisible by 2. At least one out of every three consecutive

positive integers is divisible by 3. The product of three consecutive positive

integers is divisible by 3. Since the product of three consecutive positive

integers is divisible by 2 and 3. It is divisible by 6 also. 4. Write whether the square of any positive

integer can be of the form 3m + 2, where m is a natural number. Justify your answer.

Solution: No. Justification: Let a be any positive integer. Then by Euclid’s

division lemma, we have a = bq + r, where 0 ≤ r < b For b = 3, we have a = 3q + r, where 0 ≤ r < 3 ...(i)

The numbers are of the form 3q, 3q + 1 and 3q + 2.

(3q)2 = 9q2 = 3(3q2) = 3m, where m is a integer. (3q + 1)2 = 9q2 + 6q + 1 = 3(3q2 + 2q) + 1 = 3m + 1, where m is a integer. (3q + 2)2 = 9q2 + 12q + 4, which cannot be expressed in the

form 3m + 2. Square of any positive integer cannot be

expressed in the form 3m + 2. 5. A positive integer is of the form 3q + 1, q

being a natural number. Can you write its square in any form other than 3m + 1, i.e., 3m or 3m + 2 for some integer m? Justify your answer.

Solution: No. Justification: Consider the positive integer 3q + 1, where q

is a natural number. (3q + 1)2 = 9q2 + 6q + 1 = 3(3q2 + 2q) + 1 = 3m + 1, where m is an integer. Thus (3q + 1)2 cannot be expressed in any

other form apart from 3m + 1. 6. The numbers 525 and 3000 are both

divisible only by 3, 5, 15, 25 and 75. What is HCF (525, 3000)? Justify your answer.

Solution: HCF(525, 3000) = 75 Justification: 3, 5, 15, 25 and 75 are the only common

factors of 525 and 3000. 75 is highest among the common factors. HCF(525, 3000) = 75 7. Show that the square of any positive integer

is either of the form 4q or 4q + 1 for some integer q.

Solution: Let a be the positive integer and b = 4. Then, by Euclid’s algorithm, a = 4m + r for

some integer m ≥ 0 and r = 0, 1, 2, 3 because 0 ≤ r < 4.

a = 4m or 4m + 1 or 4m + 2 or 4m + 3. (4m)2 = 16m2 = 4(4m2) = 4q, where q is some integer. (4m + 1)2 = 16m2 + 8m + 1 = 4(4m2 + 2m) + 1 = 4q + 1, where q is some integer. (4m + 2)2 = 16m2 + 16m + 4 = 4(4m2 + 4m + 1) = 4q, where q is some integer.

5


(4m + 3)2 = 16m2 + 24m + 9 = 4(4m2 + 6m + 2) + 1 = 4q + 1, where q is some integer. The square of any positive integer is either of

the form 4q or 4q + 1, where q is some integer. 8. Show that cube of any positive integer is of

the form 4m, 4m + 1 or 4m + 3, for some integer m.

Solution: Let a be the positive integer and b = 4. Then, by Euclid’s algorithm, a = 4q + r for

some integer q ≥ 0 and r = 0, 1, 2, 3 because 0 ≤ r < 4.

a = 4q or 4q + 1 or 4q + 2 or 4q + 3. (4q)3 = 64q3 = 4(16q3) = 4m, where m is some integer. (4q + 1)3 = 64q3 + 48q2 + 12q + 1 = 4(16q3 + 12q2 + 3) + 1 = 4m + 1, where m is some integer. (4q + 2)3 = 64q3 + 96q2 + 48q + 8 = 4(16q3 + 24q2 + 12q + 2) = 4m, where m is some integer. (4q + 3)3 = 64q3 + 144q2 + 108q + 27 = 4(16q3 + 36q2 + 27q + 6) + 3 = 4m + 3, where m is some integer. The cube of any positive integer is of the form

4m, 4m + 1 or 4m + 3 for some integer m. 9. Show that the square of any positive integer

cannot be of the form 5q + 2 or 5q + 3 for any integer q.

Solution: Let a be the positive integer and b = 5. Then, by Euclid’s algorithm, a = 5m + r for

some integer m ≥ 0 and r = 0, 1, 2, 3, 4 because 0 ≤ r < 5.

a = 5m or 5m + 1 or 5m + 2 or 5m + 3 or 5m + 4.

(5m)2 = 25m2 = 5(5m2) = 5q, where q is any integer. (5m + 1)2 = 25m2 + 10m + 1 = 5(5m2 + 2m) + 1 = 5q + 1, where q is any integer. (5m + 2)2 = 25m2 + 20m + 4 = 5(5m2 + 4m) + 4 = 5q + 4, where q is any integer. (5m + 3)2 = 25m2 + 30m + 9 = 5(5m2 + 6m + 1) + 4 = 5q + 4, where q is any integer. (5m + 4)2 = 25m2 + 40m + 16 = 5(5m2 + 8m + 3) + 1 = 5q + 1, where q is any integer. The square of any positive integer is of the

form 5q, 5q + 1, 5q + 4 and cannot be of the form 5q + 2 or 5q + 3 for any integer q.

10. Show that the square of any positive integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.

Solution: Let a be the positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r for

some integer q ≥ 0 and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 5.

a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5.

(6q)2 = 36q2 = 6(6q2) = 6m, where m is any integer. (6q + 1)2 = 36q2 + 12q + 1 = 6(6q2 + 2q) + 1 = 6m + 1, where m is any integer. (6q + 2)2 = 36q2 + 24q + 4 = 6(6q2 + 4q) + 4 = 6m + 4, where m is any integer. (6q + 3)2 = 36q2 + 36q + 9 = 6(6q2 + 6q + 1) + 3 = 6m + 3, where m is any integer. (6q + 4)2 = 36q2 + 48q + 16 = 6(6q2 + 7q + 2) + 4 = 6m + 4, where m is any integer. (6q + 5)2 = 36q2 + 60q + 25 = 6(6q2 + 10q + 4) + 1 = 6m + 1, where m is any integer. The square of any positive integer is of the

form 6m, 6m + 1, 6m + 3, 6m + 4 and cannot be of the form 6m + 2 or 6m + 5 for any integer m.

11. Show that the square of any odd integer is

of the form 4q + 1, for some integer q. Solution: Let a be any odd integer and b = 4. Then, by Euclid’s algorithm, a = 4m + r for

some integer m 0 and r = 0,1,2,3 because 0 r 4.

a = 4m or 4m + 1 or 4m + 2 or 4m + 3 a = 4m + 1 or 4m + 3 Here, a cannot be 4m or 4m + 2, as they are

divisible by 2. (4m + 1)2 = 16m2 + 8m + 1 = 4(4m2 + 2m) + 1 = 4q + 1, where q is some integer. (4m + 3)2 = 16m2 + 24m + 9 = 4(4m2 + 6m + 2) + 1 = 4q + 1, where q is some integer. The square of any odd integer is of the form

4q + 1, for some integer q. 12. If n is an odd integer, then show that n2 – 1

is divisible by 8. Solution: Any odd integer n is of the form 4m + 1 or

4m + 3.

6


6

n2 – 1 = (4m + 1)2 – 1 = 16m2 + 8m = 8(2m2 + m), which is divisible by 8. Also, n2 – 1 = (4m + 3)2 – 1 = 16m2 + 24m + 8 = 8(2m2 + 3m + 1), which is divisible by 8. n2 – 1 is divisible by 8 for any odd integer n. 13. Prove that if x and y are both odd positive

integers, then x2 + y2 is even but not divisible by 4.

Solution: Since x and y are odd positive integers, we

have x = 2m + 1 and y = 2n + 1 x2 + y2 = (2m + 1)2 + (2n + 1)2 = 4m2 + 4m + 1 + 4n2 + 4n + 1 = 4(m2 + n2) + 4(m + n) + 2 x2 + y2 is an even number but not divisible by 4. 14. Use Euclid’s division algorithm to find HCF

of 441, 567, 693. Solution: By Euclid’s division algorithm, 693 = 567 1 + 126 567 = 126 4 + 63 126 = 63 2 + 0 HCF(441, 63) = 63 HCF (693, 567) = 63 441 = 63 7 + 0 HCF (693, 567, 441) = 63 15. Using Euclid’s division algorithm, find the

largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3, respectively.

Solution: Since, 1, 2 and 3 are the remainders of 1251,

9377 and 15628 respectively. 1251 – 1 = 1250 is exactly divisible by the

required number, 9377 – 2 = 9375 is exactly divisible by the

required number, 15628 – 3 = 15625 is exactly divisible by the

required number. required number = HCF of 1250, 9375 and

15625. By Euclid’s division algorithm, 15625 = 9375 1 + 6250 9375 = 6250 1 + 3125 6250 = 3125 2 + 0 HCF (15625, 9375) = 3125 3125 = 1250 2 + 625 1250 = 625 2 + 0

HCF(3125, 1250) = 625 HCF (1250, 9375, 15625) = 625 the largest number is 625. 16. Show that the cube of a positive integer of

the form 6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m+r.

Solution: 6q + r is a positive integer, where q is an

integer and r = 0, 1, 2, 3, 4, 5 Then, the positive integers are of the form 6q,

6q+1, 6q+2, 6q+3, 6q+4 and 6q+5. Taking cube of each term, we have, (6q)3 = 216 q3 = 6(36q)3 + 0 = 6m + 0, where m is an integer (6q+1)3 = 216q3 + 108q2 + 18q + 1 = 6(36q3 + 18q2 + 3q) + 1 = 6m + 1, where m is an integer (6q+2)3 = 216q3 + 216q2 + 72q + 8 = 6(36q3 + 36q2 + 12q + 1) +2 = 6m + 2, where m is an integer (6q+3)3 = 216q3 + 324q2 + 162q + 27 = 6(36q3 + 54q2 + 27q + 4) + 3 = 6m + 3, where m is an integer (6q+4)3 = 216q3 + 432q2 + 288q + 64 = 6(36q3 + 72q2 + 48q + 10) + 4 = 6m + 4, where m is an integer (6q+5)3 = 216q3 + 540q2 + 450q + 125 = 6(36q3 + 90q2 + 75q + 20) + 5 = 6m + 5, where m is an integer the cube of a positive integer of the form

6q + r, q is an integer and r = 0, 1, 2, 3, 4, 5 is also of the form 6m + r.

17. Prove that one and only one out of n, n + 2

and n + 4 is divisible by 3, where n is any positive integer.

Solution: By Euclid’s division lemma, we have a = bq + r; 0 r < b For a = n and b = 3, we have n = 3q + r, …(i) where q is an integer and 0 ≤ r < 3, i.e. r = 0, 1, 2. Putting r = 0 in (i), we get n = 3q n is divisible by 3. n + 2 = 3q + 2 n + 2 is not divisible by 3. n + 4 = 3q + 4 n + 4 is not divisible by 3. Putting r = 1 in (i), we get n = 3q + 1 n is not divisible by 3. n + 2 = 3q + 3 = 3(q + 1) n + 2 is divisible by 3. n + 4 = 3q + 5

7


n + 4 is not divisible by 3. Putting r = 2 in (i), we get n = 3q + 2 n is not divisible by 3. n + 2 = 3q + 4 n + 2 is not divisible by 3. n + 4 = 3q + 6 = 3(q + 2) n + 4 is divisible by 3 Thus for each value of r such that 0 ≤ r < 3

only one out of n, n + 2 and n + 4 is divisible by 3.

18. Prove that one of any three consecutive

positive integers must be divisible by 3. Solution: Let the three consecutive positive integers be

n, n + 1 and n + 2, where n is any integer. By Euclid’s division lemma, we have a = bq + r; 0 r < b For a = n and b = 3, we have n = 3q + r …(i) Where q is an integer and 0 ≤ r < 3, i.e. r = 0, 1, 2. Putting r = 0 in (i), we get n = 3q n is divisible by 3. n + 1 = 3q + 1 n + 1 is not divisible by 3. n + 2 = 3q + 2 n + 2 is not divisible by 3. Putting r = 1 in (i), we get n = 3q + 1 n is not divisible by 3. n + 1 = 3q + 2 n + 1 is not divisible by 3. n + 2 = 3q + 3 = 3(q + 1) n + 2 is divisible by 3. Putting r = 2 in (i), we get n = 3q + 2 n is not divisible by 3. n + 1 = 3q + 3 = 3(q + 1) n + 1 is divisible by 3. n + 2 = 3q + 4 n + 2 is not divisible by 3. Thus for each value of r such that 0 ≤ r < 3

only one out of n, n + 1 and n + 2 is divisible by 3.

19. For any positive integer n, prove that n3 – n

is divisible by 6. [CBSE 2012] Solution: n3 – n = n (n2 – 1) = n (n – 1) (n + 1) n3 – n is product of three consecutive positive

integers, where n is any positive integer. Since one out of every two consecutive

integers is divisible by 2.

The product n3 – n is divisible by 2. Since one out of every three consecutive

integers is divisible by 3. The product n3 – n is divisible by 3. Any number which is divisible by 2 and 3 is

also divisible by 6. The product n3 – n is divisible by 6. 20. Show that one and only one out of n, n + 4,

n + 8, n + 12 and n + 16 is divisible by 5, where n is any positive integer.

Solution: By Euclid’s division lemma, we have a = bq + r; 0 r < b For a = n and b = 5, we have n = 5q + r …(i) Where q is an integer and 0 ≤ r < 5, i.e. r = 0, 1, 2, 3, 4. Putting r = 0 in (i), we get n = 5q n is divisible by 5. n + 4 = 5q + 4 n + 4 is not divisible by 5. n + 8 = 5q + 8 n + 8 is not divisible by 5. n + 12 = 5q + 12 n + 12 is not divisible by 5. n + 16 = 5q + 16 n + 16 is not divisible by 5. Putting r = 1 in (i), we get n = 5q + 1 n is not divisible by 5. n + 4 = 5q + 5 = 5(q + 1) n + 4 is divisible by 5. n + 8 = 5q + 9 n + 8 is not divisible by 5. n + 12 = 5q + 13 n + 12 is not divisible by 5. n + 16 = 5q + 17 n + 16 is not divisible by 5. Putting r = 2 in (i), we get n = 5q + 2 n is not divisible by 5. n + 4 = 5q + 9 n + 4 is not divisible by 5. n + 8 = 5q + 10 = 5(q + 2) n + 8 is divisible by 5. n + 12 = 5q + 14 n + 12 is not divisible by 5. n + 16 = 5q + 18 n + 16 is not divisible by 5. Putting r = 3 in (i), we get n = 5q + 3 n is not divisible by 5. n + 4 = 5q + 7 n + 4 is not divisible by 5.

8


8

n + 8 = 5q + 11 n + 8 is not divisible by 5. n + 12 = 5q + 15 = 5(q + 3) n + 12 is divisible by 5. n + 16 = 5q + 19 n + 16 is not divisible by 5. Putting r = 4 in (i), we get n = 5q + 4 n is not divisible by 5. n + 4 = 5q + 8 n + 4 is not divisible by 5. n + 8 = 5q + 12 n + 8 is not divisible by 5. n + 12 = 5q + 16 n + 12 is not divisible by 5. n + 16 = 5q + 20 = 5(q + 4) n + 16 is divisible by 5. Thus for each value of r such that 0 ≤ r < 5

only one out of n, n + 4, n + 8, n + 12 and n + 16 is divisible by 5.

Practice problems based on Exercise 1.1 1. Find the HCF of 1656 and 4025 by Euclid’s

division algorithm. [CBSE 2013] 2. Use Euclid’s division algorithm to find HCF

of 65 and 175. [CBSE 2015] 3. Find the HCF of 1620, 1725 and 255 by

Euclid’s division algorithm. [CBSE 2012] 4. Using Euclid’s division algorithm, find

whether the pair of numbers 847, 2160 are coprime or not. [CBSE 2012]

5. Find the largest number that divides 2053 and

967 and leaves remainders 5 and 7 respectively.

6. Find HCF of 81 and 237 and express it as a

linear combination of 81 and 237 i.e., HCF of 81, 237 = 81x + 237y for some x and y.

[CBSE 2012] 7. Show that the square of an odd positive

integer is of the form 8m + 1, for some whole number m.

8. Show that the square of any positive integer is

of the form 4m or 4m + 1, where m is any integer. [CBSE 2012]

9. Show that any positive odd integer is of the

form 4q + 1 or 4q + 3, where q is some integer. [CBSE 2012]

10. Prove that n2 n is divisible by 2 for every

positive integer n. [CBSE 2012]

11. An army contingent of 104 members is to march behind an army band of 96 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march? [CBSE 2011]

12. 144 cartons of Coke cans and 90 cartons of Pepsi cans are to be stacked in a canteen. If each stack is of the same height and if it contain cartons of the same drink, what would be the greatest number of cartons each stack would have? [CBSE 2011]

13. Pens are sold in pack of 8 and notepads are sold in pack of 12. Find the least number of pack of each type that one should buy so that there are equal number of pen and notepads.

[CBSE 2014] Answers 1. 23 2. 5 3. 5 4. Yes 5. 64 6. 3 11. 8 12. 18 13. 2, 3 Multiple Choice Questions 1. Euclid’s division lemma states that for two

positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy [NCERT Exemplar]

(A) 1 < r < b (B) 0 < r b (C) 0 r < b (D) 0 < r < b 2. For some integer m, every even integer is of

the form [NCERT Exemplar] (A) m (B) m + 1 (C) 2m (D) 2m + 1 3. For some integer q, every odd integer is of the

form [NCERT Exemplar] (A) q (B) q + 1 (C) 2q (D) 2q + 1 4. n2 1 is divisible by 8, if n is

[NCERT Exemplar] (A) an integer (B) a natural number (C) an odd integer (D) an even integer 5. The largest number which divides 70 and 125,

leaving remainders 5 and 8, respectively, is [NCERT Exemplar]

(A) 13 (B) 65 (C) 875 (D) 1750 6. For any positive integer a and 3, there exist

unique integers q and r such that a = 3q + r, where r must satisfy [CBSE 2012]

(A) 0 r < 3 (B) 1 < r < 3 (C) 0 < r < 3 (D) 0 < r 3

9


The Fundamental Theorem of Arithmetic Every composite number can be expressed (factorised) as a product of primes and this factorisation is unique, apart from the order in which the prime factors occur. This fundamental theorem of arithmetic help us to find HCF and LCM of the numbers. This method is also called the prime factorisation method. Example: Consider a composite number 2352. Prime factorization of 2352 is 2352 = 2 2 2 2 3 7 7 = 24 3 72 HCF and LCM: HCF = Product of the smallest power of each

common prime factor in the numbers. LCM = Product of the greatest power of each prime

factor in the numbers. Relation between HCF and LCM of any two positive integers: For any two positive integers a and b, HCF(a, b) LCM(a, b) = a b Example: Find HCF and LCM of the following pairs of integers. i. 60 and 72 ii. 12, 30 and 144 i. The prime factorization of 60 and 72 gives: 60 = 2 2 3 5 = 22 3 5 72 = 2 2 2 3 3 = 23 32 HCF = 22 3 = 12 and LCM = 23 32 51 = 360

OR LCM = 60 72

HCF

=

60 72

12

LCM = 360 ii. The prime factorization of 12, 30 and 144

gives: 12 = 2 2 3 = 22 3

30 = 2 3 5 144 = 2 2 2 2 3 3 = 24 32 HCF = 2 3 = 6 and LCM = 24 32 5 = 720 NCERT Exercise 1.2 1. Express each number as a product of its

prime factors: i. 140 ii. 156 iii. 3825 iv. 5005 v. 7429 Solution: i. 140 = 2 2 5 7 = 22 5 7 ii. 156 = 2 2 3 13 = 22 3 13 iii. 3825 = 3 3 5 5 17 = 32 52 17

140

70

35

7

2

2

5

156

39

13

2

2

3

78

3825

85

3

3

1275

425

17 5

5

2

49

2352

1176

2 588

2 294

2 147

3

7 7

Things to Remember

HCF of two numbers is always a factor oftheir LCM.

LCM is always a multiple of HCF. HCF (p, q, r) LCM(p, q, r) p q r,

where p, q, r are positive integers. However,the following results hold good for threenumbers p, q and r:

LCM(p, q, r) = p.q.r.HCF(p,q,r)

HCF(p,q).HCF(q,r).HCF(p,r)

HCF(p, q, r) = p.q.r.LCM(p,q,r)

LCM(p,q).LCM(q,r).LCM(p,r)

10


10

iv. 5005 = 5 7 11 13 v. 7429 = 17 19 23 2. Find the LCM and HCF of the following

pairs of integers and verify that LCM HCF = product of the two numbers.

i. 26 and 91 ii. 510 and 92 [CBSE 2011] iii. 336 and 54 Solution: i. 26 = 2 13 91 = 7 13 LCM(26, 91) = 2 7 13 = 182 HCF (26, 91) = 13 Verification: LCM HCF = 182 13 = 2366 Product of two numbers = 26 91 = 2366 LCM HCF = Product of two numbers ii. 510 = 2 3 5 17 92 = 2 2 23 = 22 23 LCM(510, 92) = 22 3 5 17 23 = 23460 HCF(510, 92) = 2 Verification: LCM HCF = 23460 2 = 46920 Product of two numbers = 510 92 = 46920 LCM HCF = Product of two numbers iii. 336 = 2 2 2 2 3 7 = 24 3 7 54 = 2 3 3 3 = 2 33 LCM(336, 54) = 24 33 7 = 3024 HCF(336, 54) = 2 3 = 6 Verification: LCM HCF = 3024 6 = 18144 Product of two numbers = 336 54 = 18144 LCM HCF = Product of two numbers 3. Find the LCM and HCF of the following

integers by applying the prime factorization method.

i. 12, 15 and 21 ii. 17, 23 and 29 iii. 8, 9 and 25

Solution: i. 12 = 2 2 3 = 22 3 15 = 3 5 21 = 3 7 LCM(12, 15, 21) = 22 3 5 7 = 420 HCF(12, 15, 21) = 3 ii. 17 = 1 17 23 = 1 23 29 = 1 29 LCM(17, 23, 29) = 1 17 29 23 = 11339 HCF(17, 23, 29) = 1 iii. 8 = 2 2 2 = 23 9 = 3 3 = 32 25 = 5 5 = 52 LCM(8, 9, 25) = 23 32 52 = 1800 HCF(8, 9, 25) = 1 4. Given that HCF (306, 657) = 9, find LCM

(306, 657). Solution: LCM HCF = Product of two numbers LCM 9 = 306 657

LCM = 306 657

9

= 22338

5. Check whether 6n can end with the digit 0

for any natural number n. Solution: If the number 6n, for any natural number n, ends

with the digit zero, then it is divisible by 5. That is, the prime factorization of 6n contains the prime 5. This is not possible because

prime factorisation of 6n = (2 3)n = 2n 3n; so the only primes in the factorisation of 6n are 2 and 3 and the uniqueness of the fundamental theorem of arithmetic guarantees that there are no other primes in the factorization of 6n.

So, there is no natural number n for which 6n ends with the digit zero.

6. Explain why 7 11 13 + 13 and 7 6 5 4 3 2 1 + 5 are composite

numbers. Solution: 7 11 13 + 13 = (7 11 + 1) 13 = (77 + 1) 13 = 78 13 = (2 3 13) 13 = 2 3 132 Since 7 11 13 + 13 can be expressed as a

product of primes, it is a composite number. 7 6 5 4 3 2 1 + 5 = (7 6 4 3 2 1 + 1) 5 = (1008 + 1) 5 = 1009 5 = 5 1009 Since 7 6 5 4 3 2 1 + 5 can be

expressed as a product of primes, it is a composite number.

5005

13

5

7 143

11

1001

7429

437

2319

17

11


7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

Solution: Sonia and Ravi start at the same time from the

same point and go in the same direction and at the same time, so to find the time when they will meet again at the starting point, we have to find the LCM of 12 and 18.

12 = 2 2 3 = 22 3 18 = 2 3 3 = 2 32 LCM(12, 18) = 22 32 = 36 Sonia and Ravi will meet again at the starting

point after 36 minutes. Problems based on Exercise 1.2 1. Given that HCF (306, 1314) = 18. Find

LCM (306, 1314). [CBSE 2013] Solution: LCM HCF = Product of two numbers LCM 18 = 306 1314

LCM = 306 1314

18

= 22338

2. Find the HCF and LCM of 90 and 144 by

the method of prime factorization. [CBSE 2012]

Solution: 90 = 2 3 3 5 = 2 32 5 144 = 2 2 2 2 3 3 = 24 32 HCF (90, 144) = 2 32 = 18 LCM (90, 144) = 24 32 5 = 720 3. Find the LCM of 96 and 360 by using

fundamental theorem of arithmetic. [CBSE 2012]

Solution: 96 = 2 2 2 2 2 3 = 25 3 360 = 2 2 2 3 3 5 = 23 32 5 LCM (96, 360) = 25 32 5 = 1440 4. Determine the values of p and q so that the

prime factorization of 2520 is expressible as 23 3p q 7. [CBSE 2014]

Solution: Since, 2520 = 2 2 2 3 3 5 7 = 23 32 5 7 p = 2 and q = 5 5. State Fundamental theorem of arithmetic.

Is it possible for the HCF and LCM of two numbers to be 18 and 378 respectively. Justify your answer. [CBSE 2014]

Solution: Every composite number can be expressed

(factorised) as a product of primes and this factorisation is unique, apart from the order in which the prime factors occur.

Yes. Justification: HCF of two numbers is always a factor of

their LCM. But here 18 is a factor of 378. Two numbers can have 18 as their HCF and

380 as their LCM. 6. Find the missing numbers a, b, c and d in

the given factor tree.

[CBSE 2012] Solution:

9009 = a 3003 a = 9009

3003= 3

1001 = b 143 b = 1001

143 = 7

143 = c d Since, 143 = 11 13 c = 11, d = 13 or c = 13, d = 11 7. Complete the following factor tree and find

the composite number x.

[CBSE 2012] Solution: x = 6762

x

161

2

3

3381

?

7

7

?

3

b

c

2

143

d

a

9009

1001

3003

18018

161

2

3

3381

7

7

6762

1127

23

2 3381 = 6762

7 161 = 1127

161 = 7 23

12


12

8. Explain why 7 13 11 + 11 and (7 6 5 4 3 2 1) + 3 are composite

numbers. [CBSE 2012] Solution: 7 13 11 + 11 = (7 13 + 1) 11 = (91 + 1) 11 = 92 11 = (2 2 23) 11 = 2 2 11 23 Since 7 13 11 + 11 can be expressed as a

product of primes, it is a composite number. (7 6 5 4 3 2 1) + 3 = (7 6 5 4 2 1 + 1) 3 = (1680 + 1) 3 = 1681 3 = (41 41) 3 = 3 41 41 Since (7 6 5 4 3 2 1) + 3 can be

expressed as a product of primes, it is a composite number.

9. Three bells toll at intervals of 9, 12, 15 minutes respectively. If they start tolling together, after what time they next toll together? [CBSE 2011]

Solution: 9 = 3 3 = 32 12 = 2 2 3 = 22 3 15 = 3 5 LCM (9, 12, 15) = 22 32 5 = 180 the bells will toll together after 180 minutes. 10. The HCF of 65 and 117 is expressible in the

form 65m 117. Find the value of m. Also find the LCM of 65 and 117 using prime factorization method. [CBSE 2011]

Solution: 117 = 65 1 + 52 65 = 52 1 + 13 65 = 13 5 + 0 HCF(65, 117) = 13 Since, HCF(65, 117) = 65m 117 13 = 65m 117 65m = 117 + 13 = 130

m = 130

65= 2

Now, 65 = 5 13 and 117 = 3 3 13 = 32 13 LCM(65, 117) = 32 5 13 = 585 11. If two positive integers x and y are

expressible in terms of primes as x = p2q3 and y = p3q, what can you say about their LCM and HCF. Is LCM a multiple of HCF? Explain. [CBSE 2014]

Solution: x = p2q3 and y = p3q HCF(x, y) = p2 q LCM(x, y) = p3 q3 LCM(x, y) = (p q2) HCF(x, y) LCM is a multiple of HCF.

NCERT Exemplar 1 Explain why 3 × 5 × 7 + 7 is a composite

number. Solution: 3 × 5 × 7 + 7 = (3 5 + 1) × 7 = (15 + 1) 7 = 16 × 7 = 7 16 Since (3 × 5 × 7 + 7) can be expressed as a

product of primes, it is a composite number. 2. Can two numbers have 18 as their HCF and

380 as their LCM? Give reasons. Solution: No. Justification: HCF of two numbers is always a factor of

their LCM. But here 18 is not a factor of 380. Two numbers cannot have 18 as their HCF

and 380 as their LCM. 3. Show that 12n cannot end with the digit 0 or

5 for any natural number n. Solution: If the number 12n, for any natural number n,

ends with the digit 0 or 5, then it is divisible by 5. That is, the prime factorization of 12n contains the prime 5. This is not possible because prime

factorisation of 12n = (22 3)n = 22n 3n; so the only primes in the factorisation of 12n are 2 and 3 and the uniqueness of the fundamental theorem of arithmetic guarantees that there are no other primes in the factorization of 12n.

So, there is no natural number n for which 12n ends with the digit zero.

4. On morning walk, three persons step off together and their steps measure 40 cm, 42 cm and 45 cm, respectively. What is the minimum distance each should walk so that each can cover the same distance in complete steps?

Solution: Since, the three persons start walking together. The minimum distance covered by each of

them in complete steps = LCM of the measures of their steps

= LCM (40, 42, 45) 40 = 2 2 2 5 = 23 5 42 = 2 3 7 45 = 3 3 5 = 32 5 LCM (40, 42, 45) = 23 32 5 7 = 2520 Each person should walk a minimum distance

of 2520 cm in complete steps.

13


Practice problems based on Exercise 1.2 1. Find the HCF of 960 and 432. 2. Find the HCF and LCM of 404 and 96 and

verify HCF LCM = Product of the two given numbers. [CBSE 2012]

3. Express 2120 as the product of its prime

factors. [CBSE 2012] 4. Find the HCF and LCM of 60, 120 and 288.

[CBSE 2012] 5. The HCF and LCM of two numbers are 9 and

360 respectively. If one number is 45, write the other number.

6. The HCF of 45 and 105 is 15. Write their

LCM. [CBSE 2010] 7. The LCM of 2 numbers is 14 times their HCF.

The sum of LCM and HCF is 600. If one number is 280, then find the other number.

[CBSE 2012] 8. Find the HCF of the numbers 520 and 468 by

prime factorization method. [CBSE 2014] 9. Find LCM of the numbers given below: m, 2m, 3m, 4m and 5m, where m is any

positive integer. [CBSE 2014] 10. Show that the number 4, when n is a natural

number cannot end with the digit zero. [CBSE 2012]

11. Show that 7n cannot end with the digit zero,

for any natural number n. [CBSE 2012] 12. Can two numbers have 15 as their HCF and

175 as their LCM? Give reasons. [CBSE 2012]

13. State Fundamental theorem of arithmetic. Is it possible that HCF and LCM of two

numbers be 24 and 540 respectively. Justify your answer. [CBSE 2015]

14. Explain why the number 7 5 3 2 + 3 is

not a prime number? [CBSE 2015] Answers 1. 48 2. HCF = 4, LCM = 9696 3. 23 5 53 4. HCF = 12, LCM = 1440 5. 72 6. 315 7. 80 8. 52 9. 120 m 12. No 13. No

Multiple Choice Questions 1. If the HCF of 65 and 117 is expressible in the

form 65m 117, then the value of m is [NCERT Exemplar]

(A) 4 (B) 2 (C) 1 (D) 3 2. If two positive integers a and b are written as a = x3y2 and b = xy3; x, y are prime numbers,

then HCF (a, b) is [NCERT Exemplar] (A) xy (B) xy2 (C) x3y3 (D) x2y2 3. If two positive integers p and q can be

expressed as p = ab2 and q = a3b; a, b being prime numbers, then LCM (p, q) is

[NCERT Exemplar] (A) ab (B) a2b2 (C) a3b2 (D) a3b3 4. The least number that is divisible by all the

numbers from 1 to 10 (both inclusive) is [NCERT Exemplar]

(A) 10 (B) 100 (C) 504 (D) 2520 5. Given that LCM (91, 26) = 182, then HCF

(91, 26) is [CBSE 2011] (A) 13 (B) 26 (C) 7 (D) 9 6. If the HCF of 55 and 99 is expressible in the

form 55m 99, then the value of m is [CBSE 2011]

(A) 4 (B) 2 (C) 1 (D) 3 7. The values of x and y in the given figure are

[CBSE 2012]

(A) x = 10; y = 14 (B) x = 21; y = 84 (C) x = 21; y = 25 (D) x = 10; y = 40 8. LCM of 23 32 and 22 33 is

[CBSE 2012] (A) 23 (B) 33 (C) 23 33 (D) 22 32 Revisiting Irrational Numbers Rational Number: A number r is called a rational number, if it can be

written in the formp

q, where p and q are integers and

q 0.

yx

4

7

3

14


14

Irrational Number: A number s is called a irrational number, if it cannot

be written in the form p

q, where p and q are integers

and q 0. Theorem 1.3: Let p be a prime number. If p divides a2, then p divides a, where a is a positive integer Example: Let prime number p be 5 and a = 15 a2 = 225

Here, p divides a2 i.e., 2a

p=

225

5= 45

p divides a i.e., a

p=

15

5= 3

Theorem 1.4: Prove that 2 is an irrational number: Proof: Let us assume, to the contrary, that 2 is a rational number. So, we can find coprime integers a and b(b 0) such

that 2 = a

b

Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime. 2 b = a 2b2 = a2 …(i) [Squaring both the sides] a2 is divisible by 2 a is divisible by 2 So, we can write a = 2c for some integer c. a2 = (2c)2 …[Squaring both the sides] 2b2 = 4c2 …[From (i)] b2 = 2c2 b2 is divisible by 2 b is divisible by 2 2 divides both a and b. a and b have at least 2 as a common factor. But this contradicts the fact that a and b are coprime. This contradiction arises because of our incorrect assumption that 2 is rational. 2 is an irrational number.

NCERT Exercise 1.3 1. Prove that 5 is irrational. Solution:

Let us assume, to the contrary, that 5 is rational.

So, we can find coprime integers a and b(b 0)

such that 5 = a

b

5 b = a 5b2 = a2 ….(i) [Squaring both the sides] a2 is divisible by 5 a is divisible by 5 So, we can write a = 5c for some integer c. a2 = 25c2 ….[Squaring both the sides] 5b2 = 25c2 ….[From (i)] b2 = 5c2 b2 is divisible by 5 b is divisible by 5 5 divides both a and b. a and b have at least 5 as a common factor. But this contradicts the fact that a and b are

coprime. This contradiction arises because we have

assumed that 5 is rational.

5 is irrational. 2. Prove that 3 + 2 5 is irrational. Solution:

Let us assume, to the contrary, that 3 + 2 5 is rational.

So, we can find coprime integers a and

b(b 0) such that 3 + 2 5 = a

b

2 5 = a

b 3

5 = 1 a 3b

2 b

Since a and b are integers, 1 a 3b

2 b

is

rational.

5 is rational

But this contradicts the fact that 5 is irrational.

3 + 2 5 is irrational. 3. Prove that the following are irrationals:

i. 1

2 ii. 7 5

iii. 6 + 2

Things to Remember

Any two numbers whose highest commonfactor is 1 are said to be co-prime numbers.

Eg. i. 15, 22 ii. 11, 39 The sum or difference of a rational and an

irrational number is irrational. The product and quotient of a non-zero

rational and irrational number is irrational.

15


Solution:

i. Let us assume, to the contrary, that 1

2 is

rational. So, we can find coprime integers a and b(b 0)

such that 1

2=

a

b

2 = b

a

2 a = b 2a2 = b2 ….(i) [Squaring both the sides] b2 is divisible by 2 b is divisible by 2 So, we can write b = 2c for some integer c. b2 = 4c2 ….[Squaring both the sides] 2a2 = 4c2 ….[From (i)] a2 = 2c2 a2 is divisible by 2 a is divisible by 2 2 divides both a and b. a and b have at least 2 as a common factor. But this contradicts the fact that a and b are

coprime.

1

2is irrational.

ii. Let us assume, to the contrary, that 7 5 is rational.


such that 7 5 = a

b

5 = a

7b

Since 7, a and b are integers, a

7bis rational.

5 is rational.


7 5 is irrational.

iii. Let us assume, to the contrary that 6 + 2 is rational.


such that 6 + 2 = a

b

2 = a

b 6 =

a 6b

b

Since a and b are integers, a 6b

b

is rational.

2 is rational.


6 + 2 is irrational.

Problems based on Exercise 1.3 1. Show that 3 is an irrational number.

[CBSE 2015] Solution: Let us assume, to the contrary, that 3 is


such that 3 = a

b

3 b = a 3b2 = a2 ….(i) [Squaring both the sides] a2 is divisible by 3 a is divisible by 3 So, we can write a = 3c for some integer c. a2 = 9c2 ….[Squaring both the sides] 3b2 = 9c2 ….[From (i)] b2 = 3c2 b2 is divisible by 3 b is divisible by 3 3 divides both a and b. a and b have at least 3 as a common factor. But this contradicts the fact that a and b are


assumed that 3 is rational.

3 is irrational. 2. If p is a prime number, then prove that p

is irrational. [CBSE 2014] Solution: Let us assume, to the contrary, that p is


such that p = a

b

p b = a

pb2 = a2 ….(i) [Squaring both the sides] a2 is divisible by p a is divisible by p So, we can write a = pc for some integer c. a2 = p2c2 ….[Squaring both the sides] pb2 = p2c2 ….[From (i)] b2 = pc2 b2 is divisible by p b is divisible by p p divides both a and b. a and b have at least p as a common factor. But this contradicts the fact that a and b are


assumed that p is rational.

p is irrational.

16


16

3. Show that 4 5 2 is an irrational number. [CBSE 2010]

Solution: Let us assume, to the contrary that 4 – 5 2 is


such that 4 – 5 2 = a

b

5 2 = 4 – a

b

2 = 4b a

5b

Since a and b are integers, 4b a

5b

is rational.

2 is rational.


4 – 5 2 is irrational. 4. Show that 5 + 3 2 is an irrational number.

[CBSE 2012] Solution: Let us assume, to the contrary that 5 + 3 2 is


such that 5 + 3 2 = a

b

3 2 = a

b 5

2 = a 5b

3b

Since a and b are integers, a 5b

3b

is rational.

2 is rational.


5 + 3 2 is irrational. 5. Is product of a rational number and an

irrational number, a rational number? Is product of two irrational numbers a rational number or irrational number? Justify giving examples. [CBSE 2015]

Solution: The product of a rational number and an

irrational number can be a rational number or an irrational number.

Examples: i. rational number = 0 irrational number = 2

0 2 = 0, which is a rational number. ii. rational number = 2 irrational number = 2

2 2 = 2 2 , which is an irrational number. The product of two irrational numbers can be

a rational number or an irrational number. Examples:

i. 2 2 = 4 = 2, which is a rational number

ii. 2 3 = 6 , which is an irrational number NCERT Exemplar 1. Prove that 3 5 is irrational. Solution: Let us suppose that 3 + 5 is rational.

Let 3 + 5 = a, where a is rational.

3 = a 5 Squaring on both sides, we get

2

3 = 2

a 5

3 = a2 + 5 – 2a 5

2a 5 = a2 + 2

5 = 2a 2

2a

, which is a contradiction as the

right hand side is rational number, while 5 is irrational.

3 + 5 is irrational. 2. Prove that p q is irrational, where p, q

are primes. Solution: Let us suppose that p q is rational.

Let p q = a, where a is rational.

q = a – p

Squaring on both sides, we get q = a2 + p 2a p

p =2a p q

2a

, which is a contradiction as the

right hand side is rational number, while p is

irrational. p q is irrational. Practice problems based on Exercise 1.3 1. Prove that 2 is irrational. [CBSE 2011]

2. Show that 3 2 is irrational.

3. Prove that 2 3

5 is an irrational number.

4. Prove that 13 5

7 is an irrational number.

[CBSE 2014]

5. Prove that 3 + 5 is an irrational number. [CBSE 2011]

17


6. Prove that 3 2 is an irrational number.

7. Show that 2 3 1 is an irrational number. [CBSE 2010]

8. Prove that 2 3 5 is an irrational number.

9. Prove that 5 + 2 is an irrational number. [CBSE 2012]

10. Prove that 5 is an irrational number and

hence show that 2 5 is also an irrational number. [CBSE 2011]

Multiple Choice Questions 1. The product of a non zero rational and an

irrational number is [NCERT Exemplar] (A) always irrational (B) always rational (C) rational or irrational (D) one

2. 22

7 is [CBSE 2012]

(A) a rational number (B) an irrational number (C) a prime number (D) an even number 3. The reciprocal of an irrational number is

[CBSE 2012] (A) an integer (B) a rational (C) a natural number (D) an irrational 4. The product of two irrational numbers is

[CBSE 2012] (A) always a rational (B) always an irrational (C) one (D) always a non-zero number Revisiting Rational Numbers and Their Decimal Expansions Rational number: A number whose decimal expansion is terminating or non-terminating recurring is rational. Irrational number: A number whose decimal expansion is non-terminating non-recurring is irrational. Theorem 1.5: Let x be a rational number whose decimal expansion

terminates. Then x can be expressed in the form p

q,

where p and q are co-prime and the prime factorization of q is of the form 2n5m, where n, m are non-negative integers.

Example: Consider rational number x = 0.0945

x = 945

10000=

4

945

10

= 3

4

3 5 7

(5 2)

= 3

4 4

3 5 7

5 2

x = 3

4 3

3 7

2 5

Here x is expressed in p

q form, where p and q are

co-prime numbers. Also prime factorization of q is of the form 2n5m. i.e., 2453 Theorem 1.6:

Let x = p

q be a rational number, such that the prime

factorization of q is of the form 2n5m, where n, m are non-negative integers. Then x has a decimal expansion which terminates. Example:

Consider a rational number 648

625

x = p

q=

648

625=

3 4

4

2 3

5

=

3 4 4

4 4

2 3 2

5 2

=4

10368

10= 1.0368

Theorem 1.7:

Let x = p

q be a rational number, such that the prime

factorisation of q is not of the form 2m5n, where m, n are non-negative integers. Then, x has a decimal expansion which is non-terminating repeating (recurring). Example:

Consider a rational number 133

11= 12.0909…

Here, q = 11, which is not of the form 2m5n. NCERT Exercise 1.4 1. Without actually performing the long

division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

i. 13

3125 ii.

17

8

iii. 64

455 iv.

15

1600

v. 29

343 vi.

3 2

23

2 5

18


18

vii. 2 7 5

129

2 5 7 viii.

6

15

ix. 35

50 x.

77

210

Solution:

i. 13

3125

3125 = 5 5 5 5 5 = 55 = 20 55

Since the denominator is of the form 2m 5n, the rational number has a terminating decimal expansion.

ii. 17

8

8 = 2 2 2 = 23 = 23 50 Since the denominator is of the form 2m 5n

the rational number has a terminating decimal expansion.

iii. 64

455

455 = 5 7 13 Since the denominator is not of the form

2m 5n, the rational number has a non-terminating repeating decimal expansion.

iv. 15

1600 =

5 3

5 320

= 3

320

320 = 2 2 2 2 2 2 5 = 26 51 Since the denominator is of the form 2m 5n,

the rational number has a terminating decimal expansion.

v. 29

343

343 = 7 7 7 = 73 Since the denominator is not of the form


vi. 3 2

23

2 5


vii. 2 7 5

129

2 5 7

Since the denominator is not of the form 2m 5n, the rational number has a non-terminating repeating decimal expansion.

viii. 6

15=

3 2

3 5

= 2

5

5 = 20 51


ix. 35

50=

5 7

5 10

= 7

10

10 = 2 5


x. 77

210=

7 11

7 30

= 11

30

30 = 2 3 5 Since the denominator is not of the form


2. Write down the decimal expansions of those

rational numbers in Question 1 above which have terminating decimal expansions.

Solution:

i. 13

3125=

5

13

5=

5

5 5

13 2

5 2

=5

13 32

(5 2)

= 5

416

10= 0.00416

ii. 17

8=

3

17

2=

3

3 3

17 5

2 5

= 3

17 125

(2 5)

= 3

2125

10 = 2.125

iv. 15

1600 =

4 2

15

2 10

= 4

4 4 2

15 5

2 5 10

= 4 2

15 625

(2 5) 10

= 4 2

9375

10 10 =

6

9375

10 = 0.009375

vi. 3 2

23

2 5 =

3 2

3 2 3 2

23 5 2

2 5 5 2

= 5 5

23 125 4

2 5

=5

11500

(2 5) =

5

11500

10= 0.115

viii. 6

15=

2

5=

2 2

5 2

= 4

10 = 0.4

ix. 35

50=

7

10 = 0.7

3. The following real numbers have decimal

expansions as given below. In each case, decide whether they are rational or not. If

they are rational, and of the form p

q, what

can you say about the prime factors of q? i. 43.123456789 ii. 0.120120012000120000…

iii. 43.123456789 Solution: i. Since the given number has a terminating

decimal expansion, it is a rational number of

the form p

qand q is of the form 2m 5n.

The prime factors of p and q will be either 2 or 5 or both.

19


ii. Since the decimal expansion is neither terminating nor recurring, the given number is an irrational number.

iii. Since the decimal expansion is non-terminating repeating, the given number is a

rational number of the form p

qand q is not of

the form 2m 5n. The prime factors of q will also have a factor

other than 2 or 5. Problems based on Exercise 1.4

1. Write the decimal expansion of 27

1250

without actual division. [CBSE 2015] Solution:

27

1250 =

3

27

5 10=

3

3 3

2 27

2 5 10

= 3

8 27

(2 5) 10

= 3

216

10 10=

4

216

10

= 0.0216

2. Express 15 5

4 40

as a decimal without

actual division. [CBSE 2011] Solution:

15 5

4 40 =

15 10

4 10

+5

40

= 150 5

40 40

= 155

40=

3

155

2 5

= 2

3 2

155 5

2 5 5

= 3 3

155 25

2 5

= 3

3875

(2 5)=

3

3875

10

= 3.875 3. Write the denominator of the rational

number 257

500in the form 2m 5n,

where m and n are non-negative integers. Hence write its decimal expansion without actual division. [CBSE 2012]

Solution: Denominator = 500 = 22 53

257

500=

2 3

257

2 5=

2 3

2 257

2 2 5

= 3 3

514

2 5=

3

514

(2 5)=

3

514

10= 0.514

4. If 241

4000=

m n

241

2 5, find the values of m and n

where m and n are non-negative integers. Hence, write its decimal expansion without actual division. [CBSE 2012]

Solution:

241

4000=

m n

241

2 5

5 3

241

2 5=

m n

241

2 5

m = 5 and n = 3

Now, 241

4000 =

5 3

241

2 5

= 2

5 3 2

241 5

2 5 5

= 5 5

241 25

2 5

= 5

6025

(2 5)=

5

6025

10

= 0.06025 5. What is the condition for the decimal

expansion of a rational number to terminate? Explain with the help of an example. [CBSE 2015]

Solution:

Let x = p

q be a rational number, such that the

prime factorization of q is of the form 2m5n, where n, m are non-negative integers. Then x has a decimal expansion which terminates.

Example:

2 3

49 49

500 2 5


6. Express the number 0.3178 in the form of

rational numbera

b. [CBSE 2011]

Solution:

Let x = 0.3178

x = 0.3178178178… ….(i) Multiplying both sides of (i) by 10, we get 10x = 3.178178… ….(ii) Multiplying both sides of (i) by 10000, we get 10000x = 3178.178… ….(iii) Subtracting (ii) from (iii), we get 9990x = 3175

x = 3175

9990= 635

1998

20


20

NCERT Exemplar 1. Without actually performing the long

division, find if 987

10500will have terminating

or non-terminating (repeating) decimal expansion. Give reasons for your answer.

Solution:

10500

987=

50073

4773

= 47

500=

32 52

47


2. A rational number in its decimal expansion

is 327.7081. What can you say about the prime factors of q, when this number is

expressed in the form q

p? Give reasons.

Solution:

327.7081 = 10000

3277081=

44 52

3277081

=

q

p

Here, q is of the form 2m × 5n, where m and n are natural numbers.

The prime factors of p and q will be either 2 or 5 or both.

3. Write the denominator of the rational

number 257

5000 in the form 2m 5n, where m,

n are non-negative integers. Hence, write its decimal expansion, without actual division.

Solution: Denominator = 5000 = 23 54

257

5000=

3 4

257

2 5=

3 4

2 257

2 2 5

= 4 4

514

2 5=

4

514

(2 5)=

4

514

10= 0.0514

Practice problems based on Exercise 1.4 1. Without actually performing the long division,

state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

i. 7

80 ii.

27

40

iii. 2 3 2

49

2 5 7 iv.

7294

625

v. 175

4000 vi.

22

7

vii. 13

875

2. Write down the decimal expansions of the following rational numbers:

i. 189

125 ii.

3 2

75

2 5

ii. 133

5000

3. Write down the decimal expansion of

16

3125without actual division. [CBSE 2012]

Answers 1. i. terminating ii. terminating iii. non-terminating repeating iv. terminating v. terminating vi. non-terminating repeating vii. non-terminating repeating 2. i. 1.512 ii. 0.375 iii. 0.0266 3. 0.00512 Multiple Choice Questions 1. The decimal expansion of the rational number

2

33

2 .5 will terminate after [NCERT Exemplar]

(A) one decimal place (B) two decimal places (C) three decimal places (D) more than 3 decimal places 2. The decimal expansion of the rational number

14587

1250 will terminate after

[NCERT Exemplar] (A) one decimal place (B) two decimal places (C) three decimal places (D) four decimal places

3. The decimal expansion of 17

8will terminate

after how many places of decimals? [CBSE 2011]

(A) 1 (B) 2 (C) 3 (D) will not terminate 4. From the following, the rational number

whose decimal expansion is terminating is [CBSE 2011]

(A) 2

15 (B)

11

160

(C) 17

60 (D)

6

35

21


5. The decimal expansion of is [CBSE 2011]

(A) terminating (B) non-terminating and non-recurring (C) non-terminating and recurring (D) doesn’t exist 6. Which of the following rational numbers have

a terminating decimal expansion? [CBSE 2011]

(A) 125

441 (B)

77

210

(C) 15

1600 (D)

2 2 2

129

2 5 7

7. The decimal representation of a rational

number p

q is a terminating decimal only if for

non-negative integers m and n, prime factors of q are of the form [CBSE 2012]

(A) 2m 3n (B) 3m 5n (C) 3n 7n (D) 2m 5n One Mark Questions 1. State Fundamental theorem of arithmetic. Solution: Every composite number can be expressed

(factorised) as a product of primes and this factorisation is unique, apart from the order in which the prime factors occur.

2. Express 960 as the product of its prime

factors. Solution: 960 = 2 2 2 2 2 2 3 5 = 26 3 5 3. Write the HCF of smallest composite

number and smallest prime number. Solution: The smallest composite number is 22 = 4 and the smallest prime number is 2. HCF (4, 2) = 2

4. Has the rational number 2 7 2

441

2 .5 .7, a

terminating or a non-terminating decimal representation?

Solution: Since the denominator is not of the form


5. Write whether 2 45 3 20

2 5

on

simplification gives a rational or an irrational number.

Solution:

2 45 3 20

2 5

=

2 9 5 3 4 5

2 5

= 6 5 6 5

2 5

=

12 5

2 5

= 6, which is a rational number

6. If p

qis a rational number (q 0), what is

the condition of q so that the decimal

representation of p

qis terminating?

Solution:

For a rational number p

q to have terminating

decimal representation, the prime factorisation of q should be of the form 2m 5n, where m and n are non-negative integers.

HOTS Questions 1. Show that there is no positive integer n, for

which n 1 n 1 is rational. [CBSE 2012]

Solution: Let us assume that there is a positive integer n

for which n 1 + n 1 is rational and equal

to A

B, where A and B are positive integers

(B 0). Then,

n 1 + n 1 = A

B ....(i)

B

A=

1

n 1 n 1

B

A=

n 1 n 1

n 1 n 1 n 1 n 1

B

A=

n 1 n 1

(n 1) (n 1)

= n 1 n 1

2

n 1 – n 1 = 2B

A ....(ii)

Adding (i) and (ii), we get

2 n 1 = A 2B

B A =

2 2A 2B

AB

n 1 = 2 2A 2B

2AB

Subtracting (ii) from (i), we get

n 1 = 2 2A 2B

2AB

22


22

Since, A and B are positive integers. n 1 and n 1 are rationals. But it is possible only when n + 1 and n 1

both are perfect squares. But they differ by 2 and any two perfect squares differ at least by 3.

n + 1 and n 1 cannot be perfect squares. Hence, there is no positive integer n for which

n 1 + n 1 is rational. Value Based Questions 1. For a morning walk, three persons steps off

together. Their steps measure 80 cm, 85 cm and 90 cm respectively. What is the minimum distance each should walk to show that they can cover the distance in complete steps? Which value is preferred in this situation? [CBSE 2013]

Solution: Since, the three persons start walking together. The minimum distance covered by each of

them in complete steps = LCM of the measures of their steps

= LCM (80, 85, 90) 80 = 2 2 2 2 5= 24 5 85 = 5 17 90 = 2 3 3 5 = 2 32 5 LCM (80, 85, 90) = 24 32 5 17 = 12240 Each person should walk a minimum distance

of 12240 cm in complete steps. Morning walk is good for healthy life. 2. A school library has 280 science journals

and 300 maths journals. Students were told to stack these journals in such a way that each stack contains equal number of journals. Determine the number of stacks of science and maths journals. What is the benefit of library in student life?

Solution: Since, each stack contains equal number of

journals. Number of journals in each stack

= HCF (280, 300) 280 = 2 2 2 5 7 = 23 5 7 300 = 2 2 5 5 3 = 22 52 3 HCF (280, 300) = 22 5 = 20 Number of stacks of science journals

= 280

20= 14

Number of stacks of maths journals

= 300

20 = 15

There are 14 stacks of science journals and 15 stacks of maths journals.

Library helps students to develop life-long learning skills.

23


Sample Test Paper

Total Marks: 20 1. Find HCF of 24 32 and 2 33 [1]

2. Express 140 as a product of its prime factors. [1] 3. State whether the following rational number has a terminating decimal expansion or a non-

terminating repeating expansion: 3

3125 [2]

4. Complete the following factor tree and find the numbers x and y. [2] 5. Find the HCF of the numbers 4052 and 12576 by Euclid’s division algorithm. [3] 6. Explain why 7 12 11 + 11 and 7 6 4 3 + 6 are composite numbers. [3] 7. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m

or 3m + 1 for some integer m. [4] 8. Show that 5 is an irrational number. [4]

Memory Map

y 5

425

85x

?

3

Euclid’s DivisionLemma

The Fundamental Theorem of Arithmetic

Relation between LCMand HCF of two numbers

Real Numbers

For any two positiveintegers a and b HCF (a, b) LCM(a, b) = a b

a = bq + r; 0 r < b where, ‘a’ is dividend,‘b’ is divisor ‘q’ isquotient and ‘r’ is theremainder.

Euclid’s Division Algorithm

HCF of two positive integers ‘a’ and ‘b’ (a > b): I: By Euclid’s division

lemma, find wholenumbers ‘q’ and ‘r’ where a = bq + r; 0 r < b

II: If r = 0, the HCF is b. If r 0, apply the divisionlemma to b and r.

III: Continue the process till the remainder is zero. When the remainder iszero the divisor at that stage is the required HCF.

HCF(a, b) = HCF(b, r)

Rational numbers and theirdecimal expansions

If rational number x has terminating

decimal expansion, then x = p

qand

q = 2m5n, where p and q are co-prime; m and n are non-negative integers.

If x = p

qand q = 2m5n

(where, p and q areco-prime; m and n are non-negative integers), then x hasterminating decimal expansion.

If x = p

qand q 2m5n

(where, p and q are co-prime; mand n are non-negative integers),then x has decimal expansionnon-terminating repeating.

Every compositenumber can beexpressed (factorised)as a product of primesand this factorisationis unique, apart fromthe order in which theprime factors occur. Example: 60 = 2 2 3 5

Documents

CBSE Class 10, Mathematics Term 1 - Book (guide) · proportional and the triangles are similar. iv. (Motivate) If the corresponding sides of two triangles are proportional, their