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MATHEMATICS [SOLUTIONS OF R.0. SHARMA]

1. REAL NUMBERS 1-40

EXERCISE 1.1 2-7

EXERCISE 1.2 7-15

EXERCISE 1.3 15-16

EXERCISE 1.4 16-20

EXERCISE 1.5 20-25

EXERCISE 1.6 25-28

VSAQS 28-31

MCQS 31-37

FBQS 37-40

2. POLYNOMIALS 38-90

EXERCISE 2.1 39-52

EXERCISE 2.2 52-55

EXERCISE 2.3 55-63

VSAQS 63-68

MCQS & FBQS 68-90

3. PAIR OF LINEAR EQUATIONS IN TWO VARIABLES 82-218

EXERCISE 3.1 83-88

EXERCISE 3.2 88-123

EXERCISE 3.3 123-144







EXERCISE 3.10 191-201

EXERCISE 3.11 202-210

VSAQS 211-213

MCQS & FBQS 213-218

4. QUADRATIC EQUATIONS 219-306


CONTENTS

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EXERCISE 4.10 284-286

EXERCISE 4.11 286-289

EXERCISE 4.12 289-291

EXERCISE 4.13 292-296

VSAQS 296-299

MCQS & FBQS 299-306

5. ARITHMETIC PROGRESSIONS 307-384







VSAQS 371-373

MCQS & FBQS 373-384

6. CO-ORDINATE GEOMETRY 385-489






VSAQS 467-474

MCQS & FBQS 474-489

7. TRIANGLES 490-597









REVISION EXERCISE 554-574

VSAQS 575-581

MCQS & FBQS 581-597

8. CIRCLES 598-648



VSAQS 624-629

MCQS & FBQS 629-648

9. CONSTRUCTIONS 649-663




10. TRIGONOMETRIC RATIOS 664-718

EXERCISE 10.1 666-687

EXERCISE 10.2 687-697

EXERCISE 10.3 697-705

VSAQS 705-709

MCQS & FBQS 709-718

11. TRIGONOMETRIC IDENTITIES 719-769

EXERCISE 11.1 719-751

EXERCISE 11.2 751-756

VSAQS 757-762

MCQS & FBQS 762-769

12. HEIGHTS AND DISTANCES 770-834

EXERCISE 12.1 771-816

VSAQS 816-821

MCQS & FBQS 822-834

13. AREAS RELATED TO CIRCLES 835-907

EXERCISE 13.1 836-846

EXERCISE 13.2 846-854

EXERCISE 13.3 854-860


EXERCISE 13.4 860-887

VSAQS 887-891

MCQS & FBQS 892-907

14. SURFACE AREAS AND VOLUMES 908-1031

EXERCISE 14.1 910-941

EXERCISE 14.2 941-961

EXERCISE 14.3 961-975

REVISION EXERCISE 975-1008

VSAQS 1008-1015

MCQS & FBQS 1016-1031

15. STATISTICS 1032-1118

EXERCISE 15.1 1032-1039

EXERCISE 15.2 1039-1044

EXERCISE 15.3 1045-1062

EXERCISE 15.4 1062-1080

EXERCISE 15.5 1080-1095

EXERCISE 15.6 1095-1105

VSAQS 1105-1108

MCQS & FBQS 1108-1118

16. PROBABILITY 1119-1153

EXERCISE 16.1 1119-1141

EXERCISE 16.2 1141-1144

VSAQS 1144-1146

MCQS & FBQS 1146-1153


1 Arundeep Math X, C.B.S.E.

Points to Remember :

1. Real Numbers— Numbers, rational and

irrational together, are called real numbers

such as 0, 3, –4, 3

2, 3 etc. These number

can be represented on a number line also.

2. Divisibility— A non-zero integer ‘a’ is said

to divide an integer ‘b’ there exists an integer

c such that b = ac. The integer a is called

dividend, b is called divisor and c is called

quolient. a | b means b is divisible by ‘a’ and

a | b means b is not divisible by a.

3. Properties of divisibility—

(i) + divides every non-zero integer i.e. + a

1

for every non-zero integer a.

(ii) 0 is divided by every non-zero integer a

i.e. 0

a for every non zero integer a.

(iii) 0 does not divide any integer.

(iv) If a is a non-zero integer and b is any integer,

then a|b a|–b, –a|b and –a|–b.

(v) If a and b are non-zero integers, then a|b

and b|a a = +b.

4. Euclid’s Division Lemma— Let a and b be

any two positive integers, then there exists

unique integers q and r, such that

a = bq + r, o < r < b.

If b|a, then r = o, other wise r satisfies the

stronger inequality o < r < b.

Remarks—

(i) The above Lemma is nothing but a

restatement of the long division process we

have been doing all these years and that the

integers q and r are called the quotient and

remainder respectively.

(ii) The above Lemma has been stated for

Chapter — 1

REAL NUMBERS

positive integers only. But it can be extended

to all integers as stated below :

Let a and b be any two integers with b o,

then there exists unique integers q and r such

that a = bq + r where o < r < |b|.

5. H.C.F. or G.C.D.— The largest or greatest

divisor among the common divisors of two

or more integers is called the Greatest

Common Divisor (G.C.D.) or Highest

Common Factor (H.C.F.) of the given

integers.

6. Some results of Theorems—

Theorem 1. If a and b are positive integers

such that a = bq + r, then every common

divisor of a and b is a common divisor of b

and r and vice-versa.

Theorem 2. Linear combination— H.C.F.

(say a) of two positive integers a and b can

be expressed as a linear combination of a

and b i.e. d = xa + yb, for some integers x

and y.

This represculation is not unique, because

d = xa + yb

= xa + yb + ab – ab

= xa + ab + yb – ab

= (x + b) a + (y – a) b

Note— To represent the H.C.F. as a linear

combination of the given two numbers, we

start from the last but one stop and

successively diminate the previous

remainder.

7. How to find H.C.F. and L.C.M. of given

two numbers a and b

(i) Factorise each of the given positive integers

and express them as a product of powers of

prime in ascending order of magnitude of

primes.

(ii) To find the H.C.F. identify the common

prime factors and find the smallest (least)



exponent of these common factors. Now

raise these common prime factors to their

smallest exponents and multiply them to get

the H.C.F.

(iii) To find the L.C.M., list all prime factors

(once only) accuring in the prime

factorization of the given positive integers.

For each of these factors, find the greatest

exponent and raise each prime factor to the

greatest exponent and multiply them to get

the L.C.M.

Note— Product of two positive integers a

and b

= the product of their H.C.F. and L.C.M.

i.e. a × b = H.C.F. × L.C.M.

8. Rational numbers and Irrational

numbers— The numbers which can be

written in the form of q

p where p and q are

integers and q 0, are called rational

numbers. For example 6

5,

3

2,

7

1,

1

5 etc.

Those numbers which cannot be written in

q

p form, are called irrational numbers. For

example 2 , 3 , 5 etc.

is also an irrational number.

9. Terminating and non-terminating

repeating decimal expansions of a

rational number— Every rational number

can be express in decimals either terminating

or non-terminating repeating decimals.

(a) Terminating Decimals— A rational

number whose denominator can be

factorise in the form of 2m × 5n where m

and n are non-negative integers are

terminating decimals.

(b) Non-terminating repeating decimals— If

in a rational number, the denominator is not

factorise in the form of 2m × 5n, where m

and n are non-negative integers, the rational

number has non-terminating repeating

decimals.

EXERCISE 1.1

1. If a and b are two odd positive integers

such that a > b, then prove that one of

the two numbers 2

ba and

2

ba is odd

and the other is even.

Solution—

a and b are two odd numbers such that a > b

Let a = 2n + 1, then b = 2n + 3

Now 2

ba =

2

3212 nn =

2

44 n

= 2n + 2

= 2 (n + 1) which is even

and 2

ba =

2

3212 nn =

2

2 = –1

Which is odd Hence proved.

2. Prove that the product of two consecutive

positive integers is divisible by 2.

Solution—

Let n and n + 1 are two consecutive positive

integer

We know that n is of the form

n = 2q and n + 1 = 2q + 1

n (n + 1) = 2q (2q + 1) = 2 (2q2 + q)

Which is divisible by 2

If n = 2q + 1, then

n (n + 1) = (2q + 1) (2q + 2)

= (2q + 1) × 2 (q + 1)

= 2 (2q + 1) (q + 1)

Which is also divisible by 2

Hence the product of two consecutive

positive integers is divisible by 2

3. Prove that the product of three

consecutive positive integer is divisible

by 6.

Solution—

Let n be the positive any integer

Then n (n + 1) (n + 2) = (n2 + n) (n + 2)



= n3 + 2n2 + n2 + 2n

= n3 + 3n2 + 2n

Now if n = 6q, then

n3 + 3n2 + 2n = (6q)3 + 3(6q)2 + 2 (6q)

= 216q3 + 108q2 + 12q

= 6q (36q2 + 18q + 2q)


If n = 6q + 1, then

n3 + 3n2 + 2n = (6q + 1)3 + 3 (6q + 1)2 + 2

(6q + 1)

= 216q3 + 108q2 + 18q + 1 + 3 (36q2 + 12q

+ 1) + 2 (6q + 1)

= 216q3 + 108q2 + 18q + 1 + 108q2 + 36q +

3 + 12q + 2

= 216q3 + 216q2 + 66q + 6

= 6 (36q2 + 36q2 + 11q + 1)


If n = 6q + 2, then

n3 + 3n2 + 2n = (6q + 2)3 + 3 (6q + 2)2 + 2

(6q + 2)

= 216q3 + 216q2 + 72q + 8 + 3 (36q2 + 24q

+ 4) + 2 (6q + 2)

= 216q3 + 216q2 + 72q + 8 + 108q2 + 72q +

12 + 12q + 4

= 216q3 + 324q2 + 156q + 24

= 6 (36q3 + 54q2 + 26q + 4)


Hence the product of three consecutive

positive integers is divisible by 6

4. For any positive integer n, prove that

n3 – n is divisible by 6.

Solution—

Let n = 6q or 6q + 1, 6q + 2, 6q + 3 ... 6q + 5

If n = 6q, then

Then n3 – n = (6q)3 – 6q = 216q3 – 6q

= 6 (36q3 – q)


If n = 6q + 1, then

n3 – n = (6q + 1)3 – (6q + 1)

= 216q3 + 108q2 + 18q + 1 – 6q – 1

= 216q3 + 108q2 + 12q

= 6(36q3 + 18q2 + 2q)


If n = 6q + 2, then

n3 – n = (6q + 2)3 – (6q + 2)

= 216q3 + 216q2 + 72q + 8 – 6q – 2

= 216q3 + 216q2 + 66q + 6

= 6 (36q3 + 36q2 + 11q + 1)


Hence we can similarly, prove that

n2 – n is divisible by 6 for any positive integer

n. Hence proved.

5. Prove that if a positive integer is of the

form 6q + 5, then it is of the form 3q + 2

for some integer q, but not conversely.

Solution—

Let n = 6q + 5, where q is a positive integer

We know that any positive integer is of the

form 3k or 3k + 1 or 3k + 2, 1

q = 3k or 3k + 1 or 3k + 2

If q = 3k, then

n = 6q + 5

= 6 (3k) + 5 = 18k + 5

= 18k + 3 + 2 = 3 (6k + 1) + 2

= 3m + 2, where m = 6k + 1

If q = 3k + 1, then

n = 6q + 5 = 6 (3k + 1) + 5

= 18k + 6 + 5 = 18k + 11

= 18k + 9 + 2 = 3 (6k + 3) + 2

= 3m + 2 where m = 6k + 3

If q = 3k + 2, then

n = 6q + 5 = 6 (3k + 2) + 5

= 18k + 12 + 5 = 18k + 17

= 18k + 15 + 2 = 3 (6k + 5) + 2

= 3m + 2 where m = 6k + 5

Hence proved.

6. Prove that the square of any positive

integer of the form 5q + 1 is of the same

form.

Solution—

Let a be any positive integer

Then a = 5m + 1

a2 = (5m + 1)2

= 25m2 + 10m + 1



= 5 (5m2 + 2m) + 1

= 5q + 1 where q = 5m2 + 2m

Which is of the same form as given

Hence proved.


integer is of the form 3m or, 3m + 1 but

not of the form 3m + 2.

Solution—

Let a be any positive integer

Let it be in the form of 3m or 3m + 1

Let a = 3q, then

a2 = 9q2 = 3(3q2) = 3m

Where m = 3q2

Which is odd

Let a = 3q + 1, then

a2 = (3q + 1)2 = 9q2 + 6q + 1

= 3 (3q2 + 2q) + 1

= 3m + 1 where m = 3q2 + 2q

Which is odd

If a = 3q + 2, then

a2 = (3q + 2)2 = 9q2 + 12q + 4

= 9q2 + 12q + 3 + 1

= 3 (3q2 + 4q + 1) + 1 = 3m + 1

Which is odd

Hence proved.


integer is of the form 4q or 4q + 1 for

some integer q.

Solution—

Let a be the positive integer and

Let a = 4m

a2 = (4m)2 = 16m2 = 4(4m2)

= 4q where q = 4m2

and let a = 4m + 1

a2 = (4m + 1)2

= 16m2 + 8m + 1

= 4 (4m2 + 2m) + 1

= 4q + 1 where q = 4m2 + 2m

Hence proved.


integer is of the form 5q, 5q + 1, 5q + 4

for some integer q.

Solution—

Let a be the positive integer, and

Let a = 5m, then

a2 = (5m)2 = 25m2

= 5 (5m2) = 5q

Where q = 5m2

and a = (5m + 1) then

a2 = (5m + 1)2

= 25m2 + 10m + 1

= 5 (5m2 + 2m) + 1

= 5q + 1 where q = 5m2 + 2m

and let a = 5m + 1, then

a2 = (5m + 4)2 = 25m2 + 40m + 16

= 25m2 + 40m + 15 + 1

= 5 (5m2 + 8m + 3) + 1

= 5q + 1 where q = 5m2 + 8m + 3

and a = 5m + 2, then

a2 = (5m + 2)2

= 25m2 + 20m + 4

= 5 (5m2 + 4m) + 4

= 5q + 4 where q = 5m2 + 4m

and a = 5m + 3, then

a2 = (5m + 3)2 = 25m2 + 30m + 9

= 25m2 + 30m + 5 + 4

= 5 (5m2 + 6m + 1) + 4

= 5q + 4 where q = 5m2 + 6m + 1

Hence proved.

10. Show that the square of an odd positive

integer is of the form 8q + 1, for some

integer q.

Solution—

Let n is any positive odd integer

Let n = 4p + 1, then

(4p + 1)2 = 16p2 + 8p + 1

n2 = 8p (2p + 1) + 1

= 8q + 1 where q = p (2p + 1)

Hence proved.

11. Show that any positive odd integer is of

the form 6q + 1 or 6q + 3 or 6q + 5, where

q is some integer.



Solution—

Let n be any positive odd integer

and let n = 6q + r

6q + r, b = 6, and 0 < r < 6

or r = 0, 1, 2, 3, 4, 5

If n = 6q = 2 × 3q

But it is not odd

When n = 6q + 1 which is odd

When n = 6q + 2 which is not odd

= 2 (3q + 1)

When n = 6q + 3 which is odd

When n = 6q + 4 = 2 (3q + 2) which is not

odd

When n = 6n + 5, which is odd

Hence 6q + 1 or 6q + 3 or 6q + 5 are odd

numbers.

12. Show that the square of any positive

integer cannot be of the form 6m + 2 or

6m + 5 for any integer m.

[NCERT Exemplar]

Solution—

Let a be an arbitrary positive integer, then

by Euclid’s division algorithm, corresponding

to the positive integers a and 6, there exist

non-negative integers q and r such that

a = 6q + r, where 0 < r < 6

a2 = (6q + r)2 = 36q2 + r2 + 12qr

[_ (a + b)2 = a2 + 2ab + b2]

a2 = 6(6q2 + 2qr) + r2 ...(i)

Where, 0 < r < 6

Case I : Where r = 0, then putting r = 0 in

Eq. (i), we get

a2 = 6(6q2) = 6m

Where, m = 6q2 is an integer.

Case II : When r = 1, then putting r = 1 in

Eq. (i), we get

a2 = 6(6q2 + 2q) + 1 = 6m + 1

Where, m = (6q2 + 2q) is an integer.

Case III : When r = 2, then putting r = 2 in

Eq. (i), we get

a2 = 6(6q2 + 4q) + 4 = 6m + 4

Where, m = (6q2 + 4q) is an integer.

Case IV : When r = 3, then putting r = 3 in

Eq. (i), we get

a2 = 6(6q2 + 6q) + 9

= 6(6q2 + 6q) + 6 + 3

a2 = 6(6q2 + 6q + 1) + 3 = 6m + 3

Where, m = (6q2 + 6q + 1) is an integer.

Case V : When r = 4, then putting r = 4 in

Eq. (i), we get

a2 = 6(6q2 + 8q) + 16

= 6(6q2 + 8q) + 12 + 4

a2 = 6(6q2 + 8q + 2) + 4 = 6m + 4

Where, n = (6q2 + 8q + 2) is an integer.

Case VI : When r = 5, then putting r = 5 in

Eq. (i), we get

a2 = 6(6q2 + 10q) + 25

= 6(6q2 + 10q) + 24 + 1

a2 = 6(6q2 + 10q + 4) + 1 = 6m + 1

Where, m = (6q2 + 10q + 1) is an integer.

Hence, the square of any positive integer

cannot be of the form 6m + 2 or 6m + 5 for

any integer m.

13. Show that the cube of a positive integer

is of the form 6q + r, where q is an integer

and r = 0, 1, 2, 3, 4, 5.

[NCERT Exemplar]

Solution—

Let a be an arbitrary positive integer. Then,

by Euclid’s division algorithm, corresponding

to the positive integers ‘a’ and 6, there exist

non-negative integers q and r such that

a = 6q + r, where, 0 < r < 6

a3 = (6q + r)3 = 216q3 + r3 + 3.6q.r(6q + r)

[_ (a + b)3 = a3 + b3 + 3ab(a + b)]

a3 = (216q3 + 108q2r + 18qr2) + r3 ...(i)

Where, 0 < r < 6

Case I : When r = 0, then putting r = 0 in

Eq. (i), we get

a3 = 216q3 = 6(36q3) = 6m

Where, m = 36q3 is an integer.

Case II : When r = 1, then putting r = 1 in

Eq. (i), we get

a3 = (216q3 + 108q3 + 18q) + 1 = 6(36q3 +



18q3 + 3q) + 1

a3 = 6m + 1,

Where m = (36q3 + 18q3 + 3q) is an integer.

Case III : When r = 2, then putting r = 2 in

Eq. (i), we get

a3 = (216q3 + 216q2 + 72q) + 8

a3 = (216q3 + 216q2 + 72q + 6) + 2

a3 = 6(36q3 + 36q2 + 12q + 1) + 2 = 6m + 2

Where, m = (36q3 + 36q2 + 12q + 1) is an

integer.

Case IV : When r = 3, then putting r = 3 in

Eq. (i), we get

a3 = (216q3 + 324q2 + 162q) + 27

= (216q3 + 324q2 + 162q + 24) + 3

= 6(36q3 + 54q2 + 27q + 4) + 3 = 6m + 3

Where, m = (36q3 + 64q2 + 27q + 4) is an

integer.

Case V : When r = 4, then putting r = 4 in

Eq. (i), we get

a3 = (216q3 + 432q2 + 288q) + 64

a3 = 6(36q3 + 72q2 + 48q) + 60 + 4

a3 = 6(36q3 + 72q2 + 48q + 10) + 4 = 6m + 4

Where, m = (36q3 + 72q2 + 48q + 10) is an

integer.

Case VI : When r = 5, then putting r = 5 in

Eq. (i), we get

a3 = (216q3 + 540q2 + 450q) + 125

a3 = (216q3 + 540q2 + 450q) + 120 + 5

a3 = 6(36q3 + 90q2 + 75q + 20) + 5

a3 = 6m + 5

Where, m = (36q3 + 90q2 + 75q + 20) is an

integer.

Hence, 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 forms 6m, 6m + 1, 6m + 3,

6m + 3, 6m + 4 and 6m + 5 i.e., 6m + r.

14. 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.

[NCERT Exemplar]

Solution—

Given numbers are n(n + 4), (n + 8),

(n + 12) and (n + 16), where n is any positive

integer.

Then, let n = 5q, 5q + 1, 5q + 2, 5q + 3, 5q

+ 4 for q N [By Euclid’s algorithm]

Then, in each case if we put the different

values of n in the given numbers. We

definitely get one and only one of given

numbers is divisible by 5.

Hence, one and only one out of n, n + 4, n +

8, n + 12 and n + 16 is divisible by 5.

Alternate Method

On dividing on n by 5, let q be the quotient

and r be the remainder.

Then n = 5q + r, where 0 < r < 5.

n = 5q + r, where r = 0, 1, 2, 3, 4

n = 5q or 5q + 1 or 5q + 2 or 5q + 3 or 5q + 4

Case I : If n = 5q, then n is only divisible by

5.

Case II : If n = 5q + 1, then n + 4 = 5q + 1

+ 4 = 5q + 5 = 5(q + 1), which is only

divisible by 5.

So, in this case, (n + 4) is divisible by 5.

Case III : If n = 5q + 3, then n + 2 = 5q +

3 + 12 = 5q + 15 = 5(q + 3), which is

divisible by 5.

So, in this case (n + 12) is only divisible by

5.

Case IV : If n = 5q + 4, then n + 16 = 5q +

4 + 16 = 5q + 20 = 5(q + 4), which is

divisible by 5.

So, in this case, (n + 16) is only divisible

by 5.

Hence, 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.

15. Show that the square of an odd positive

integer can be of the form 6q + 1 or 6q + 3

for some integer q.

[NCERT Exemplar]

Solution—

We know that any positive integer can be of

the form 6m, 6m + 1, 6m + 2, 6m + 3, 6m + 4



or 6m + 5, for some integer m.

Thus, an odd positive integer can be of the

form 6m + 1, 6m + 3, or 6m + 5

Thus we have:

(6m + 1)2 = 36m2 + 12m + 1 = 6(6m2 + 2m)

+ 1 = 6q + 1, q is an integer

(6m + 3)2 = 36m2 + 36m + 9 = 6(6m2 + 6m +

1) + 3 = 6q + 3, q is an integer

(6m + 5)2 = 36m2 + 60m + 25 = 6(6m2 + 10m

+ 4) + 1 = 6q + 1, q is an integer.

Thus, the square of an odd positive integer

can be of the form 6q + 1 or 6q + 3.

16. 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,

3m or 3m + 2 for some integer m? Justify

your answer.

Solution—

No, by Euclid’s Lemma, b = aq + r, 0 < r < a

Here, b is any positive integer

a = 3, b = 3q + r for 0 < r < 3

So, this must be in the form 3q, 3q + 1 or

3q + 2

Now, (3q)2 = 9q2 = 3m [here, m = 3q2]

and (3q + 1)2 = 9q2 + 6q + 1

= 3(3q2 + 2q) + 1 = 3m + 1

[where, m = 3q2 + 2q]

Also, (3q + 2)2 = 9q2 + 12q + 4

= 9q2 + 12q + 3 + 1

= 3(3q2 + 4q + 1) + 1

= 3m + 1 [here, m = 3q2 + 4q + 1]

Hence, square of a positive integer is of the

form 3q + 1 is always in the form 3m + 1

for some integer m.

17. Show that the square of any positive

integer cannot be of the form 3m + 2,

where m is a natural number.

Solution—

By Euclid’s lemma, b = aq + r, 0 < r < a

Here, b is any positive integer,

a = 3, b = 3q + r for 0 < r < 2

So, any positive integer is of the form 3k,

3k + 1 or 3k + 2

Now, (3k)2 = 9k2 = 3m [where, m = 3k2]

and (3k + 1)2 = 9k2 + 6k + 1

= 3(3k2 + 2k) + 1 = 3m + 1

[where, m = 3k2 + 2k]

Also, (3k + 2)2 = 9k2 + 12k + 4

[_ (a + b)2 = a2 + 2ab + b2]

= 9k2 + 12k + 3 + 1

= 3(3k2 + 4k + 1) + 1

= 3m + 1 [where, m = 3k2 + 4k + 1]

Which is in the form of 3m + 1. Hence,

square of any positive number cannot be of

the form 3m + 2.

EXERCISE 1.2

1. Define H.C.F. of two positive integers and

find the H.C.F. of the following pairs of

numbers.

(i) 32 and 54 (ii) 18 and 24

(iii) 70 and 30 (iv) 56 and 88

(v) 475 and 495 (vi) 75 and 243

(vii) 240 and 6552 (viii) 155 and 1385

(ix) 100 and 190 (x) 105 and 120

Solution—

Defination : The greatest among the

common divisor of two or more integers is

the Greatest Common Divisor (G.C.D.) or

Highest Common Factor (H.C.F.) of the

given integers.

(i) HC.F. of 32 and 54

Factors 32 = 1, 2, 4, 8, 16, 32

and factors of 54 = 1, 2, 3, 6, 9, 18, 27, 54

H.C.F. = 2

(ii) H.C.F. of 18 and 24

Factors of 18 = 1, 2, 3, 6, 9, 18

and factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24

Highest common factor = 6

H.C.F. = 6

(iii) H.C.F. of 70 and 30

Factors of 70 = 1, 2, 5, 7, 10, 14, 35, 70

and factors of 30 = 1, 2, 3, 5, 6, 10, 15, 30

H.C.F. = 10



(iv) H.C.F. of 56 and 88

Factors of 56 = 1, 2, 4, 7, 8, 14, 28, 56

and factors of 88 = 1, 2, 4, 8, 11, 22, 44, 88

H.C.F. = 8

(v) H.C.F. of 475 and 495

Factors of 475 = 1, 5, 25, 19, 95, 475

and factors of 495 = 1, 3, 5, 9, 11, 15, 33,

45, 55, 99, 165, 495

H.C.F. = 5

(vi) H.C.F. of 75 and 243

Factors of 75 = 1, 3, 5, 15, 25, 75

Factors of 243 = 1, 3, 9, 27, 81, 243

H.C.F. = 3

(vii) H.C.F. of 240 and 6552

Factors of 240 = 1, 2, 3, 4, 5, 6, 8, 10, 12,

15, 16, 20, 24, 48, 60, 80, 120, 240

Factors of 6552 = 1, 2, 3, 4, 6, 7, 8, 9, 12,

13, 14, 18, 21, 24, 26, 28, 36, 39, 42, 52,

56, 63, 72, 91, 104, 117, 126, 156, 168,

182, 234, 252, 273, 312, 364, 488, 504, 546,

728, 819, 936, 1092, 1638, 2184, 3276, 6552

H.C.F. = 24

(viii)H.C.F. of 155 and 1385

Factors of 155 = 1, 5, 31, 155

Factors of 1385 = 1, 5, 277, 1385

H.C.F. = 5

(ix) 100 and 190

1 190 100 100 90

9 90 10 90 0

1

H.C.F. of 100 and 190 = 10

(x) 105 and 120

7 105 120 1105 105 0 15

H.C.F. of 105 and 120 = 15

2. Use Euclid’s division algorithm to find the

H.C.F. of

(i) 135 and 225 (ii) 196 and 38220

(iii) 867 and 255 (iv) 184, 230 and 276

(v) 136, 170 and 255 (vi) 1260 and 7344

(vii) 2048 and 960

Solution—

(i) H.C.F. of 135 and 225

_ 135 < 225

225 = 135 × 1 + 90

135 = 90 × 1 + 45

45 = 45 × 2 + 0

_ Last remainder = 0

and last divisor = 45

H.C.F. = 45

(ii) H.C.F. of 196 and 38220

_ 196 < 38220

38220 = 196 × 195 + 0



H.C.F. = 196

(iii) H.C.F. 867 and 255

255 < 867

867 = 255 × 3 + 102

255 = 102 × 2 + 51

102 = 51 × 2 + 0



H.C.F. = 51

(iv) H.C.F. of 184, 230 and 276

Let us find the highest common factor

(H.C.F.) of 184 and 230

184)230(1 184 46)184(4 184 ×

Hence, H.C.F. of 184 and 230 = 46

Now, find the H.C.F. of 276 and 46

Hence, H.C.F. of 276 and 46 = 46

Required H.C.F. of 184, 230 and 276 = 46

(v) H.C.F. of 136, 170 and 255

)867( 765 102

255 3

)255( 204 51

102 2

)102( 102 ×

51 2

)225( 135 90

135 1

)135( 90 45

90 1

)90( 90 0

45 2

)38220( 196 1862 1764 980 980 ×

196 195

46)276(4 276 ×



Let us find the highest common factor

(H.C.F.) of 136 and 70

136)170(1 136 34)136(4 136 ×

Hence, H.C.F. of 136 and 170 = 34

Now, find the H.C.F. of 34 and 255

Hence, highest common factor of 34 and

255 = 17

Required H.C.F. of 136, 170 and 255 = 17

(vi) Given integers are 1260 and 7344

and 7344 > 1260

Applying Euclid division algorithm, we have

7344 = 1260 × 5 + 1044

since remainder = 1044 0.

So, applying Euclid lemma to divisor 1260

and remainder = 1044

1260 = 1044 × 1 + 216

Since remainder = 216 0

again applying Euclid lemma to divisor 1044

and remainder 216 to get,

1044 = 216 × 4 + 180

Since remainder = 180 0

1260 ) 73446300

( 5

1044

1044 ) 12601044

( 1

216

216 ) 1044864

( 4

180

So applying division lemma to divisor 216

and remainder 180

Then 216 = 180 × 1 + 36

again remainder = 36 0

Then we apply division lemma to divisor 180

and remainder 36 to get 180 = 36 × 5 + 0

remainder = 0 Thus divisor 36 be the HCF

of 1260 and 7344.

(vii) 2048 and 960 (CBSE 2019)

Sol. Given integers are 2048 and 960

since 2048 > 960

Apply Euclid division algorithm, we have

2048 = 960 × 2 + 128

since remainder = 128 × 0, so apply

the division lemma to divisor 960 and

remainder 128, to get 960 = 128 × 7 + 64

Since remainder = 64 0, So apply division

lemma to divisor 128 and remainder = 64

to get, 128 = 64 × 2 + 0

960 ) 20481920

( 2

128

128 ) 960896

( 7

64

Since remainder = 0 divisor 64 be the

HCF of 2048 and 960.

3. Find the H.C.F. of the following pairs of

integers and express it as a linear

combinations of them.

(i) 963 and 657 (ii) 592 and 252

(iii) 506 and 1155 (iv) 1288 and 575

Solution—

(i) 963 and 657

)963( 657 306 )657( 2 612 45 )306( 6 270 36 ) 45 ( 1 36 9 ) 36 ( 4 36 ×

657 1

H.C.F. = 9

9 = 45 – 36 × 1 = 45 – (306 – 45 × 6)

= 45 – 306 + 45 × 6

34)255(7 238 17)34(2 34 ×



= 45 × 7 – 306

= (657 – 306 × 2) × 7 – 306

= 657 × 7 – 14 + 306 – 306 × 1

= 657 × 7 – 15 × 306

= 657 × 7 + (–15) 306 = xa + yb

Here x = 7, y = –15

(ii) HCF of 592 and 252

252)592(2 504 88

88)252(2 176 76

592 = 252 × 2 + 88

252 = 88 × 2 + 76

88 = 76 × 1 + 12

76 = 12 × 6 + 4

12 = 4 × 3 + 0

HCF = 4

Now, 4 = 76 – 12 × 6

4 = 76 – (88 – 76 × 1) × 6

4 = 76 – (88 × 6 – 76 × 6)

4 = 76 × 7 – 88 × 6

4 = 76 × 7 + (–88 × 6)

76x + 88y

Where x = 7, y = –6

(iii) 506 and 1155

H.C.F. = 11

11 = 77 – 66

= 77 – (143 – 77)

= 77 – 143 + 77

= 77 × 2 – 143

= (506 – 143 × 3) × 2 – 143

= 506 × 2 – 143 × 6 – 143

= 506 × 2 – 143 × 7

= 506 × 2 – (1155 – 2 × 506) × 7

= 506 × 2 – 1155 × 7 + 14 × 506

= 506 × 16 + (–7) × 1155

= ax + by

x = 16, y = –7

(iv) 1288 and 575

H.C.F. = 23

Now, 23 = 575 – 138 × 4

= 575 – (1288 – 575 × 2) × 4

= 575 – 1288 × 4 + 575 × 8

= 575 × 9 + 1288 × (–4)

= ax + by

x = 9, y = –4

4. Find the largest number which divides

615 and 963 leaving remainder 6 in each

case.

Solution—

The given numbers are 615 and 963

Remainder in each case = 6

1 609 957 348 609

3 261 348 1261 261 0 87

1

Remainder H.C.F

615 – 6 = 609 and 963 – 6 = 957 are divisible

by the required number which is the H.C.F.

of 609 and 957 = 87

Hence the required largest number = 87

5. If the H.C.F. of 408 and 1032 is

expressible in the form 1032m – 408 × 5,

find m.

Solution—

408, 1032

H.C.F. = 24

24 = (216 – 192)

= 216 – (408 – 216)

= 216 – 408 + 216

= 216 × 2 – 408

= [1032 – 408 × 2] × 2 – 408

= 1032 × 2 – 408 × 4 – 408

= 1032 × 2 – 408 × 5

= 1032 × 2 – 408 × 5

)1155( 1012 143 )506( 3 429 77 )143( 1 77 66 ) 77 ( 1 66 11 ) 66 ( 1 66 ×

506 2

)1288( 1150 138 )575( 4 552 23 )138( 6 138 ×

575 2

)1032( 816 216 )408( 1 216 192 )216( 1 192 24 ) 192 ( 8 192 ×

408 2



Which is in the form of 1032m – 408 × 5

comparing, we get

m = 2

6. If the H.C.F. of 657 and 963 is expressible

in the form 657x + 963 × (–15), find x.

Solution—

657 and 963

H.C.F. = 9

)963( 657 306 )657( 2 612 45 )306( 6 270 36 ) 45 ( 1 36 9 ) 36 ( 4 36 ×

657 1

9 = 45 – 36

= 45 – (306 – 45 × 6)

= 45 – 306 + 45 × 6

= 45 × 7 – 306 = [657 – (306 × 2)] × 7 – 306

= 657 × 7 – 306 × 14 – 306

= 657 × 7 – 306 × 15

= 657 × 7 – (963 – 657) × 15

= 657 × 7 – 963 × 15 + 657 × 15

= 657 × 22 – 963 × 15

= 657 × 22 + 963 × (–15)

= 657 × x + 963 × (–15)

Comparing, we get

x = 22

7. 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—

The required number of columns will be the

H.C.F. of 616 and 32

4 32 616 32 608

9 0 8

19

Remainder H.C.F.

Using Euclid’s division

We get H.C.F. = 4

Number of columns = 4

8. A merchant has 120 litres of oil of one

kind, 180 litres of another kind and 240

litres of third kind. He wants to sell the

oil by filling the three kinds of oil in tins

of equal capacity. What should be the

greatest capacity of such a tin ?

Solution—

Quantity of oil of one kind = 120 l

and quantity of second kind = 180 l

and third kind of oil = 240 l

Maximum capacity of oil in each tin

= H.C.F. of 120 l, 180 l and 240 l

H.C.F. of 120 and 180 = 60

2 120 180 120 120 0 60

1

and H.C.F. of 60 and 240 = 60

60 240 4 240 0

Greatest capacity of each tin = 60 litres

9. During a sale, colour pencils were being

sold in packs of 24 each and crayons in

packs of 32 each. If you want full packs

of both and the same number of pencils

and crayons, how many of each would you

need to buy ?

Solution—

Number of pencils in each pack = 24

and number of crayons pack = 32

3 24 32 124 24 0 8

Remainder H.C.F.

H.C.F.



Highest number of pencils and crayons in

packs will be = H.C.F. of 24 and 32 = 8

Number of pencil’s pack = 8

24 = 3

and number of crayon’s pack = 8

32 = 4

10. 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 is to contain cartons of the

same drink, what would be the greatest

number of cartons each stack would

have ?

Solution—

Number of Coke Cans Cartons = 144

and number Pepsi Cartons = 90

1 144 90 90 54

1 54 36 1 36 36 18 0

1

Required greatest number of cartons of each

= H.C.F. of 144 and 90 = 18

11. Find the greatest number which divides

285 and 1249 leaving remainders 9 and 7

respectively.

Solution—


Remainder are 9 and 7 respectively

285 – 9 = 276

and 1249 – 7 = 1242 are divisible by required

number

2 276 1242 4276 1104 0 138

Required number = H.C.F. of 276 and 1242

Now, H.C.F. of 276 and 1242 = 138

Required number = 138

12. Find the largest number which exactly

divides 280 and 1245 leaving remainders

4 and 3, respectively.

Solution—


Remainder are 4 and 3 respectively

280 – 4 = 276 and 1245 – 3 = 1242 are

divisible by a number

The required number = H.C.F. of 276 and

1242

2 276 1242 4276 1104 0 138

H.C.F. of 276 and 1242 = 138

Hence required number = 138

13. What is the largest number that divides

626, 3127 and 15628 and leaves remain-

ders of 1, 2 and 3 respectively.

Solution—

Given numbers are 626, 3127 and 15628

and remainders are 1, 2 and 3 respectively

626 – 1 = 625

625 3125 5 3125 0

3127 – 2 = 3125 and

15628 – 3 = 15625 are divisible by a required

greatest number

625 15625 25 15625 0

The greatest number will be the H.C.F. of

625, 3125 and 15625

H.C.F. of 625 and 3125 = 625

and H.C.F. of 625 and 15625 = 625

The required number = 625

14. Find the greatest numbers that will divide

445, 572 and 699 leaving remainders 4, 5

and 6 respectively.

Solution—

Given numbers are 445, 572 and 699

and remainders are 4, 5, 6 respectively

H.C.F.

Remainder H.C.F.

Remainder H.C.F.

H.C.F. Remainder

H.C.F. Remainder



445 – 4 = 441

572 – 5 = 567

3 441 567 378 441 63 126 2 126 0

1

699 – 6 = 693 are exactly divisible by a

certains number which is the H.C.F. of these

numbers

H.C.F. of 441 and 567 = 63

and H.C.F. of 63 and 693 = 63

63 693 11 693 0

The required number = 63

15. Find the greatest number which divides

2011 and 2623 leaving remainders 9 and

5 respectively.

Solution—

The given numbers are 2011 are 2623 and

remainders are 9 and 5 respectively

2011 – 9 = 2002 and

2623 – 5 = 2618 are divisible by a greatest

number which is the H.C.F. of 2002 and

2618

1 2618 2002 32002 1848

4 616 154 616 0

H.C.F. = 2002 and 2618 = 154

The required nunber = 154

16. Using Euclid’s division algorithm, find the

largest number that divides 1251, 9377

and 15628 leaving remainders 1, 2 and 3

respectively. [NCERT Exemplar]

Solution—

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

9377 and 15628, respectively. Thus, after

subtracting these remainders from the

numbers.

We have the numbers, 1251 – 1 = 1250,

9377 – 2 = 9375 and 15628 – 3 = 15625

which is divisible by the required number.

Now, required number = HCF of 1250, 9375

and 15625 [for the largest number]

By Euclid’s division algorithm,

a = bq + r ...(i)

[_ dividend = divisor × quotient + remainder]

For largest number, put a = 15625

and b = 9375

15625 = 9375 × 1 + 6250 [From Eq. (i)]

9375 = 6250 × 1 + 3125

6250 = 3125 × 2 + 0

HCF (15625, 9375) = 3125

Now, we take c = 1250 and d = 3125, then

again using Euclid’s division algorithm,

d = cq + r [from Eq. (i)]

3125 = 1250 × 2 + 625

1250 = 625 × 2 + 0

HCF (1250, 9375, 15625) = 625

Hence, 625 is the largest number which

divides 1251, 9377 and 15628 leaving

remainder 1, 2 and 3, respectively.

17. Two brands of chocolates are available in

packs of 24 and 15 respectively. If I need

to buy an equal number of chocolates of

both kinds, what is the least number of

boxes of each kind I would need to buy ?

Solution—

Number of chocolates of first kind = 24

and of second kind = 15

Number of chocolates to be bought equally

of both kinds = H.C.F. of 24 and 15

1 24 15 115 9

1 9 6 2 6 6 3 0

= 3 chocolates

Least number of boxes of first kind = 3

24 = 8

H.C.F. Remainder

H.C.F. Remainder

H.C.F. Remainder

Remainder H.C.F.



and of second kind = 3

15 = 5

18. A mason has to fit a bathroom with

square marble tiles of the largest possible

size. The size of the bathroom is 10 ft. by

8 ft. What would be the size in inches of

the tile required that has to be cut and

how many such tiles are required ?

Solution—

Size of bathroom = 10 ft. × 8 ft.

1 10 8 4 8 8 2 0

Largest size of tile = H.C.F. of 10 ft. and 8

ft. = 2 ft.

= 2 × 12 = 24 inches

(1 ft. = 12 inches)

19. 15 pastries and 12 biscuit packets have

been donated for a school fete. These are

to be packed in several smaller identical

boxes with the same number of pastries

and biscuit packets in each. How many

biscuit packets and how many pastries

will each box contain ?

Solution—

Number of pastries = 15

and number of biscuit packets = 12

The number of pastries and pack of biscuits

to be packed in smaller identicals boxes

H.C.F. of 15 and 12

1 15 12 412 12 3 0

H.C.F. = 3

Each box will contain

= 3

15 pastries and

3

12 pack of biscuits

= 5 pastries and 4 pack of biscuits

20. 105 goats, 140 donkeys and 175 cows have

to be taken across a river. There is only

one boat which will have to make many

trips in order to do so. The lazy boatman

has his own conditions for transporting

them. He insists that he will take the

same number of animals in every trip

and they have to be of the same kind. He

will naturally like to take the largest

possible number each time. Can you tell

how many animals went in each trip ?

Solution—

The required number of animals will be the

H.C.F. of 105 goats, 140 donkeys, 175 cows

4 140 175 140 140 0 35

1

H.C.F. of 175 and 140 = 35

and H.C.F. of 35 and 105 = 35

35 105 3 105 0

The required number of animals = 35

21. The length, breadth and height of a room

are 8m 25 cm, 6 m 75 cm and 4 m 50 cm

respectively. Determine the longest rod

which can measure the three dimensions

of the room exactly.

Solution—

Length = 8m 25 cm = 825 cm

Breadth = 6 m 75 cm = 675 cm

Height = 4 m 50 cm = 450 cm

1 825 675 4675 600

2 150 75150 0

The required measure will be the H.C.F. of

these three dimensions

H.C.F. of 825 and 675 = 75

and H.C.F. of 75 and 450 = 75

75 450 6 450 0

The required length = 75 cm

22. Express the H.C.F. of 468 and 222 as 468x

H.C.F. Remainder

H.C.F. Remainder

H.C.F. Remainder

H.C.F. Remainder

Remainder H.C.F.



+ 222y where x, y are integers in two

different ways.

Solution—

468 and 222

H.C.F. = 6

)468( 444 24 )222( 9 216 6 )24( 4 24 ×

222 2

6 = (222 – 24 × 9)

= 222 – (468 – 222 × 2) × 9

= 222 – 468 × 9 + 222 × 18

= 222 × 19 + 468 × (–9) = 468 (9) + 222 × 19

Which is in the form of 468x + 222y

Similarly we can write it in the following

form also

6 = 468 × 213 + 222 × (–449)

EXERCISE 1.3

1. Express each of the following integers as

a product of its prime factors :

(i) 420 (ii) 468

(iii) 945 (iv) 7325

Solution—

(i) 420

= 2 × 2 × 3 × 5 × 7

= 22 × 3 × 5 × 7

420

×2 210

×2

×2

×2

2 × 105

2 ×

2 ×

3 × 35

3 × 5 × 7

(ii) 468

= 2 × 2 × 3 × 3 × 13

= 22 × 32 × 13

468

×2 234

×2

×2

×2

2 × 117

2 ×

2 ×

3 × 32

3 × 3 × 13

(iii) 945

= 3 × 3 × 3 × 5 × 7

= 33 × 5 × 7

945

×3 315

×3

×3

×3

3 × 105

3 ×

3 ×

3 × 35

3 × 5 × 7

(iv) 7325

= 5 × 5 × 293

= 52 × 293

7325

×5 1465

×5 5 × 293

2. Determine the prime factorisation of each

of the following positive integer :

(i) 20570 (ii) 58500

(iii) 45470971

Solution—

(i) 20570

2 205705 10285

11 205711 18717 17

1

20570 = 2 × 5 × 11 × 11 × 17

= 2 × 5 × 112 × 17



(ii) 58500

58500 = 2 × 2 × 3 × 3 × 5

× 5 × 5 × 13

= 22 × 32 × 53 × 13

(iii) 45470971

7 454709717 6495853

13 92797913 7138317 549117 32319 19

1

45470971 = 7 × 7 × 13 × 13 × 17 × 17 × 19

= 72 × 132 × 172 × 19

3. Explain why 7 × 11 × 13 + 13 and 7 × 6 ×

5 × 4 × 3 × 2 × 1 + 5 are composite

numbers ?

Solution—

We know that a composite number is that

number which can be factorise. It has more

factors other than itself and one

Now, 7 × 11 × 13 + 13 = 13 (7 × 11 + 1)

= 13 × 78

Which is composite number

Similarly,

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

= 5 (7 × 6 × 4 × 3 × 2 × 1 + 1)

= 5 × 1009

Which is a composite number

Hence proved.

4. Check whether 6n can end with the digit

0 for any natural number n.

Solution—

No, 6n can’t end with the digit 0 as the

number ending 0 can be factorise of the type

2n × 5m only but 6n = (2 × 3)n = 2n × 3n

Which does not has 5m as factors.

5. Explain why 3 × 5 × 7 + 7 is a composite

number. [NCERT Exemplar]

Solution—

We have, 3 × 5 × 7 + 7 = 105 + 7 = 112

Now, 112 = 2 × 2 × 2 × 2 × 7 = 24 × 7

So, it is the product of prime factors 2 and

7. i.e., it has more than two factors. Hence,

it is a composite number.

EXERCISE 1.4

1. Find the L.C.M. and H.C.F. of the

following pairs of integers and verify that

L.C.M. × H.C.F. = Product of the integers.

(i) 26 and 91 (ii) 510 and 92

(iii) 336 and 54 (iv) 404 and 96

Solution—

(i) 26 and 91

26 = 2 × 13

91 = 7 × 13

H.C.F. = 13

and L.C.M. = 2 × 7 × 13 = 182

Now, L.C.M. × H.C.F. = 182 × 13 = 2366

and 26 × 91 = 2366

L.C.M. × H.C.F. = Product of integers

(ii) 510 and 92

2 922 46

23 231

2 5103 2555 85

17 171

92 = 2 × 2 × 23 = 22 × 23

510 = 2 × 3 × 5 × 17

H.C.F. = 2

and L.C.M. = 2 × 2 × 3 × 5 × 17 × 23

= 23460

Now L.C.M. × H.C.F. = 2 × 23460 = 46920

and product of integers = 510 × 92 = 46920

L.C.M. × H.C.F. = Product of numbers

(iii) 336 and 54

2 543 273 93 3

1

2 3362 1682 842 423 217 7

1

54 = 2 × 3 × 3 × 3 = 2 × 33

336 = 2 × 2 × 2 × 2 × 3 × 7 = 24 × 3 × 7

2 585002 292503 146253 48755 16255 3255 65

13 131



H.C.F. = 2 × 3 = 6

L.C.M. = 24 × 33 × 7

= 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 = 3024

Now, H.C.F. × L.C.M. = 6 × 3024 = 18144

and product of integer = 336 × 54 = 18144

H.C.F. × L.C.M. = Product of integers

(iv) 96, 404

2 962 482 242 122 63 3

1

2 4042 202

101 1011

96 = 2 × 2 × 2 × 2 × 2 × 3 = 25 × 3

404 = 2 × 2 × 101 = 22 × 101

HCF = 22 = 2 × 2 = 4

LCM = 25 × 3 × 101 = 32 × 3 × 101 = 9696

Now HCF × LCM = 4 × 9696 = 38784

and product of two numbers = 96 × 404

= 38784

HCF × LCM = Product of given two

numbers

2. Find the L.C.M. and H.C.F. of the

following integers by applying the prime

factorisation method :

(i) 12, 15 and 21 (ii) 17, 23 and 29

(iii) 8, 9 and 25 (iv) 40, 36 and 126

(v) 84, 90 and 120 (vi) 24, 15 and 36

Solution—

(i) 12, 15 and 21

12 = 2 × 2 × 3 = 22 × 3

15 = 3 × 5

21 = 3 × 7

H.C.F. = 3

L.C.M. = 3 × 22 × 5 × 7 = 420

Hence H.C.F. = 3, L.C.M. = 420

(ii) 17, 23 29

17 = 1 × 17

23 = 1 × 23

29 = 1 × 29

H.C.F. = 1

L.C.M. = 17 × 23 × 29 = 11339

Hence H.C.F. = 1, L.C.M. = 11339

(iii) 8, 9 and 25

8 = 2 × 2 × 2 = 23

9 = 3 × 3 = 32

25 = 5 × 5 = 52

H.C.F. = 1

L.C.M. = 23 × 32 × 52

= 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800

Hence H.C.F. = 1, L.C.M. = 1800

(iv) 40, 36, 126

2 402 202 105 5

1

2 362 183 93 3

1

2 1263 633 217 7

1

40 = 2 × 2 × 2 × 5 = 23 × 5

36 = 2 × 2 × 3 × 3 = 22 × 32

126 = 2 × 3 × 3 × 7 = 2 × 32 × 7

H.C.F. = 2

L.C.M. = 23 × 5 × 32 × 7

= 2×2×2×5×3×3×7 = 2520

H.C.F. = 2 and L.C.M. = 2520

(v) 84, 90 and 120

2 842 423 217 7

1

2 903 453 155 5

1

2 1202 602 303 155 5

1

84 = 2 × 2 × 3 × 7 = 22 × 3 × 7

90 = 2 × 3 × 3 × 5 = 2 × 32 × 5

120 = 2 × 2 × 2 × 3 × 5 = 23 × 3 × 5

H.C.F. = 2 × 3 = 6

L.C.M. = 2 × 2 × 2 × 3 × 3 × 5 × 7

= 23 × 32 × 7 × 5 = 2520

H.C.F. = 6 and L.C.M. = 2520

(vi) 24, 15 and 36

2 242 122 63 3

1

3 155 5

1

2 362 183 93 3

1



24 = 2 × 2 × 2 × 3 = 23 × 3

15 = 3 × 5

36 = 2 × 2 × 3 × 3 = 22 × 32

H.C.F. = 31 = 3

L.C.M. = 23 × 32 × 5 = 360

H.C.F. = 3 and L.C.M. = 360

3. (i) Given that HCF (306, 657) = 9, Find

LCM (306, 657). [NCERT]

Solution—

(i) HCF of 306, 657 = 9

LCM = HCF

numberSecondnumberFirst

= 9

657306 = 306 × 73 = 22338

(ii) Write the smallest number which is

divisible by both 306 and 657.

Solution—

(ii) The smallest number which is divisible by

both 306 and 657 = LCM (306, 657)

= 2 × 3 × 3 × 17 × 73 = 22338

3 306, 657

3 102, 219

2 34, 73

17, 73

4. Can two numbers have 16 as their H.C.F.

and 380 as their L.C.M. ? Give reason.

Solution—

H.C.F. of two numbers = 16

and their L.C.M. = 380

)380( 32 60 48 12

16 23

We know the H.C.F. of two numbers is a

factor of their L.C.M. but 16 is not a factor

of 380 or 380 is not divisible by 16

It can not be possible.

5. The H.C.F. of two numbers is 145 and

their L.C.M. is 2175. If one number is

725, find the other.

Solution—

First number = 725

Let second number = x

Their H.C.F. = 145

and L.C.M. = 2175

Second number (x) = numberFirst

H.C.F.L.C.M.

= 725

1452175 = 435

Second number = 435

6. The H.C.F. of two numbers is 16 and their

product is 3072. Find their L.C.M.

Solution—

H.C.F. of two numbers = 16

and product of two numbers = 3072

Their L.C.M. = H.C.F.

numberstwoofProduct

= 16

3072 = 192

L.C.M. = 192

7. The L.C.M. and H.C.F. of two numbers

are 180 and 6 respectively. If one of the

number is 30, find the other number.

Solution—

First number = 30

Let x be the second number

Their L.C.M. = 180 and H.C.F. = 6

We know that first number × second number

= L.C.M. × H.C.F.

30 × x = 180 × 6

x = 30

6180 = 36

Second number = 36

8. Find the smallest number which when

increased by 17 is exactly divisible by both

520 and 468.

Solution—

L.C.M. of 520 and 468

2 520, 4682 260, 234

13 130, 11710, 9

= 2 × 2 × 9 × 10 × 13 = 4680



The number which is increased = 17

Required number 4680 – 17 = 4663

9. Find the smallest number which leaves

remainders 8 and 12 when divided by 28

and 32 respectively.

Solution—

Dividing by 28 and 32, the remainders are 8

and 12 respectively

28 – 8 = 20

32 – 12 = 20

Common difference = 20

Now, L.C.M. of 28 and 32

= 2 × 2 × 7 × 8 = 224

Required smallest number = 224 – 20 = 204

10. What is the smallest number that, when

divided by 35, 56 and 91 leaves

remainders of 7 in each case ?

Solution—

L.C.M. of 35, 56, 91

= 5 × 7 × 8 × 13 = 3640

Remainder in each case = 7

The required smallest number = 3640 + 7

= 3647

11. A rectangular courtyard is 18 m 72 cm

long and 13 m 20 cm broad. It is to be

paved with square tiles of same size. Find

the least possible number of such tiles.

Solution—

)1872( 1320 552 )1320( 2 1104 216 )552( 2 432 120 ) 216 ( 1 120 96 ) 120 ( 1 96 24 ) 96 ( 4 96 ×

1320 1

Length of rectangle = 18 m 72 cm = 1872 cm

and breadth = 13 m 20 cm = 1320 cm

Side of the greatest size of square tile =

H.C.F. of 1872 and 1320

Now H.C.F. of 1872 and 1320 = 24

Length of greatest square tile = 24 cm

Now number of tiles = 2tile)squareof(side

breadthlength

= 2424

13201872

= 4290

12. Find the greatest number of 6 digits

exactly divisible by 24, 15 and 36.

Solution—

Greatest number of 6 digits = 999999

Now L.C.M. of 24, 15 and 36

2 24, 15, 362 12, 15, 183 6, 15, 9

2, 5, 3

= 2 × 2 × 2 × 3 × 3 × 5 = 360

Now dividing 999999 by 360

)999999( 720 2799 2520 2799 2520 2799 2520 279

360 2777

We get quotient = 2777

and remainder = 279

Required number = 999999 – 279 = 999720

13. Determine the number nearest to 110000

but greater than 100000 which is exactly

divisible by each of 8, 15 and 21.

Solution—

L.C.M. of 8, 15, 21

= 3 × 5 × 7 × 8 = 840

But the required number is nearest to 110000

but greatest than 100000

)110000( 840 2600 2520 800

840 130

2 28, 322 14, 16

7, 8

7 35, 56, 912 5, 8, 13

3 8, 15, 218, 5, 7



Required number will be = 110000 – 800

= 109200

14. Find the least number that is divisible by

all the numbers between 1 to 10 (both

inclusive).

Solution—

The required least number which is divisible

by 1 to 10 will be the L.C.M. of 1 to 10

L.C.M. of 1 to 10

2 1, 2, 3, 4, 5, 6, 7, 8, 9, 102 1, 1, 3, 2, 5, 3, 7, 4, 9, 53 1, 1, 3, 1, 5, 3, 7, 2, 9, 55 1, 1, 1, 1, 5, 1, 7, 2, 3, 6

1, 1, 1, 1, 1, 1, 7, 2, 3, 1

= 2 × 2 × 3 × 5 × 7 × 2 × 3 = 2520

15. A circular field has a circumference of

360 km. Three cyclists start together and

can cycle 48, 60 and 72 km a day, round

the field. When will they meet again ?

Solution—

Circumference of a circular field = 360 km

Three cyclist start together who can cycle

48, 60 and 72 km per day round the field

L.C.M. of 48, 60, 72

2 48, 60, 722 24, 30, 362 12, 15, 183 6, 15, 9

2, 5, 3

= 2 × 2 × 2 × 2 × 3 × 3 × 5 = 720

They will meet again after 720 km distance

16. In a morning walk, three persons step

off together, their steps measure 80 cm,

85 cm and 90 cm respectively. What is

the minimum distance each should walk

so that they can cover the distance in

complete steps ?

Solution—

Measures of steps of three persons

= 80 cm, 85 cm and 90 cm

Minimum required distance

covered by them

= L.C.M. of 80 cm, 85 cm,

90 cm

= 2 × 5 × 8 × 9 × 17 = 12240 cm

= 122.40 m = 122 m 40 cm

17. On a morning walk, three persons step

out together and their steps measure

30 cm, 36 cm and 40 cm respectively.

What is the minimum distance each

should walk so that each can cover the

same distance in complete steps ?

Sol. Now, each person will cover the same

distance in complete steps

if the distance covered by each person in

cm is the LCM of 30, 36 and 40.

Now 30 = 2 × 3 × 5

36 = 2 × 2 × 3 × 3 = 22 × 32

40 = 2 × 2 × 2 × 5 = 23 × 51

LCM of 30, 36, 40 = 23 × 32 × 5

= 8 × 9 × 5 = 360

Thus, minimum distance each should walk

= 360 cm

18. Find the largest number which on dividing

1251, 9377 and 15628 leaves remainders

1, 2 and 3 respectively. (CBSE 2019)

Sol. It is given that on dividing 1251 by required

number leaves remainder 1.

Therefore 1251 – 1 = 1250 is exactly divisible

by required number.

Thus, required no. be the divisor of 1250.

Similarly required number is the factor of

9377 – 2 = 9375 and 15628 – 3 = 15625

Clearly required number must be the HCF

of 1256, 9375 and 15625.

1250 = 2 × 54

9375 = 3 × 55

15625 = 5 × 5 × 5 × 5 × 5 × 5 = 56

Clearly HCF (1250, 9375, 15625)

= 54 = 625

Hence the largest required number be 625.

2 1250

5 625

5 125

5 25

5

3 9375

5 3125

5 625

5 125

5 25

5

5 15625

5 3125

5 625

5 125

5 25

5

2 80, 85, 905 40, 85, 45

8, 17, 9



EXERCISE 1.5

1. Show that the following numbers are

irrational

(i)2

1(ii) 7 5

(iii) 6 + 2 (iv) 3 – 5

Solution—

(i) Let 2

1 is a rational number and let

2

1 =

b

a where a and b are co-primes

2

1 = 2

2

b

a(Squaring both sides)

2a2 = b2

2|b2 ( 2|2a2)

2|b ....(i)

Let b = 2c for some positive integer c

2a2 = b2 2a2 = 4c2 a2 = 2c2

2|a2 (_ 2|2c2)

2|a ....(ii)

From (i) and (ii)

2|b and 2|a

But it contradics because a and b are co-

primes

Our supposition is wrong

Hence 2

1 is an irrational

(ii) Let 7 5 is a rational number and let

7 5 = b

a 5 =

b

a

7

_

b

a

7 is a rational number


But it contradicts become 5 is an irrational

Hence 7 5 is an irrational

(iii) Let 6 + 2 is not an irrational

Let 6 + 2 = b

a where a and b are co-

primes

Then 2 = b

a – 6 =

b

ba 6

But b

ba 6 is a rational number


Which contradicts because 2 is an

irrational

Hence 6 + 2 is irrational

(iv) Let 3 – 5 is not an irrational

Let 3 – 5 = b

a

Where a and b are co-primes

Then 5 = 3 – b

a =

b

ab 3

_

b

ab 3 is a rational number

5 is also a rational number

But it contradics that because 5 is

irrational

3 – 5 is irrational

2. Prove that following numbers are

irrationals :

(i)7

2(ii)

52

3

(iii) 4 + 2 (iv) 5 2

Solution—

(i) Let 7

2 is a rational number and

7

2 =

b

a

where a and b are co-primes

2

7 =

a

b 7 =

a

b2



_

a

b2 is a rational number

7 is also a rational number

But it contradict because 7 is an irrational

number

7

2 is an irrational number

(ii) Let 52

3 is not an irrational number

Then 52

3=

b


3

52 =

a

b 5 =

a

b

2

3

But a

b

2



But it contradicts because 5 is an irrational

number

Hence 52

3 is an irrational number

(iii) Let 4 + 2 be a rational number

and 4 + 2 = b


primes

2 = b

a – 4 =

b

ba 4

But b



But this contradicts because 2 is an

irrational

4 + 2 is an irrational number

(iv) Let 5 2 is not a rational number

and let 5 2 = b


primes

2 = b

a

5

But b

a




number

5 2 is irrational number

3. Show that 2 – 3 is an irrational number..

(C.B.S.E. 2008)

Solution—

Let 2 – 3 is not an irrational number

and let 2 – 3 = b

a

3 = 2 – b

a =

b

ab 2

But b

ab 2 is a rational number



number

2 – 3 is an irrational number

Hence proved.

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

Solution—

Let 3 + 2 is a rational number



and let 3 + 2 = b

a where a and b are

positive integers

b

a – 3 = 2 b

ba 3 = 2

But b


and 2 is irrational

But our suppositon is wrong

3 + 2 is an irrational number

5. Prove that 4 – 5 2 is an irrational

number. [CBSE 2010]

Solution—

Let 4 – 5 2 is not are irrational number

and let 4 – 5 2 is a rational number

and 4 – 5 2 = b

a where a and b are positive

prime integers

4 – b

a = 5 2 b

ab 4 = 5 2

b

ab

5

4 = 2


But 2 is an irrational number


4 – 5 2 is an irrational number

6. Show that 5 – 2 3 is an irrational

number.

Solution—

Let 5 – 2 3 is a rational number

Let 5 – 2 3 = b

a where a and b are

possitive integers

b

a – 5 = –2 3 5 –

b

a = 2 3

b

ab 5 = 2 3

b

ab

2

5 = 3

But b

ab

2


and 3 is a rational number


5 – 2 3 is an irrational number

7. Prove that 2 3 – 1 is an irrational

number. [CBSE 2010]

Solution—

Let 2 3 – 1 is not an irrational number

and let 2 3 – 1 a ration number

and then 2 3 – 1 = b

a where a, b are

positive prime integers

2 3 = b

a + 1 2 3 =

b

ba

3 = b

ba

2




2 3 – 1 is an irrational number

8. Prove that 2 – 3 5 is an irrational

number. [CBSE 2010]

Solution—

Let 2 – 3 5 is not an irrational number

and let 2 – 3 5 is a rational number

Let 2 – 3 5 = b

a where a and b are positive

prime integers

2 – b

a = 3 5

b

ab 2 = 3 5



b

ab

3

2 = 5

5 is a rational



2 – 3 5 is an irrational

9. Prove that 5 + 3 is irrational.

Solution—

Let 5 + 3 is a rational number

and let 5 + 3

= b


Then 5 = b

a – 3

5 =

2

3

b

a(Squaring both sides)

5 = 2

2

b

a + 3 – 2 3

b

a

2 3 b

a = –5 + 3 + 2

2

b

a

2 3 b

a = –2 + 2

2

b

a = 2

222

b

ab

= 2

22 2

b

ba

3 = 2

22 2

b

ba ×

a

b

2 =

ab

ba

2

2 22

But ab

ba

2

2 22 is a rational number


But it contradics as 3 is irrational number

5 + 3 is irrational

10. Prove that 2 + 3 is an irrational

number.

Solution—

Let us suppose that 2 + 3 is rational.

Let 2 + 3 = a, where a is rational.

Therefore, 2 = a – 3

Squaring on both sides, we get

2 = a2 + 3 – 2a 3

Therefore,

3 = a

a

2

12 which is a contradiction as

the right hand side is a rational number while

3 is irrational.

Hence, 2 + 3 is irrational.

11. Given that 2 is irrational, prove that

(5 3 2) is an irrational number..

[CBSE 2018]

Sol. Let us assume that (5 + 3 2 ) is rational.

Then there exists co-prime positive integers

a and b such that

5 3 2a

b

3 2 5a

b

5

23

a b

b

2 is rational

[ a, b are integers, 5

3

a b

b

is rational]

This contradicts the fact that 2 is irrational

So, our assumption is wrong.

Hence (5 3 2) is an irrational number..



12. Prove that 2 3

5

is an irrational

number, given that 3 is an irrational

number.

(CBSE 2019)

Sol. Given 3 is an irrational number

Let 2 3

5

be a rational number

2 3

5

a

b

5 5 2

3 2a a b

b b

3 is rational

[ a, b are integers

5 2a b

b

is a rational]


Hence our supposition is wrong.

Thus, 2 3

5

be an irrational number..

13. Prove that 2 5 3 is an irrational

number, given that 3 is an irrational

number.

(CBSE 2019)

Sol. Let 2 5 3 be an irrational number..

2 5 3 ;a

b where a, b are integers and

b 0

5 3 2a

b

2

35

a b

b

since a and b are integers

2

5

a b

b

be rational

Thus 3 be a rational number which is a

contradiction to fact that 3 is an irrational

number.

Thus our supposition is wrong.

Hence 2 5 3 is an irrational number..

EXERCISE 1.6

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)8

23 (ii)

441

125

(iii) 50

35(iv)

210

77

(v) 1772 752

129

(vi)

10500

987

Solution—

(i)8

23 = 03 52

23

_ Denominator 8 = 23 × 50 which is in the form

of 2m × 5n (_ 50 = 1)



8

23 is terminating decimal expansion.

(ii)441

125

3 4413 1477 497 7

1

= 22 73

125

_ Denominator 441 is not in the form of 2m ×

5n

441

125 is not terminating repeating decimal

expansion.

(iii)50

35

2 505 255 5

1

= 21 52

35

_ The denominator 50 is in the form of 2m × 5n

50

35 is in terminating decimal expansion.

(iv)210

77

= 7210

777

=

30

11

2 303 155 5

1

= 111 532

11

_ 210 or 30 is not in the form of 2m × 5n

210

77 or

30

11 is non-terminating repeating

decimal expansion.

(v) 1772 752

129

_ Denominator 22 × 52 × 717 is not in the form

of 2m × 5n

1772 752

129

is non-termination repeating

decimal expansion.

(vi)10500

987 =

50073

4773

=

500

47 = 32 52

47

Since, the denominator is of the form 2m ×

5n, the rational number has a terminating

decimal expansion.

2. Write down the decimal expansions of the

following rational numbers by writing

their denominators in the form 2m × 5n,

where m and n are non-negative

integers:

(i)8

3(ii)

125

13

(iii)80

7(iv)

625

14588

(v) 72 52

129



Solution—

(i)8

3 = 03 52

3

(_ 50 = 1 and 8 = 2 × 2 × 2 = 23)

= 33

3

52

53

= 310

5553

{Multiplying and dividing by 53}

= 1000

375 = 0.375

(ii)125

13 = 35

13(125 = 5 × 5 × 5 = 53)

= 33

3

25

213

= 310

22213


= 1000

104 = 0.104

(iii)80

7 = 14 52

7

(_ 80 = 16 × 5 = 2 × 2 × 2 × 2 × 5 = 24 × 51)

= 314

3

552

57

= 44

3

52

57

= 410

5557


= 10000

875 = 0.0875

(iv)625

14588

5 6255 1255 255 5

1

= 45

14588

= 44

4

25

214588

= 410

222214588


= 10000

233408 = 23.3408

(v) 72 52

129

= 752

5

522

2129

= 77 52

22222129


= 710

4128 =

10000000

4128

= 0.0004128

3. Write the denominator of the rational

number 5000

257 in the form 2m × 5n, where

m, n are non-negative integers. Hence,

write the decimal expansion, without

actual division.

Solution—

Denominator of the rational number 5000

257

is 5000.

Now, factors of 5000 = 2 × 2 × 2 × 5 × 5 ×

5 × 5 = (2)3 × (5)4, which is of the type 2m ×

5n, where m = 3 and n = 4 are non-negative

integers.

Rational number = 5000

257 = 43 52

257

×

2

2



[Since, multiplying numerator and

denominater by 2]

= 44 52

514

= 4)10(

514

= 10000

514 = 0.0514

Hence, which is the required decimal

expansion of the rational 5000

257 and it is also

a terminating decimal number.

4. What can you say about the prime

factorisations of the denominators of the

following rationals :

(i) 43.123456789 (ii) 43.123456789

(iii) 14285727. [CBSE 2010]

(iv) 0.120120012000120000 ..... [NCERT]

Solution—

(i) 43.123456789

_ This decimal fraction is terminating

Its denominator will be factorised in the form

of 2m × 5n where m and n are non-negative

integers.

(ii) 43.123456789

This decimal fraction is non-terminating

repeating decimals.

The denominator of their fraction will be not

in the form of 2m × 5n where m and n are

non-negative integers.

(iii) 14285727.

This decimal fraction is terminating

Its denominator will be factorised in the form

of 2m × 5n where m and n are non-negative

integers

(iv) 0.120120012000120000 .....

_ This decimal fraction in non-terminating non-

recurring

Its denominator will not be factorised in the

form of 2m × 5n where m and n are non-

negative integers

5. 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. [NCERT Exemplar]

Solution—

327.7081 is terminating decimal number. So,

it represents a rational number and also its

denominator must have the form 2m × 5n.

Thus, 327.7081 = 10000

3277081 =

q

p(say)

q = 104 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5

= 24 × 54 = (2 × 5)4

Hence, the prime factors of q is 2 and 5.

VERY SHORT ANSWER QUESTIONS

Answer each of the following questions

either in one word or one sentence or as

per requirement of the questions :

1. State Euclid’s division lemma.

Solution—

Euclid’s division lemma :

Let a and b be any two positive integers,

then there exist unique integers q and r such

that

a = bq + r, 0 < r < b

If b|a, then r = 0, otherwise r satisfies the

stronger inequality 0 < r < b.

2. State Fundamental Theorem of Arith-

metic.

Solution—

Fundamental Theorem of Arithmetics :

Every composite number can be expressed



(factorised) as a product of primes and this

factorization is unique except for the order

in which the prime factors occur.

3. Write 98 as product of its prime factors.

Solution—

2 987 497 7

1 = 2 × 7 × 7 = 21 × 72

4. Write the exponent of 2 in the prime

factorization of 144.

Solution—

2 1442 722 362 183 93 3

1 = 2 × 2 × 2 × 2 × 3 × 3 = 24 × 32

Exponent of 2 is 4

5. Write the sum of the exponents of prime

factors in the prime factorization of 98.

Solution—

98 = 2 × 7 × 7 = 21 × 72

Sum of exponents = 1 + 2 = 3

6. If the prime factorization of a natural

number n is 23 × 32 × 52 × 7, write the

number of consecutive zeros in n.

Solution—

n = 23 × 32 × 52 × 7

Number of zeros will be 52 × 22 = 102 two

zeros

7. If the product of two numbers is 1080

and their H.C.F. is 30, find their L.C.M.

Solution—

Product of two numbers = 1080

H.C.F. = 30

L.C.M. = H.C.F.

numberstwoofProduct

= 30

1080 = 36

8. Write the condition to be satisfied by q

so that a rational number q

p has a

terminating decimal expansion.

(C.B.S.E. 2008)

Solution—

In the rational number q

p, the factorization

of denominator q must be in form of 2m × 5n

where m and n are non-negative integers.

9. Write the condition to be satisfied by q

so that a rational number q

p has a non-

terminating decimal expansion.

Solution—

In the rational number q

p, the factorization

of denominator q, is not in the form of 2m ×

5n where m and n are non-negative integers.

10. Complete the missing entries in the

following factor tree.

2

3

7

(C.B.S.E. 2008)

Solution—

2

3

7

42

21

3 × 7 = 21

21 × 2 = 42

Missing entries are 42 and 21

11. The decimal expression of the rational

number 34 52

43

will terminate after how

many places of decimals.(C.B.S.E. 2009)

Solution—

The denominator of 34 52

43

is 24 × 53 which



is in the form of 2m × 5n where m and n are

positive integers

34 52

43

has terminating decimals

The decimal expansion of 34 52

43

terminates after 4 (the heighest power is 4)

decimal places

12. Has the rational number 275 752

441

a

terminating or a nonterminating decimal

representation ? [CBSE 2010]

Solution—

Because the denominator 22 × 57 × 72 is not

in the form of 2m × 5n

275 752

441

is not terimating repeating

decimal expansion

13. Write whether 52

203452 on

simplification gives a rational or an

irrational number. [CBSE 2010]

Solution—

On simplifying 52

203452 , we get

52

203452 ×

5

5

(multiplying and dividing by 5 )

= 52

10032252

=

102

103152

= 20

3030 =

20

60 = 3

Which is a rational number

14. What is an algorithm ?

Solution—

Algorithm : An algorithm is a series of well

defined slips which gives a procedure for

solving a type of problem.

15. What is a lemma ?

Solution—

A lemma is a proven statement used for

proving another statement.

16. If p and q are two prime numbers, then

what is their HCF ?

Solution—

If p and q are two primes, then their HCF

will be 1 as they have no common factor

except 1.

17. If p and q are two prime numbers, then

what is their LCM ?

Solution—

If p and q are two primes, their LCM will be

their product.

18. What is the total number of factors of a

prime number ?

Solution—

Total number of factors of a prime number

are 2, first 1 and second the number itself.

19. What is a composite number ?

Solution—

A composite number is a number which can

be factorised into more than two factors.

20. What is the HCF of the smallest

composite number and the smallest

prime number ?

Solution—

We know that 2 is the smallest prime number

and 4 is the smallest composite number

HCF of 2 and 4 = 2

21. HCF of two numbers is always a factor of

their LCM (True / False).

Solution—

True.

22. is an irrational number (True / False).

Solution—

True as value of is neither terminating norrepeating.



23. The sum of two prime numbers is always

a prime number (True / False).

Solution—

False. Sum of two prime numbers can be a

composite number e.g. 3 and 5 are prime

numbers but their sum 3 + 5 = 8 is a

composite number.

24. The product of any three consecutive

natural numbers is divisible by 6 (True /

False).

Solution—

True.

25. Every even integer is of the form 2m,

where m is an integer (True / False).

Solution—

True, as 2m is divisible by 2.

26. Every odd integer is of the form 2m – 1,

where m is an integer (True / False).

Solution—

True, as 2m is an even number but if we

subtract 1 from it, it will be odd number.

27. The product of two irrational numbers is

an irrational number (True / False).

Solution—

False, as it is not always possible that the

product of two irrational number be also an

irrational number, it may be a rational number

for example

3 × 3 = 3, 7 × 7 = 7

28. The sum of two irrational numbers is an

irrational number (True / False).

Solution—

False, as it is not always possible that the

sum of two irrational is also an irrational

number, it may be rational number also. For

example

(2 + 3 ) + (2 – 3 ) = 2 + 3 + 2 – 3

= 4

29. For what value of n, 2n × 5n ends in 5.

Solution—

In 2n × 5n,

There is no such value of n, which satisifies

the given condition.

30. If a and b are relatively prime numbers,

then what is their HCF ?

Solution—

_ a and b are two prime numbers

Their HCF = 131. If a and b are relatively prime numbers,

then what is their LCM ?

Solution—

_ a and b are two prime numbers

Their LCM = a × b32. Two numbers have 12 as their HCF and

350 as their LCM (True / False).

Solution—

HCF of two numbers = 12

and LCM is 350

False, as the HCF of two numbers is a factor

of their LCM and 12 is not a factor of 350

33. Find after how many places of decimal

the decimal form of the number

3 4 2

27

2 .5 . 3 will terminate. (CBSE 2019)

Sol. Let a 3 4 2 3 4

27 3

2 ·5 ·3 2 ·5

We know that, Let p

xq

be a rational

number.

s.t q = 2m × 5n where m, n are non-negative

integers.

Then x has a terminating decimal expansion.

Thus given number will terminate after 4

places. [Here k = larger (3, 4) = 4]

34. Express 429 as the product of its prime

factors. (CBSE 2019)

Sol. By prime factorisation,

we have

429 = 3 × 11 × 13

35. Two positive integers a and b can be

written as a = x3y2 and b = xy3, where x,

y are prime numbers. Find LCM (a, b).

(CBSE 2019)

Sol. Given a = x3y2 and b = xy3

LCM (x, y) = product of each prime factorsto greater exponent = x3y3

3 429

11 143

13



36. If HCF (336, 54) = 6, find LCM (336,

54). (CBSE 2019)

Sol. Given HCF (336, 54) = 6

We know that,

HCF (a, b) × LCM (a, b) = ab

LCM (336, 54) 336 54

30246

MULTIPLE CHOICE QUESTIONS

1. The exponent of 2 in the prime

factorisation of 144, is

(a) 4 (b) 5

(c) 6 (d) 3

Solution—

144 = 24 × 32

Exponant of 2 is 4 (a)

2. The LCM of two numbers is 1200. Which

of the following cannot be their HCF ?

(a) 600 (b) 500

(c) 400 (d) 200

Solution—

LCM of two number = 1200

Their HCF of these two numbers will be thefactor of 1200

500 cannot be its HCF (b)

3. If n = 23 × 34 × 44 × 7, then the number of

consecutive zeroes in n, where n is a

natural number, is

(a) 2 (b) 3

(c) 4 (d) 7

Solution—

Because it has four factors

n = 23 × 34 × 44 × 7

It has 4 zeroes (c)

4. The sum of the exponents of the prime

factors in the prime factorisation of 196,

is

(a) 1 (b) 2

(c) 4 (d) 6

Solution—

Prime factors of 196

= 2 × 2 × 7 × 7

= 22 × 72

Sum of exponents = 2 + 2 = 4

(c)

5. The number of decimal places after which

the decimal expansion of the rational

number 52

232

will terminate, is

(a) 1 (b) 2

(c) 3 (d) 4

Solution—

Decimal expansion of 52

232

= 20

23

= 520

523

=

100

115 = 1.15

Number of decimal places = 2 (b)6. If p

1 and p

2 are two odd prime numbers

such that p1 > p

2, then p2

1 – p2

2 is

(a) an even number (b) an odd number

(c) an odd prime number

(d) a prime number

Solution—

p1 and p

2 are two odd prime numbers such

that p1 > p

2, then

p1 – p

2 > 0

p1 – p

2 > 0

Now p21 – p2

2 = (p

1 + p

2) (p

1 – p

2)

But p1 – p

2 and p

1 + p

2 are both even number

as difference and sum of two odd numbers

are even number

p21 – p2

2 is an even number (a)

7. If two positive integers a and b are

expressible in the form a = pq2 and b =

p3q; p, q being prime numbers, then LCM

(a, b) is

(a) pq (b) p3q3

(c) p3q2 (d) p2q2

Solution—

a and b are two positive integers an

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