# RD SHARMA Solutions for Class 10 Maths Chapter 1 - Real Numbers

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## Chapter 1 - Real Numbers Exercise Ex. 1.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

For any positive integer n, prove that n3 - n divisible by 6.

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Prove that the square of any positive integer is of the form 5q, 5q + 1, 5q + 4 for some integer q.

Solution 9

Question 10

Solution 10

Question 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 11

Let a be any odd positive integer we need to prove that a is of the form 6q+1 , or 6q+3 , or 6q+5 , where q is some integer.

Since a is an integer consider b = 6 another integer applying Euclid's division lemma  we get
a = 6q + r f or some integer q  0, and r = 0, 1, 2, 3, 4, 5  since
r < 6.
Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
However since a is odd so a cannot take the values 6q, 6q+2 and 6q+4
(since all these are divisible by 2)

Also, 6q + 1 = 2 x 3q + 1 = 2k1 + 1, where k1 is a positive integer
6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an integer
6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3 +  1, where k3 is an integer
Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.
Therefore, 6q + 1, 6q + 3, 6q + 5 are odd numbers.

Therefore, any odd integer can be expressed is of the form
6q + 1, or 6q + 3, or 6q + 5 where q is some integer

Concept Insight:  In order to solve such problems  Euclid's division lemma is applied to two integers a and b the integer b must be taken in accordance with what is to be proved, for example here the integer b was taken 6 because a must  be of the form 6q + 1, 6q + 3, 6q + 5.

Basic definition of even (divisible by 2) and odd numbers (not divisible by 2) and the fact that addition and  multiplication of integers is always an integer are applicable here.

Question 12

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

[NCERT EXEMPLER]

Solution 12

Let x be any positive integer.

When we divide x by 6, the remainder is either 0 or 1 or 2 or 3 or 4 or 5.

So, x can be written as

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

Thus, we have the following cases:

CASE I:

When x = 6a,

x2 = 36a2 = 6(6a2) = 6m, where m = 6a2

CASE II:

When x = 6a + 1,

x2 = (6a + 1)2 = 36a2 + 12a + 1 = 6(6a2 + 2a) + 1 = 6m + 1, where m = 6a2 + 2a

CASE III:

When x = 6a + 2,

x2 = (6a + 2)2 = 36a2 + 24a + 4 = 6(6a2 + 4a) + 4 = 6m + 4, where m = 6a2 + 4a

CASE IV:

When x = 6a + 3,

x2 = (6a + 3)2 = 36a2 + 36a + 9 = (36a2 + 36a + 6) + 3 = 6(6a2 + 6a + 1) + 3 = 6m + 3, where m = 6a2 + 6a + 1

CASE V:

When x = 6a + 4,

x2 = (6a + 4)2 = 36a2 + 48a + 16 = (36a2 + 48a + 6) + 10 = 6(6a2 + 8a + 1) + 10 = 6m + 10, where m = 6a2 + 8a + 10

CASE VI:

When x = 6a + 5,

x2 = (6a + 5)2 = 36a2 + 60a + 25 = (36a2 + 60a + 6) + 19 = 6(6a2 + 10a + 1) + 19 = 6m + 19, where m = 6a2 + 10a + 19

Here, x is of the form 6m or 6m + 1 or 6m + 3 or 6m + 4 or 6m + 10 or 6mn + 19.

So, it cannot be of the form 6m + 2 or 6m + 5.

Question 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.

Solution 13

Let x be any positive integer.

Then, it is of the form 6n, 6n + 1, 6n + 2, 6n + 3, 6n + 4 or 6n + 5.

So, we have the following cases:

CASE I:

When x = 6n,

x3 = (6n)3 = 216n3 = 6(36n3) = 6q, where q = 36n3

CASE II:

When x = 6n + 1,

x3 = (6n + 1)3 = 216n3 + 108n2 + 18n + 1 = 6(36n3 + 18n2 + 3n) + 1 = 6q + 1, where q = 36n3 + 18n2 + 3n

CASE III:

When x = 6n + 2,

x3 = (6n + 2)3 = 216n3 + 216n2 + 72n + 8 = 216n3 + 216n2 + 72n + 6 + 2 =6(36n3 + 36n2 + 12n + 1) + 2 = 6q + 2, where q = 36n3 + 54n2 + 12n + 1

CASE IV:

When x = 6n + 3,

x3 = (6n + 3)3 = 216n3 + 324n2 + 162n + 27 = 216n3 + 324n2 + 162n + 24 + 3 =6(36n3 + 54n2 + 27n + 4) + 3 = 6q + 3, where q = 36n3 + 54n2 + 27n + 4

CASE V:

When x = 6n + 4,

x3 = (6n + 4)3 = 216n3 + 432n2 + 288n2 + 64 = 216n3 + 432n2 + 288n + 60 + 4 =6(36n3 + 72n2 + 48n + 10) + 4 = 6q + 4, where q = 36n3 + 72n2 + 48n + 10

CASE VI:

When x = 6n + 5,

x3 = (6n + 5)3 = 216n3 + 540n2 + 450n2 + 125 = 216n3 + 540n2 + 450n + 120 + 5 =6(36n3 + 90n2 + 75n + 20) + 5 = 6q + 5, where q = 36n3 + 72n2 + 48n + 10

Thus, the cube of any positive integer is of the form 6q + r, where q is an integer and r = 0, 1, 2, 3, 4, 5.

Question 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.

Solution 14

Given numbers are n, n + 4, n + 8, n + 12 and n + 16.

Let n = 5q + r, where 0 ≤ r < 5

n = 5q, 5q + 1, 5q + 2, 5q + 3 or 5q + 4 for any natural number q.

So, we have the following cases:

CASE I:

When n = 5q

n = 5q is divisible by 5

n + 4 = 5q + 4 is not divisible by 5

n + 8 = 5q + 8 = 5q + 5 + 3 = 5(q + 1) + 3 is not divisible by 5

n + 12 = 5q + 12 = 5q + 10 + 2 = 5(q + 2) + 2 is not divisible by 5

n + 16 = 5q + 16 = 5q + 15 + 1 = 5(q + 3) + 1 is not divisible by 5

CASE II:

When n = 5q + 1

n = 5q + 1 is not divisible by 5

n + 4 = 5q + 1 + 4 = 5q + 5 = 5(q + 1) is divisible by 5

n + 8 = 5q + 1 + 8 = 5q + 9 = 5q + 5 + 4 = 5(q + 1) + 4 is not divisible by 5

n + 12 = 5q + 1 + 12 = 5q + 13 = 5q + 10 + 3 = 5(q + 2) + 3 is not divisible by 5

n + 16 = 5q + 1 + 16 = 5q + 17 = 5q + 15 + 2 = 5(q + 3) + 2 is not divisible by 5

CASE III:

When n = 5q + 2

n = 5q + 2 is not divisible by 5

n + 4 = 5q + 2 + 4 = 5q + 6 = 5q + 5 + 1 = 5(q + 1) + 1 is not divisible by 5

n + 8 = 5q + 2 + 8 = 5q + 10 = 5(q + 2) is divisible by 5

n + 12 = 5q + 2 + 12 = 5q + 14 = 5q + 10 + 4 = 5(q + 2) + 4 is not divisible by 5

n + 16 = 5q + 2 + 16 = 5q + 18 = 5q + 15 + 3 = 5(q + 3) + 3 is not divisible by 5

CASE IV:

When n = 5q + 3

n = 5q + 3 is not divisible by 5

n + 4 = 5q + 3 + 4 = 5q + 7 = 5q + 5 + 2 = 5(q + 1) + 2 is not divisible by 5

n + 8 = 5q + 3 + 8 = 5q + 11 = 5(q + 2) + 1 is not divisible by 5

n + 12 = 5q + 3 + 12 = 5q + 15 = 5(q + 3) is divisible by 5

n + 16 = 5q + 3 + 16 = 5q + 19 = 5q + 15 + 4 = 5(q + 3) + 4 is not divisible by 5

CASE V:

When n = 5q + 4

n = 5q + 4 is not divisible by 5

n + 4 = 5q + 4 + 4 = 5q + 8 = 5q + 5 + 3 = 5(q + 1) + 3 is not divisible by 5

n + 8 = 5q + 4 + 8 = 5q + 12 = 5(q + 2) + 2 is not divisible by 5

n + 12 = 5q + 4 + 12 = 5q + 16 = 5q + 15 + 1 = 5(q + 3) + 1 is not divisible by 5

n + 16 = 5q + 4 + 16 = 5q + 20 = 5(q + 4) is divisible by 5

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

Question 15

Show that the square of an odd positive integer can be of the form 6q + 1 or 6q + 3 for some integer q.

Solution 15

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 x can be of the form 6m + 1, 6m + 3 or 6m + 5.

Thus, we have

CASE I:

x = 6m + 1

x2 = (6m + 1)2 = 36m2 + 12m + 1 = 6(6m2 + 2m) + 1 = 6q + 1, where q = 6m2 + 2m

CASE II:

x = 6m + 3

x2 = (6m + 3)2 = 36m2 + 36m + 9 = 36m2 + 36m + 6 + 3 = 6(6m2 + 6m + 1) + 3 = 6q + 3, where q = 6m2 + 6m + 1

CASE III:

x = 6m + 5

x2 = (6m + 5)2 = 36m2 + 60m + 25 = 36m2 + 60m + 24 + 1 = 6(6m2 + 10m + 4) + 1 = 6q + 1, where q = 6m2 + 10m + 4

Thus, the square of an odd positive integer can be of the form 6q + 1 or 6q + 3.

Question 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 16

By Euclid's Lemma,

a = bq + r, 0 ≤ r < b

Here, a is any positive integer and b = 3,

a = 3q + r

So, this must be in the form 3q, 3q + 1 or 3q + 2.

Now,

(3q)2 = 9q2 = 3m

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

(3q + 2)2 = 9q2 + 12q + 4 = 9q2 + 12q + 3 + 1 = 3(3q2 + 4q + 1) + 1 = 3m + 1

Thus, square of a positive integer of the form 3q + 1 is always of the form 3m + 1 or 3m for some integer m.

Question 17

Show that the square of any positive integer cannot be of the form 3m + 2, where m is a natural number.

Solution 17

Let x be any positive integer.

So, x can be written as

x = 3a or x = 3a + 1 or x = 3a + 2

Thus, we have the following cases:

CASE I:

When x = 3a,

x2 = 9a2 = 3(3a2) = 3m, where m = 3a2

CASE II:

When x = 3a + 1,

x2 = (3a + 1)2 = 9a2 + 6a + 1 = 3(3a2 + 2a) + 1 = 3m + 1, where m = 3a2 + 2a

CASE III:

When x = 3a + 2,

x2 = (3a + 2)2 = 9a2 + 12a + 4 = 9a2 + 12a + 3 + 1 = 3(3a2 + 4a + 1) + 1 = 3m + 1, where m = 3a2 + 4a + 1

Thus, the square of any positive integer cannot be of the form 3m + 2, where m is a natural number.

## Chapter 1 - Real Numbers Exercise Ex. 1.2

Question 1(i)
Find H.C.F. of 32 and 54
Solution 1(i)

Question 1(ii)
Solution 1(ii)
Question 1(iii)
Solution 1(iii)
Question 1(iv)
Solution 1(iv)
Question 1(v)
Solution 1(v)
Question 1(vi)
Solution 1(vi)
Question 1(vii)
Solution 1(vii)
Question 1(viii)
Solution 1(viii)
Question 1(ix)
Find H.C.F. of 100 and 190
Solution 1(ix)

Question 1(x)
Solution 1(x)
Question 2(i)

Use Euclid's division algorithm to find the HCF of:

135 and 225

Solution 2(i)

135 and 225

Step 1: Since 225 > 135, apply Euclid's division lemma, to a =225 and b=135 to find q and r such that 225 = 135q+r, 0 r

On dividing 225 by 135 we get quotient as 1 and remainder as 90
i.e 225 = 135 x 1 + 90

Step 2: Remainder r which is 90 0, we apply Euclid's division lemma to a = 135 and b = 90 to find whole numbers q and r such that
135 = 90 x q + r 0 r<90
On dividing 135 by 90 we get quotient as 1 and remainder as 45
i.e 135 = 90 x 1 + 45

Step 3: Again remainder r = 45 0 so we apply Euclid's division lemma to a = 90 and b = 45 to find q and r such that
90 = 45 x q + r 0 r<45
On dividing 90 by 45 we get quotient as 2 and remainder as 0
i.e 90 = 2 x 45 + 0

Step 4: Since the remainder is zero, the divisor at this stage will be HCF of (135, 225).

Since the divisor at this stage is 45, therefore, the HCF of 135 and 225 is 45.

Question 2(iv)

Use Euclid's division algorithm to find the HCF of

184, 230 and 276

Solution 2(iv)

Question 2(v)

Use Euclid's division algorithm to find the HCF

136, 170 and 255

Solution 2(v)

Question 3(i)
Solution 3(i)
Question 3(ii)
Find H.C.F. of 592 and 252 and express it as a linear combination of them.
Solution 3(ii)

Question 3(iii)
Solution 3(iii)
Question 3(iv)
Solution 3(iv)
Question 4

Find the largest number which divides 615 and 963 leaving remainder 6 in each case.

Solution 4

Question 5
Solution 5
Question 6

Solution 6
Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Using Euclid's division algorithm, find the largest number that divides 1251, 9377 and 15628 leaving remainders 1, 2 and 3 respectively.

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 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 20

Question 21

Solution 21

Question 22

Solution 22

Question 2(ii)

Use Euclid's division algorithm to find the HCF of:

196 and 38220

Solution 2(ii)

196 and 38220

Step 1: Since 38220 > 196, apply Euclid's division lemma
to a =38220 and b=196 to find whole numbers q and r such that
38220 = 196 q + r, 0 r < 196
On dividing 38220 we get quotient as 195 and remainder r as 0
i.e 38220 = 196 x 195 + 0
Since the remainder is zero, divisor at this stage will be HCF
Since divisor at this stage is 196 , therefore, HCF of 196 and 38220 is 196.

NOTE: HCF( a,b) = a if a is a factor of b. Here, 196 is a factor of 38220 so HCF is 196.

Question 2(iii)

Use Euclid's division algorithm to find the HCF of:

867 and 255

Solution 2(iii)

867 and 255

Step 1: Since 867 > 255, apply Euclid's division lemma, to a =867 and b=255 to find q and r such that 867 = 255q + r, 0 r<255
On dividing 867 by 255 we get quotient as 3 and remainder as 102
i.e 867 = 255 x 3 + 102

Step 2: Since remainder 102 0, we apply the division lemma to a=255 and b= 102 to find whole numbers q and r such that
255 = 102q + r where 0 r<102
On dividing 255 by 102 we get quotient as 2 and remainder as 51
i.e 255 = 102 x 2 + 51

Step 3: Again remainder 51 is non zero, so we apply the division lemma to a=102 and b= 51 to find whole numbers q and r such that
102 = 51 q + r where 0 r < 51

On dividing 102 by 51 quotient is 2 and remainder is 0
i.e 102 = 51 x 2 + 0
Since the remainder is zero, the divisor at this stage is the HCF
Since the divisor at this stage is 51,therefore, HCF of 867 and 255 is 51.

Concept Insight: To crack such problem remember to apply the Euclid's division Lemma which states that "Given positive integers a and b, there exists unique integers q and r satisfying a = bq + r, where 0 r < b" in the correct order.

Here, a > b.

Euclid's algorithm works since Dividing 'a' by 'b', replacing 'b' by 'r' and 'a' by 'b' and repeating the process of division till remainder 0 is reached, gives a number which divides a and b exactly.

i.e HCF(a,b) =HCF(b,r)

Note that do not find the HCF using prime factorisation in this question when the method is specified and do not skip steps.

lemma to a=135 and b=

## Chapter 1 - Real Numbers Exercise Ex. 1.3

Question 1
Solution 1

Question 2
Solution 2

Question 3
Explain why 7 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 are composite numbers.
Solution 3
Numbers are of two types - prime and composite. Prime numbers has only two factors namely 1 and the number itself  whereas composite numbers have factors other than 1 and itself.

It can be observed that
7 x 11 x 13 + 13 = 13 x (7 x 11 + 1) = 13 x (77 + 1)
= 13 x 78
= 13 x 13 x 6

The given expression has 6 and 13  as its factors. Therefore, it is a composite number.
7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 = 5 x (7 x 6 x  4 x 3 x 2 x 1 + 1)
= 5 x (1008 + 1)
= 5 x 1009

1009 cannot be factorised further. Therefore, the given expression has 5 and 1009 as its factors. Hence, it is a composite number.

Concept Insight: Definition of prime numbers and composite numbers is used. Do not miss the reasoning.

Question 4
Check whether 6n can end with the digit 0 for any natural number n.
Solution 4
If any number ends with the digit 0, it should be divisible by 10 or in other words its prime factorisation must include primes 2 and  5 both
Prime factorisation of 6n = (2 x 3)n

By Fundamental Theorem of Arithmetic Prime factorisation of a number is unique. So 5 is not a prime factor of 6n.
Hence, for any value of n, 6n will not be divisible by 5.
Therefore, 6n cannot end with the digit 0 for any natural number n.

Concept Insight: In order solve such problems the concept used is if a number is to end with zero then it must be divisible by 10 and the prime factorisation of a number is unique.

Question 5

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

Solution 5

Numbers are of two types - prime and composite. Prime numbers has only two factors namely 1 and the number itself whereas composite numbers have factors other than 1 and itself. It can be observed that 3 × 5 × 7+ 7 = 7 × (3 × 5 + 1) = 7 × (15 + 1) = 7 × 16

The given expression has 7 and 16 as its factors. Therefore, it is a composite number.

## Chapter 1 - Real Numbers Exercise Ex. 1.4

Question 1
Solution 1
Question 2
Find the LCM and HCF 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 2

Concept Insight: HCF is the product of common prime factors of all three numbers  raised to least power, while LCM is product of prime factors of all here  raised to highest power.  Use the fact that HCF is always a factor of the LCM to verify the answer. Note HCF of (a,b,c)  can also be calculated by taking two numbers at a time i.e HCF (a,b) and then HCF (b,c) .
Question 3

Given that HCF (306, 657) = 9, find LCM (306, 657).

Solution 3

Concept Insight: This problem must be solved using product of two numbers = HCF x LCM rather than prime factorisation

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Find the least number that is divisible by all the numbers between 1 and 10 (both inclusive).

Solution 14

Question 15

Solution 15

Question 16

Solution 16

## Chapter 1 - Real Numbers Exercise Ex. 1.5

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 1(iv)

Solution 1(iv)

Question 2(i)

Solution 2(i)

Question 2(ii)

Solution 2(ii)

Question 2(iii)

Solution 2(iii)

Question 2(iv)

Solution 2(iv)

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Prove that  is an irrational number.

Solution 10

Question 11

Solution 11

Question 12

Solution 12

## Chapter 1 - Real Numbers Exercise Ex. 1.6

Question 1(i)

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.

Solution 1(i)

Question 1(ii)

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.

Solution 1(ii)

Question 1(iii)

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.

Solution 1(iii)

Question 1(iv)

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.

Solution 1(iv)

Question 1(v)

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.

Solution 1(v)

Question 1(vi)

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.

Solution 1(vi)

Question 2

Solution 2

Question 3

Write the denominator of the rational number  in the form 2m × 5n, where m, n are non-negative integers. Hence, write the decimal expansion, without actual division.

Solution 3

Question 4

Solution 4

Question 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? Give reasons.

Solution 5

## Chapter 1 - Real Numbers Exercise 1.59

Question 1

The exponent of 2 in the prime factorisation of 144, is

(a) 4

(b) 5

(c) 6

(d) 3

Solution 1

Factorisation of 144 can be done as follows

so 144 = 2 × 2 × 2 × 2 × 3 × 3

= 24 × 32

Exponent of 2 in the prime factorisation of 144 is 4.

So, the correct option is (a).

Question 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 2

We know that LCM of two numbers is divisible of HCF of these two numbers.

Hence

(a) 1200 is divisible by 600. So 600 can be the HCF.

(b) 500 cannot be the HCF because 1200 is not divisible by 500.

(c) 400 can be the HCF because 1200 is divisible by 400.

(d) 200 can be the HCF because 1200 is divisible by 200.

So, the correct option is (b).

Question 3

If n = 2× 34 × 54 × 7, then the number of consecutive zeros in n, where n is a natural number, is

(a) 2

(b) 3

(c) 4

(d) 7

Solution 3

n can also be written as

34 × 23 × 53 × 5 × 7

34 × (2 × 5)3 × 5 × 7

34 × 5 × 7 × 103

exponent of 10 in n is 3.

Hence number of consecutive zeros in n is 3.

So, the correct option is (b).

Question 4

The sum of the exponents of the prime factors in the prime factorisation of 196, is

(a) 1

(b) 2

(c) 4

(c) 6

Solution 4

Factorisation of 196 is

so 196 = 2 × 2 × 7 × 7

= 22 × 72

exponent of 2 is 2

exponent of 7 is 2

Hence sum of exponents is 4.

So, the correct option is (c).

Question 5

(a) 1

(b) 2

(c) 3

(d) 4

Solution 5

So, the correct option is (b).

Question 6

(a) an even number

(b) an odd number

(c) an odd prime number

(d) a prime number

Solution 6

So, the correct option is (a).

Question 7

If two positive integers a and b are expressible in the form a = pqand b = p3q ; p, q being prime numbers,

then LCM (a, b) is

(a) pq

(b)

(c)

(d)

Solution 7

LCM (a, b) is

LCM (a, b) = p × q × q × p2

= p3q2

So, the correct option is (c).

Question 8

In Q. no. 7, HCF (a, b) is

(a) pq

(b)

(c)

(d)

Solution 8

HCF (a, b) is

No further common division is possible

Hence HCF (a, b) = p × q

= pq

So, the correct option is (a).

## Chapter 1 - Real Numbers Exercise 1.60

Question 9

If two positive numbers m and n are expressible in the form m = pq3 and n = p3q2, where p, q are prime numbers,

then HCF (m, n) =

(a) pq

(b) pq2

(c) p3q3

(d) p2q3

Solution 9

HCF of m, n is

No further division is possible

Hence HCF is p × q= pq2

So, the correct option is (b).

Question 10

If the LCM of a and 18 is 36 and the HCF of a and 18 is 2, then a =

(a) 2

(b) 3

(c) 4

(d) 1

Solution 10

We know,

LCM (a, b) × HCF (a, b) = a × b

So LCM (a, 18) × HCF (a, 18) = a × 18

36 × 2 = a × 18

a = 4

So, the correct option is (c).

Question 11

The HCF of 95 and 152, is

(a) 57

(b) 1

(c) 19

(d) 38

Solution 11

HCF (95, 152)

No further common division is possible.

Hence HCF (95, 152) is 19.

So, the correct option is (c).

Question 12

If HCF (26, 169) = 13, then LCM (26, 169) =

(a) 26

(b) 52

(c) 338

(d) 13

Solution 12

We know

LCM (a, b) × HCF (a, b) = a × b

so LCM (26, 169) × HCF (26, 169) = 26 × 169

So, the correct option is (c).

Question 13

If a = 23 × 3, b = 2 × 3 × 5, c = 3n × 5 and LCM (a, b, c) = 23 × 32 × 5 then n =

(a) 1

(b) 2

(c) 3

(d) 4

Solution 13

LCM (a, b, c) is

LCM (a, b, c) = 3 × 2 × 5 × 22 × 3n - 1

= 23 × 3n × 5            ........(1)

given that

LCM (a, b, c) = 23 × 32 × 5             ........(2)

from (1) & (2)

n = 2

So, the correct option is (b).

Question 14

(a) one decimal place

(b) two decimal places

(c) three decimal places

(d) four decimal places

Solution 14

So, the correct option is (d).

Question 15

If p and q are co - prime numbers, then p2 and q2 are

(a) coprime

(b) not coprime

(c) even

(d) odd

Solution 15

If p and q are co-prime numbers then

HCF (p, q) = 1

After squaring the numbers, we get pand q

If two numbers have HCF = 1 then after squaring the numbers their HCF remains equal to 1.

Hence HCF (p2 , q2) = 1

so p2 and q2 are co - prime numbers.

Ex : 2 and 3 are co - prime numbers.

HCF (2, 3) = 1

after squaring

HCF (4, 9) = 1

Hence, 4, 9 are also co - prime.

So, squares of two co - prime numbers are also co - prime.

So, the correct option is (a).

Question 16

(a) (i) and (ii)

(b) (ii) and (iii)

(c) (i) and (iii)

(d) (i) and (iv)

Solution 16

So, the correct option is (d).

*Note: Since the book has a typo error, the question has been modified.

Question 17

If 3 is the least prime factor of number a and 7 is the least prime factor of number b,

then least prime factor of a + b is

(a) 2

(b) 3

(c) 5

(d) 10

Solution 17

It is given that 3 is the least prime factor of number a so a can be 3 (least possible value)

It is given that 7 is the least prime factor of number b so least possible value of b is 7.

Hence a + b = 10 (least possible value)

prime factors of 10 are 2 and 5

Hence the least prime factor of a + b is 2.

So, the correct option is (a).

Question 18

(a) an integer

(b) a rational number

(c) a natural number

(d) an irrational number

Solution 18

We know that decimal expansion of a rational number is either terminating or non-terminating and recurring.

So the correct option is (b).

Question 19

(a)

(b)

(c)

(d) 3

Solution 19

Question 20

Solution 20

Question 21

If n is a natural number, then 92n - 42n is always divisible by

(a) 5

(b) 13

(c) both 5 and 13

(d) None of these

Solution 21

We know a2n  - b2n is always divisible by a - b and a + b

On comparing with 92n - 42n, we get a = 9 & b = 4

Hence 92n - 42n  is divisible by 9 - 4 & 9 + 4

= 5 & 13

So, the correct option is (c).

## Chapter 1 - Real Numbers Exercise 1.61

Question 22

If n is any natural number, then 6 -  5 always ends with

(a) 1

(b) 3

(c) 5

(d) 7

Solution 22

6always ends with 6

5n always ends with 5

Hence 6- 5n always with 6 - 5 = 1

So, the correct option is (a).

Question 23

The LCM and HCF of two rational numbers are equal, then the numbers must be

(a) prime

(b) co-prime

(c) composite

(d) equal

Solution 23

(a) If two numbers are prime then their HCF must be 1 but LCM can't be 1

Example: 2, 3

LCM (2, 3) = 6

HCF (2, 3) = 1

(b) If two numbers are co - prime then their HCF must be 1 but LCM can't be 1.

(c) If two numbers are composite then their LCM and HCF can only be equal if the two numbers are same.

(d) If the numbers are equal.

Example: 6, 6

LCM (6, 6) = 6

HCF (6, 6) = 6

LCM = HCF

So, the correct option is (d).

Question 24

If sum of LCM and HCF of two numbers is 1260 and their LCM is 900 more than their HCF, then the product of two numbers is

(a) 203400

(b) 194400

(c) 198400

(d) 205400

Solution 24

Let numbers be a, b

It is given that LCM (a, b) + HCF (a, b) = 1260           ..........(1)

LCM (a, b) - HCF (a, b) = 900             ..........(2)

Adding equations (1) and (2), we get 2LCM (a, b) = 2160

Subtracting equations (1) and (2), we get 2HCF (a, b) = 360

So, LCM (a, b) = 1080 and

HCF (a, b) = 180

We know LCM (a, b) × HCF (a, b) = ab

ab = 1080 × 180

= 194400

So, the correct option is (b).

Question 25

The remainder when the square of any prime number greater than 3 is divided by 6, is

(a) 1

(b) 3

(c) 2

(d) 4

Solution 25

Question 26

For some integer m, every even integer is of the form

1. m
2. m + 1
3. 2m
4. 2m + 1
Solution 26

m is an integer.

m = ….., -2, -1, 0, 1, 2, …..

2m = ……., -4, -2, 0, 2, 4, ……

Hence, correct option is (c).

Question 27

For some integer q, every odd integer is of the form

1. q
2. q + 1
3. 2q
4. 2q + 1
Solution 27

q is an integer.

q = ….., -2, -1, 0, 1, 2, …..

2q + 1 = ……., -3, -1, 0, 3, 5, ……

Hence, correct option is (d).

Question 28

n2 - 1 is divisible by 8, if n is

1. an integer
2. a natural number
3. an odd integer
4. an even integer
Solution 28

Let a = n2 - 1

Now, when n is odd, i.e. n = 2k + 1, we have

a = (2k + 1)2 - 1 =4k2 + 4k + 1 - 1 = 4k(k + 1)

At k = -1, we get

a = 4(-1)(-1 + 1) = 0, which is divisible by 8.

At k = 0, we get

a = 4(0)(0 + 1) = 0, which is divisible by 8.

At k = 1, we get

A = 4(1)(1 + 1) = 4(2) = 8, which is divisible by 8.

Hence, correct option is (c).

Question 29

The decimal expansion of the rational number  will terminate after

1. one decimal place
2. two decimal places
3. three decimal places
4. more than 3 decimal places
Solution 29

Question 30

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

1. xy
2. xy2
3. x3y3
4. x2y2
Solution 30

Question 31

The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is

1. 10
2. 100
3. 504
4. 2520
Solution 31

Question 32

The largest number which divides 70 and 125, leaving remainders 5 and 8, respectively, is

1. 13
2. 65
3. 875
4. 1750
Solution 32

Question 33

If the HCF of 65 and 117 is expressible in the form 65m - 117, then the value of m is

1. 4
2. 2
3. 1
4. 3
Solution 33

Question 35

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

1. 1 < r < b
2. 0 < r ≤ b
3. 0 ≤ r < b
4. 0 < r < b
Solution 35

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 0 r < b.

Hence, correct option is (c).

## Chapter 1 - Real Numbers Exercise 1.62

Question 34

The decimal expansion of the rational number  will terminate after:

1. One decimal place
2. Two decimal places
3. Three decimal places
4. Four decimal places
Solution 34