RD SHARMA SOLUTION CHAPTER-1 Real Numbers| CLASS 10TH MATHEMATICS-EDUGROWN

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
0  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

Prove that one of every three consecutive positive integers is divisible by 3.Solution 12

Let the three consecutive positive integers be m, (m + 1) and (m + 2).

By Euclid’s division lemma, when m is divided by 3, we have

m = 3q + r for some integer q ≥ 0 and r = 0, 1, 2

Case 1: When m = 3q

In this case, clearly, m is divisible by 3.

But, (m + 1) and (m + 2) are not divisible by 3.

Case 2: When m = 3q + 1

In this case, m + 2 = 3(q + 1), which is divisible by 3.

But, m and (m + 1) are not divisible by 3.

Case 3: When m = 3q + 2

In this case, m + 1 = 3(q + 1), which is divisible by 3.

But, m and (m + 2) are not divisible by 3.

Hence, one of m, (m + 1) and (m + 2) is always divisible by 3.Question 13

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 13

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 14

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 14

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 15

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 15

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 16

Show that the square of an odd positive integer can be of the form 6q + 1 or 6q + 3 for some integer q.Solution 16

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 17

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 17

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 18

Show that the square of any positive integer cannot be of the form 3m + 2, where m is a natural number.Solution 18

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 190Solution 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 276Solution 2(iv)

Question 2(v)

Use Euclid’s division algorithm to find the HCF

136, 170 and 255Solution 2(v)

Question 2(vi)

Use Euclid’s division algorithm to find the HCF of

1260 and 7344Solution 2(vi)

Step 1: Since 7344 > 1260, apply Euclid’s division lemma, to a = 7344 and b = 1260 to find q and r such that 7344 = 1260q + r, 0 ≤ r < 1260

On dividing 7344 by 1260 we get quotient as 5 and remainder as

i.e. 7344 = 1260 x 5 + 1044

Step 2: Remainder r which is 1044, we apply Euclid’s division lemma to a = 1260 and b = 1044 to find whole numbers q and r such that

1260 = 1044 x q + r, 0 ≤ r < 1044

On dividing 1260 by 1044 we get quotient as 1 and remainder as 216

i.e. 1260 = 1044 x 1 + 216

Step 3: Again remainder r = 216, so we apply Euclid’s division lemma to a = 1044 and b = 216 to find q and r such that

1044 = 216 x q + r, 0 ≤ r < 216

On dividing 1044 by 216 we get quotient as 4 and remainder as 180

i.e. 1044 = 216 x 4 + 180

Step 4: Again remainder r = 180, so we apply Euclid’s division lemma to a = 216 and b = 180 to find q and r such that

216 = 180 x q + r, 0 ≤ r < 180

On dividing 216 by 180 we get quotient as 1 and remainder as 36

i.e. 216 = 180 x 1 + 36

Step 5: Again remainder r = 36, so we apply Euclid’s division lemma to a = 180 and b = 36 to find q and r such that

180 = 36 x q + r, 0 ≤ r < 36

On dividing 180 by 36 we get quotient as 5 and remainder as 0

i.e. 180 = 36 x 5 + 0

Step 6: Since the remainder is zero, the divisor at this stage will be HCF of (7344, 1260).

Since the divisor at this stage is 36, therefore, the HCF of 7344 and 1260 is 36.Question 2(vii)

Use Euclid’s division algorithm to find the HCF of

2048 and 960Solution 2(vii)

Step 1: Since 2048 > 960, apply Euclid’s division lemma, to a = 2048 and b = 960 to find q and r such that 2048 = 960q + r, 0 ≤ r < 960

On dividing 2048 by 960 we get quotient as 5 and remainder as

i.e. 2048 = 960 x 2 + 128

Step 2: Remainder r which is 128, we apply Euclid’s division lemma to a = 960 and b = 128 to find whole numbers q and r such that

960 = 128 x q + r, 0 ≤ r < 128

On dividing 960 by 128 we get quotient as 7 and remainder as 216

i.e. 960 = 128 x 7 + 64

Step 3: Again remainder r = 64, so we apply Euclid’s division lemma to a = 128 and b = 64 to find q and r such that

128 = 64 x q + r, 0 ≤ r < 64

On dividing 128 by 64 we get quotient as 2 and remainder as 0

i.e. 128 = 64 x 2 + 0

Step 4: Since the remainder is zero, the divisor at this stage will be HCF of (2048, 960).

Since the divisor at this stage is 64, therefore, the HCF of 2048 and 960 is 64.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 255Solution 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 3Explain 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 3Numbers 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 4Check whether 6ⁿ can end with the digit 0 for any natural number n.Solution 4If 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 6_ⁿ = (2 x 3)_ⁿ

By Fundamental Theorem of Arithmetic Prime factorisation of a number is unique. So 5 is not a prime factor of 6^_n.
Hence, for any value of n, 6n will not be divisible by 5.
Therefore, 6^_n 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 1(iv)

Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = Product of the integers: 

404 and 96Solution 1(iv)

404 = 2 × 2 × 3 × 37 = 2² × 101

96 = 2 × 2 × 2 × 2 × 2 × 3 = 2⁵ × 3

H.C.F. = 2² = 4

L.C.M. = 2⁵ × 3 × 101 = 9696

Now, H.C.F. × L.C.M. = 4 × 9696 = 38784

Product of numbers = 404 × 96 = 38784

Hence, H.C.F. × L.C.M. = Product of numbers.Question 2Find 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 factorisationQuestion 3(ii)

Write the smallest number which is divisible by both 306 and 657.Solution 3(ii)

The smallest number divisible by both 306 and 657 is the LCM if these numbers.

306 = 2 × 3 × 3 × 17 = 2 × 3² × 17

657 = 3 × 3 × 73 = 3² × 73

L.C.M. = 2 × 3² × 17 × 73 = 22338

Hence, the smallest number which is divisible by both 306 and 657 is 22338. 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

Question 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?Solution 17

Required minimum distance each should walk so that they can cover the distance in complete step is the L.C.M. of 30 cm, 36 cm and 40 cm.

30 = 2 × 3 × 5;

36 = 2²× 3²;

40 = 2³× 5

∴ LCM (30, 36, 40) = 2³× 3²× 5

∴ LCM = 2³× 3²× 5 = 360 cm.Question 18

Find the largest number which on dividing 1251, 9377 and 15628 leaves remainders 1, 2 and 3 respectively.Solution 18

Clearly, the required number is the H.C.F of the numbers

 1251 – 1 = 1250, 9377 – 2 = 9375, and 15628 – 3 = 15625.

First we’ll find the H.C.F of 1250 and 9375 by Euclid’s algorithm as given below:

9375 = 1250 × 7 + 625

1250 = 625 × 2 + 0

Clearly, H.C.F of 1250 and 9375 is 625.

Let us now find the H.C.F of 625 and the third number 15625 by Euclid’s algorithm:

15625 = 625 × 25 + 0

Hence, the required number is 625.

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

Given that   is irrational, prove that   is an irrational number.Solution 11

Let us assume that   is a rational number.

So, there exist co-prime positive integers a and b such that

As  is rational, so   is rational.

Then,   is also rational.

Thus,   is rational.

But this is a contradiction to the fact that   is an irrational number.

Hence,   is an irrational number.Question 12

Prove that   is an irrational number, given that   is an irrational number.Solution 12

Let us assume that   is a rational number.

So, there exist co-prime positive integers a and b such that

As  is rational, so   is rational.

Then,   is also rational.

Thus,   is rational.

But this is a contradiction to the fact that   is an irrational number.

Hence,   is an irrational number.Question 13

Prove that   is an irrational number, given that   is an irrational number.Solution 13

Let us assume that   is a rational number.

So, there exist co-prime positive integers a and b such that

As  is rational, so   is rational.

Then,   is also rational.

Thus,   is rational.

But this is a contradiction to the fact that   is an irrational number.

Hence,   is an irrational number.Question 14

Solution 14

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 2^m × 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) 3Solution 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) 200Solution 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 = 2³ × 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) 7Solution 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) 6Solution 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) 4Solution 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 numberSolution 6

So, the correct option is (a).Question 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) 

(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) p²q³Solution 9

HCF of m, n is

No further division is possible

Hence HCF is p × q² = 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) 1Solution 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) 38Solution 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) 13Solution 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) 4Solution 13

LCM (a, b, c) is

LCM (a, b, c) = 3 × 2 × 5 × 2² × 3^{n – 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 placesSolution 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) oddSolution 15

If p and q are co-prime numbers then

HCF (p, q) = 1

After squaring the numbers, we get p² and q² 

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) 10Solution 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 numberSolution 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) 3Solution 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 theseSolution 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 6ⁿ  –  5ⁿ  always ends with

(a) 1

(b) 3

(c) 5

(d) 7Solution 22

6ⁿ always ends with 6

5ⁿ always ends with 5

Hence 6ⁿ – 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) equalSolution 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) 205400Solution 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) 4Solution 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 largest number which divides 70 and 125, leaving remainders 5 and 8, respectively, is

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

Solution 31

Question 34

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 34

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).Question 32

The product of a non-zero rational number and an irrational number is

(a) always irrational

(b) always rational

(c) rational or irrational

(d) oneSolution 32

The product of a non-zero rational number and an irrational number is always irrational. Question 33

The HCF and LCM of 12, 21, 15 respectively are

(a) 3, 140

(b) 12, 420

(c) 3, 420

(d) 420, 3Solution 33

Here, 12 = 2²× 3, 21 = 3 × 7 and 15 = 3 × 5

HCF = 3

LCM = 2²× 3 × 5 × 7 = 420