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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