NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers

NCERT
SOLUTIONS
CLASS - 10th

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Class : 10th
Subject : Maths
Chapter : 1
Chapter Name : Real Numbers

Exercise 1.1

Q1 Use Euclid’s division algorithm to nd the HCF of :
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255

Answer. (i) 135 and 225
Since 225 > 135, we apply the division lemma to 225 and 135 to obtain
225 = 135 x 1 + 90
Since remainder 90 ≠ 0, we apply the division lemma to 135 and 90 to
obtain
135 = 90 x 1 + 45
We consider the new divisor 90 and new remainder 45, and apply the
division lemma to obtain
90 = 2 x 45 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 45,
Therefore, the HCF of 135 and 225 is 45.

(ii) 196 and 38220
Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain
38220 = 196 x 195 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 196,
Therefore, HCF of 196 and 38220 is 196.

(iii) 867 and 255
Since 867 > 255, we apply the division lemma to 867 and 255 to obtain
867 = 255 x 3 + 102
Since remainder 102 ≠ 0 we apply the division lemma to 2S5 and 102 to obtain
255 = 102 x 2+ 51

We consider the new divisor 102 and new remainder 51, and apply the division lemma to
obtain
102 = 51 x 2+0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 51,
Therefore, HCF of 867 and 255 is 51.

Page : 7 , Block Name : Exercise 1.1

Q2 Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is
some integer?

Answer. Let a be any positive integer and b = 6. Then, by Euclid's algorithm,
a = 6q + r for some integer q 0, and r = 0, 1, 2, 3, 4, 5 because 0 ≤ r < 6.
Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5
Also, 6q + 1 = 2 x 3q +1 = 2k + 1, where k is a positive integer
1 1

6q + 3 = (6q + 2) + 1 = 2(3q + 1) + 1 = 2k + 1, where k is an integer
2 2

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 not exactly divisible by 2.
Hence, these expressions of numbers are odd numbers.
And therefore, any odd integer can be expressed in the form 6q + 1,
or 6q + 3,
or 6q + 5

Page : 7 , Block Name : Exercise 1.1

Q3 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?

Answer. HCF (616, 32) will give the maximum number of columns in which they can march.
We can use Euclid's algorithm to nd the HCF.
616 = 32 x 19 + 8
32 = 8 x 19 +8
The HCF (616, 32) is 8.
Therefore, they can march in 8 columns each.

Page : 7 , Block Name : Exercise 1.1

Q4 Use Euclid’s division lemma to show that the square of any positive integer is either of the
form 3m or 3m + 1 for some integer m.
[Hint : Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each
of these and show that they can be rewritten in the form 3m or 3m + 1.]

Answer. Let a be any positive integer and b = 3.
Then a = 3q + r for some integer q ≥ 0
And r = 0, 1, 2 because 0 ≤ r < 3
Therefore, a = 3q or 3q + 1 or 3q + 2
Or,
2 2 2 2
a = (3q) or (3q + 1) or (3q + 2)

2 2 2 2
a = (9q ) or 9q + 6q + 1 or 9q + 12q + 4

2 2 2
= 3 × (3q ) or 3 (3q + 2q) + 1 or 3 (3q + 4q + 1) + 1

= 3k1 or 3k2 + 1 or 3k3 + 1

Where k , k , and k are some positive integers
1 2 3

Hence, it can be said that the square of any positive integer is either of
the form 3m or 3m + 1

Page : 7 , Block Name : Exercise 1.1

Q5 Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m,
9m + 1 or 9m + 8.

Answer. Let a be any positive integer and b = 3
a = 3q + r, where q ≥ 0 and 0 ≤ r < 3

∴ a = 3q or 3q + 1 or 3q + 2

Therefore, every number can be represented as these three forms.
There are three cases.
Case 1: When a = 3q,
a = (3q) = 27q = 9 (3q ) = 9m,
3 3 3 3

Where m is an integer such that m = 3q 3

Case 2 : when a = 3q + 1

3 3
a = (3q + 1)

3 3 2
a = 27q + 27q + 9q + 1

3 3 2
a = 9 (3q + 3q + q) + 1

3
a = 9m + 1

3 2
Where m is an integer such that m = (3q + 3q + q)

Case 3 : when a = 3q + 2

3 3
a = (3q + 2)

3 3 2
a = 27q + 54q + 36q + 8

3 3 2
a = 9 (3q + 6q + 4q) + 8

3
a = 9m + 8

3 2
Where m is an integer such that m = (3q + 6q + 4q)

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

Page : 7 , Block Name : Exercise 1.1

Exercise 1.2

Q1 Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429

Answer.
2
(i) 140 = 2 × 2 × 5 × 7 = 2 × 5 × 7

2
(ii) 156 = 2 × 2 × 3 × 13 = 2 × 3 × 13

2 2
(iii) 3825 = 3 × 3 × 5 × 5 × 17 = 3 × 5 × 17

(iv) 5005 = 5 × 7 × 11 × 13

(v) 7429 = 17 × 19 × 23

Page : 11 , Block Name : Exercise 1.2

Q2 Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF =
product of the two numbers.
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54

Answer. (i) 26 and 91
26 = 2 x 13
91 = 7 x 13
HCF = 13
LCM = 2x7x13 = 182
Product of the two numbers 26 x 91 = 2366
HCF x LCM = 13 x 182 = 2366
Hence, product of two numbers = HCF x LCM

(ii) 510 and 92
510 = 2 x 3 x 5 x 17
92 = 2 x 2 x 23
HCF = 2
LCM = 2 x 2 x 3 x 5 x 17 x 23 = 23460

Product of the two numbers 510 x 92 = 46920
HCF x LCM = 2 x 23460
= 46920
Hence, product of two numbers = HCF x LCM M

(iii) 336 and 54
336 = 2 x 2 x 2 x 2 x 3 x 7
4
336 = 2 × 3 × 7

54 = 2 × 3 × 3 × 3

3
54 = 2 × 3

HCF = 2 × 3 = 6

4 3
LCM = 2 × 3 × 7 = 3024

Product of the two numbers = 336 × 54 = 18144

HCF × LCM = 6 × 3024 = 18144

Hence, product of two numbers = HCF x LCM

Page : 11 , Block Name : Exercise 1.2

Q3 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?

Answer. (i) 12, 15 and 21
2
12 = 2 × 3

15 = 3 x 5
21 = 3 x 7
HCF = 3
LCM = 2 × 3 × 5 × 7 = 420
2

(ii) 17,23 and 29
17 = 1 x 17
23 = 1 x 23
29 = 1 x 29
HCF = 1
LCM = 17 x 23 x 29 = 11339

(iii) 8,9 and 25
8=2x2x2
9=3x3
25 = 5 x 5
HCF = 1
LCM = 2 x 2 x 2 x 3 x 3 x 5 x 5= 1800

Page : 11 , Block Name : Exercise 1.2

Q4 Given that HCF (306, 657) = 9, nd LCM (306, 657).

Answer. H.C.F (306, 657) = 9
We know that, LCM x HCF = Product of two numbers
Therefore, LCM x HCF = 306 x 657
306×657 306×657
LCM = =
HCF 9

LCM − 22338

Page : 11 , Block Name : Exercise 1.2

Q5 Check whether 6 can end with the digit 0 for any natural number n?
n

Answer. If any number ends with the digit 0, it should be divisible by 10 or in
other words, it will also be divisible by 2 and 5 as 10 = 2 x 5
Prime factorisation of 6 = (2 × 3)
n n

It can be observed that 5 is not in the prime factorisation of 6
n

Hence, for any value of n, 6 will not be divisible by 5.
n

Therefore, 6 cannot end with the digit 0 for any natural number n.
n

Page : 11 , Block Name : Exercise 1.2

Q6 Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers?

Answer. Numbers are of two types - prime and composite. Prime numbers can
be divided by 1 and only 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
= 3 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.

Page : 11 , Block Name : Exercise 1.2

Q7 There is a circular path around a sports eld. Sonia takes 18 minutes to drive one round of

the eld, while Ravi takes 12 minutes for the same. Suppose they both start at the same point
and at the same time, and go in the same direction. After how many minutes will they meet
again at the starting point?

Answer. It can be observed that Ravi takes lesser time than Sonia for completing 1 round of
the circular path. As they are going in the same direction, they will meet again at the same
time when Ravi will have completed 1 round of that circular path with respect to Sonia. And
the total time taken for completing this 1 round of circular path will be the LCM of time taken
by Sonia and Ravi for completing 1 round of circular
path respectively i.e., LCM of 18 minutes and 12 minutes.
18 = 2 x 3 x 3
And, 12 = 2 x 2 x 3
LCM of 12 and 18 = 2 x 2 x 3 x 3 = 36
Therefore, Ravi and Sonia will meet together at the starting point after
36 minutes.

Page : 11 , Block Name : Exercise 1.2

Exercise 1.3

Q1 Prove that √5 is irrational?

Answer. Let √5 is a rational number.
Therefore, we can nd two integers a, b (b ≠ 0) such that √5 =
a

b

Let a and b have a common factor other than 1. Then we can divide
them by the common factor, and assume that a and b are co-prime.
a = √5b

2 2
a = 5b

Therefore, a is divisible by 5 and it can be said that a is divisible by 5.
2

Let a = 5k, where k is an integer.
2 2
(5k) = 5b

b
2
This means that b is divisible by 5 and hence, b is divisible by 5.
= 5k
2 2

This implies that a and b have S as a common factor.
And this is a contradiction to the fact that a and b are co-prime.
p
Hence, √5 cannot be expressed as or it can be said that √5 is
q

irrational.

Page : 14 , Block Name : Exercise 1.3

Q2 Prove that 3 + 2√5 is irrational?

Answer. Let 3 + 2√5 is rational.
Therefore, we can find two integers a, b(b ≠ 0) such that

a
3 + 2 √5 =
b

a
2 √5 = − 3
b

1 a
√5 = ( − 3)
2 b

Since a and b are integers, will also be rational and therefore ,√5 is rational.
1 a
( − 3)
2 b

This contradicts the fact that √5 is irrational. Hence, our assumption
that3 + 2√5 is rational is false. Therefore, 3 + 2√5 is irrational.

Page : 14 , Block Name : Exercise 1.3

Q3 Prove that the following are irrationals :
(i)
1

√2

(ii) 7√5
(iii) 6 + √2

Answer. (i) 1

√2

Let is rational.
1

√2

Therefore, we can nd two integers a, b (b ≠ 0) such that
1 a
=
√2 b

√2 =
b
.
a

b
is rational as a and b are integers.
a

Therefore,√2 is rational which contradicts to the fact that √2 irrational.
Hence, our assumption is false and is irrational.1

√2

(ii) 7√5
Let 7√5 is rational.
Therefore, we can find two integers a, b(b ≠ 0) such that

a
7 √5 = for some integers a and b
b

a
∴ √5 =
7b

is rational as a and b are integers.
a

7b

Therefore, √5 should be rational .
This contradicts the fact that √5 is irrational. Therefore ,our assumption that 7√5 is
irrational .

(iii) 6 + √2
Let 6 + √2 be rational .
Therefore we can nd two integers a, b (b ≠ 0) such that

a
6 + √2 =
b

a
√2 = − 6
b

Since a and b are integers, is also rational and hence,√2 should
a
− 6
b

be rational. This contradicts the fact that √2 is irrational. Therefore,
our assumption is false and hence, 6 + √2 is irrational.

Page : 14 , Block Name : Exercise 1.3

Exercise 1.4

Q1 Without actually performing the long division, state whether the following rational
numbers will have a terminating decimal expansion or a non-terminating repeating decimal
expansion:
(i) 13

3125

(ii)
17

8

(iii)
64

455

(iv)
15

1600

(v) 29

343
(
vi)
23
3 2
2 5

(vii) 2
129

1 5
2 5 7

(viii) 6

15

(ix)
35

50

(x)
77

210

Answer.
(i) 13

3125
5
3125 = 5

The denominator is of the form 5 m

Hence, the decimal expansion of 13

3125
is terminating.

(ii) 17

8
3
8 = 2

The denominator is of the form 2 m

Hence, the decimal expansion of 17

8
is terminating.

(iii) 64

455

455 = 5 x 7 x 13

Since the denominator is not in the form 2 × 5 , and it also contains
m n

7 and 13 as its factors, its decimal expansion will be non-terminating repeating.

15

(iv) 1600

6 2
1600 = 2 × 5

The denominator is of the form 2 m
× 5
n

Hence, the decimal expansion of 1600
15
is terminating.

29

(v) 343

3
343 = 7

Since the denominator is not in the form 2 × 5 , and it has 7 as its factor, the decimal
m n

expansion of is non-terminating repeating.
29

343

23

3 2
(vi) 2 ×5

The denominator is of the form 2 .
m n
× 5

Hence, the decimal expansion of is terminating.
23
3 2
2 ×5

(vii) 2
129
7 5
2 ×5 ×7

Since the denominator is not of the form 2 , and it also has 7 as
m n
× 5

its factor, the decimal expansion of 2
129

7 5
is non-terminating repeating.
2 ×5 ×7

(viii)
6 2×3 2
= =
15 3×5 5

The denominator is of the form 5 . n

Hence, the decimal expansion of is terminating.
6

15

35 7×5 7
= =
(ix) 50 10×5 10

10 = 2 × 5

The denominator is of the form 2 m
× 5 .n

Hence, the decimal expansion of 35

50
is terminating.

77 11×7 11
= =
(x) 210 30×7 30

30 = 2 × 3 × 5

Since the denominator is not of the form 2 × 5 , and it also has 3 as
m n

its factors, the decimal expansion of is non-terminating repeating.
77

210

Page : 17 , Block Name : Exercise 1.4

Q2 Write down the decimal expansions of those rational numbers in Question 1 above which
have terminating decimal expansions.

Answer. (i)
13
= 0.00416
3125

(ii)
17
= 2.125
8

(iv)
15
= 0.009375
1600

(vi) 3
23
2
=
23

200
= 0.115
2 ×5

(viii) 6

15
=
2×3

3×5
=
2

5
= 0.4

(ix) 35

50
= 0.7

Page : 18 , Block Name : Exercise 1.4

Q3 The following real numbers have decimal expansions as given below. In each case, decide
whether they are rational or not. If they are rational, and of the form , p q what can you say
about the prime factors of q?
(i) 43.123456789
(ii) 0.120120012000120000. . .
(iii) 43.123456789
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Answer. (i) 43.123456789
p
Since this number has a terminating decimal expansion, it is a rational number of the form q

and q is of the form 2 × 5
m n

i.e., the prime factors of q will be either 2 or 5 or both.

(ii) 0.120120012000120000
The decimal expansion is neither terminating nor recurring. Therefore, the given number is an
irrational number.

(iii) 43.123456789
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Since the decimal expansion is non-terminating recurring, the given number is a rational
number of the form and q is not of the form and q is not of the form 2 × 5 i.e., the prime
p m n

q

factors of q will also have a factor other than 2 or 5.

Page : 18 , Block Name : Exercise 1.4