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NCERT Solutions for class 10 Maths chapter 1 – Real Numbers

Back Exercise

Exercise 1.1

Question 1.
Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255.
Solutions:
(i) Given numbers are 135 and 225.
On applying Euclid’s division algorithm, we have
225 = 135 x 1 + 90
Since the remainder 90 ≠ 0, so again we apply Euclid’s division algorithm to 135 and 90, to get
135 = 90 x 1 + 45
Since the remainder 45 ≠ 0, so again we apply Euclid’s division algorithm to 90 and 45, to get
90 = 45 x 2 + 0
The remainder has now become zero, so we stop.
∵ At the last stage, the divisor is 45
∴ The HCF of 135 and 225 is 45.

(ii) Given numbers are 196 and 38220
On applying Euclid’s division algorithm, we have
38220 = 196 x 195 + 0
Since we get the remainder zero in the first step, so we stop.
∵ At the above stage, the divisor is 196
∴ The HCF of 196 and 38220 is 196.

(iii) Given numbers are 867 and 255
On applying Euclid’s division algorithm, we have
867 = 255 x 3 + 102
Since the remainder 102 ≠ 0, so again we apply Euclid’s division algorithm to 255 and 102. to get
255 = 102 x 2 + 51
Since the remainder 51 ≠ 0, so again we apply Euclid’s division algorithm to 102 and 51, to get
102 = 51 x 2 + 0
We find the remainder is 0 and the divisor is 51
∴ The HCF of 867 and 255 is 51.

Question 2.
Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.
Solutions:
Let ‘a’ be any positive integer and b = 6.
∴ By Euclid’s division algorithm, we have
a = bq + r, 0 ≤ r ≤ b
a = 6q + r, 0 ≤ r ≤ b [ ∵ b = 6] where q ≥ 0 and r = 0,1, 2, 3, 4,5
Now, ‘a’ may be of the form of 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4, 6q + 5
If ‘a’ is of the form 6q, 6q + 2, 6q + 4 then ‘a’ is an even.

In above we can see clearly that the numbers of the form 6q, 6q + 2, 6q + 4 are having the factor 2.
∴ The numbers of the form 6q, 6q + 2, 6q + 4 are even.
If ‘a’ is of the form 6q +1,6q +3, 6q + 5 then ‘a’ is an odd.
As if

∵ We know that the number of the form 2k + 1 is odd.
∴ The numbers of the form 6q + 1, 6q + 3, 6q + 5 are odd.

Question 3.
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?
Solutions:
Maximum number of columns = HCF of (616, 32)
For finding the HCF we should apply Euclid’s division algorithm
Given numbers are 616 and 32
On applying Euclid’s division algorithm, we have
616 = 32 x 19 + 8
Since the remainder 8 ≠ 0, so again we apply Euclid’s division algorithm to 32 and 8, to get
32 = 8 x 4 + 0

The remainder has now become zero, so we stop,
∵ At the last stage, the divisor is 8
∴ The HCF of 616 and 32 is 8.
Therefore, the maximum number of columns in which an army contingent of 616 members can march behind an army band of 32 members in a parade is 8.

Question 4.
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.
Solutions:
Let ‘a’ be any positive integer and b = 3.
∴ By Euclid’s division algorithm, we have a = 3q + r, 0 ≤ r < b
a = 3q + r, 0 ≤ r < 3 [ ∵ b = 3] where q ≥ 0 and r = 0,1, 2
∴ a = 3q or 3q + 1 or 3q + 2
Now

Thus, the square of any positive integer is either of the form 3m or 3m + 1.

Question 5.
Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Solutions:
Let ‘a’ be any positive integer and b = 3.
∴ By Euclid’s division algorithm, we have a = bq + r,0 ≤ r ≤ b
a = 3q + r,0 ≤ r < 3 [ ∵ b = 3] where q ≥ 0 and r = 0. 1, 2
∴ a = 3q or 3q + 1 or 3q + 2
Now

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

Exercise 1.2

Question 1.
Express each number as a product of its prime factors:
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429.
Solutions:
(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429.
So, 7429 = 17 x 19 x 23

Question 2.
Find the LCM and HCF of the following pairs of integers and verify that LCM x HCF = product of two numbers,
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54
Solutions:
(i) 26 and 91

(ii) 510 and 92

(iii) 336 and 54

Question 3.
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
Solutions:
(i) 12,15 and 21

(ii) 17,23 and 29

(iii) 8, 9 and 25

Question 4.
Given that HCF (306,657) = 9, find LCM (306, 657).
Solutions:
Given that HCF (306, 657) = 9
We know that LCM x HCF = Product of two numbers

Question 5.
Check whether 6ⁿ can end with the digit 0 for any natural number n.
Solutions:
Since prime factorisation of 6ⁿ is given by 6ⁿ = (2 x 3)ⁿ = 2ⁿ x 3ⁿ
Prime factorisation of 6ⁿ contains only prime numbers 2 and 3.
6ⁿ may end with the digit 0 for some ‘n’ if 5 must be in its prime factorisation which is not present.
So, there is no natural number VT for which 6ⁿ ends with the digit zero.

Question 6.
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.
Solutions:

Both N₁ and N₂ are expressed as a product of primes. Therefore, both are composite numbers.

Question 7.
There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, 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?
Solutions:
By taking LCM of time taken (in minutes) by Sonia and Ravi, we can get the actual number of minutes after which they meet again at the starting point after both start at same point and of the same time, and go in the same direction.

Therefore, both Sonia and Ravi will meet again at the starting point after 36 minutes.

Exercise 1.3

Question 1.
Prove that √5 is irrational.
Solutions:
Let us assume that √5 is rational.
∴ There exists co-prime integers a and b (b ≠ 0) such that
√5 = $\frac { a }{ b }$ ⇒ √5b= a
Squaring on both sides, we get
5b²= a²…… (i)
⇒ 5 divides a² ⇒ 5 divides a
So, we can write a = 5c for some integer c.
From (i) and (ii)
5b² = 25c²
⇒ b² = 5c²
⇒ 5 divides b²
⇒ 5 divides b
∴ 5 is a common factor of a and b.
But this contradicts the fact that a and b are co-primes.
This contradiction has arisen because of our incorrect assumption that √5 is rational.
Hence, √5 is irrational.

Question 2.
Prove that 3 + 2√5 is irrational.
Solutions:
Let us assume that 3 + 2√5 is rational.
∴ There exists co-prime integers a and b(b ≠ 0) such that

But this contradicts the fact that √5 is irrational.
This contradiction has arisen because of our incorrect assumption that 3 + 2√5 is rational. Hence, we conclude that 3 + 2√5 is irrational.

Question 3.
Prove that the following are irrationals.

Solutions:

Exercise 1.4

Question 1.
Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

Solutions:

Question 2.
Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions.
Solutions:
In Question 1, (i), (ii), (iv), (vi), (viii) and (ix) are having terminating decimal expansion.

Question 3.
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 $\frac { p }{ q }$, what can you say about the prime factors of q?
(i) 43.123456789
(ii) 0.120120012000120000…
(iii) 43. $\overline { 123456789 }$
Solutions:
(i) 43.123456789
Since the decimal expansion terminates,so the given real number is rational and therefore of the form $\frac { p }{ q }$

Here, q = 29 x 59, So prime factorisation of q is of the form 2ⁿ x 5^m.

(ii) 0.120120012000120000…
Since the decimal expansion is neither terminating nor non-terminating repeating, therefore the given real number is not rational.

(iii) 43. $\overline { 123456789 }$
Since the decimal expansion is non-terminating repeating, therefore the given real number is rational and therefore of the form $\frac { p }{ q }$
Let x = 43. $\overline { 123456789 }$ = 43.123456789123456789… ….(i)
Multiply both sides of (i) by 1000000000, we get