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Class 10 Mathematics · Chapter 1

1. Real Numbers

Step-by-step solutions to Exercise 1.1 & 1.2. NCERT Mathematics Textbook for Class X (CBSE).

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Exercise 1.1

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

(i) 140

140 = 2 × 70 = 2 × 2 × 35 = 2 × 2 × 5 × 7

140 = 22 × 5 × 7

(ii) 156

156 = 2 × 78 = 2 × 2 × 39 = 2 × 2 × 3 × 13

156 = 22 × 3 × 13

(iii) 3825

3825 = 3 × 1275 = 3 × 3 × 425 = 3 × 3 × 5 × 85 = 3 × 3 × 5 × 5 × 17

3825 = 32 × 52 × 17

(iv) 5005

5005 = 5 × 1001 = 5 × 7 × 143 = 5 × 7 × 11 × 13

5005 = 5 × 7 × 11 × 13

(v) 7429

7429 = 17 × 437 = 17 × 19 × 23

7429 = 17 × 19 × 23

2Find 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

(i) 26 and 91

26 = 2 × 13 ; 91 = 7 × 13
HCF = 13 ; LCM = 2 × 7 × 13 = 182
LCM × HCF = 182 × 13 = 2366
Product of the numbers = 26 × 91 = 2366

HCF = 13, LCM = 182. Verified: LCM × HCF = 26 × 91 = 2366

(ii) 510 and 92

510 = 2 × 3 × 5 × 17 ; 92 = 22 × 23
HCF = 2 ; LCM = 22 × 3 × 5 × 17 × 23 = 23460
LCM × HCF = 23460 × 2 = 46920
Product of the numbers = 510 × 92 = 46920

HCF = 2, LCM = 23460. Verified: LCM × HCF = 510 × 92 = 46920

(iii) 336 and 54

336 = 24 × 3 × 7 ; 54 = 2 × 33
HCF = 2 × 3 = 6 ; LCM = 24 × 33 × 7 = 3024
LCM × HCF = 3024 × 6 = 18144
Product of the numbers = 336 × 54 = 18144

HCF = 6, LCM = 3024. Verified: LCM × HCF = 336 × 54 = 18144

3Find 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

(i) 12, 15 and 21

12 = 22 × 3 ; 15 = 3 × 5 ; 21 = 3 × 7
HCF = 3 ; LCM = 22 × 3 × 5 × 7 = 420

HCF = 3, LCM = 420

(ii) 17, 23 and 29

17, 23 and 29 are all prime numbers.
HCF = 1 ; LCM = 17 × 23 × 29 = 11339

HCF = 1, LCM = 11339

(iii) 8, 9 and 25

8 = 23 ; 9 = 32 ; 25 = 52
There is no common prime factor, so HCF = 1.
LCM = 23 × 32 × 52 = 1800

HCF = 1, LCM = 1800

4Given that HCF (306, 657) = 9, find LCM (306, 657).

We know that HCF × LCM = product of the two numbers.

LCM = 306 × 657HCF = 306 × 6579 = 34 × 657 = 22338

LCM (306, 657) = 22338

5Check whether 6n can end with the digit 0 for any natural number n.

A number ending with the digit 0 is divisible by 10, i.e., by both 2 and 5.

So, its prime factorisation must contain the prime 5.

6n = (2 × 3)n = 2n × 3n

The only primes in the factorisation of 6n are 2 and 3. The prime 5 is not there.

By the Fundamental Theorem of Arithmetic, this factorisation is unique.

No, 6n cannot end with the digit 0 for any natural number n.

6Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × 78 = 1014

It has factors other than 1 and itself (13 and 78). So, it is a composite number.

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1) = 5 × 1009 = 5045

It has factors other than 1 and itself (5 and 1009). So, it is also a composite number.

A composite number has at least one factor other than 1 and the number itself.

7There 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?

They will meet again at the starting point after a time which is the LCM of 18 and 12.

18 = 2 × 32 ; 12 = 22 × 3
LCM = 22 × 32 = 36

They will meet again at the starting point after 36 minutes

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Exercise 1.2

1Prove that √5 is irrational.

Let us assume, to the contrary, that √5 is rational.

Then we can find coprime integers a and b (b ≠ 0) such that √5 = ab.

So, b√5 = a

Squaring both sides:

5b2 = a2 ... (1)

So, 5 divides a2. Hence, 5 divides a (by Theorem 1.2).

So, we can write a = 5c for some integer c.

Substituting in (1):

5b2 = 25c2, i.e., b2 = 5c2

So, 5 divides b2. Hence, 5 divides b.

Therefore, 5 is a common factor of a and b.

But this contradicts the fact that a and b are coprime.

This contradiction has arisen because of our wrong assumption.

So, we conclude that √5 is irrational.

2Prove that 3 + 2√5 is irrational.

Let us assume, to the contrary, that 3 + 2√5 is rational.

Then we can find coprime integers a and b (b ≠ 0) such that 3 + 2√5 = ab.

2√5 = ab − 3
√5 = a − 3b2b

Since a and b are integers, a − 3b2b is rational. So, √5 is rational.

But this contradicts the fact that √5 is irrational.

This contradiction has arisen because of our wrong assumption.

So, we conclude that 3 + 2√5 is irrational.

3Prove that the following are irrationals: (i) 1√2 (ii) 7√5 (iii) 6 + √2

(i) 1√2

Let us assume, to the contrary, that 1√2 is rational.

Then we can find coprime integers a and b (b ≠ 0) such that 1√2 = ab.

So, √2 = ba (a ≠ 0)

Since a and b are integers, ba is rational. So, √2 is rational.

But this contradicts the fact that √2 is irrational.

Hence, 1√2 is irrational.

(ii) 7√5

Let us assume, to the contrary, that 7√5 is rational.

Then we can find coprime integers a and b (b ≠ 0) such that 7√5 = ab.

So, √5 = a7b

Since a and b are integers, a7b is rational. So, √5 is rational.

But this contradicts the fact that √5 is irrational.

Hence, 7√5 is irrational.

(iii) 6 + √2

Let us assume, to the contrary, that 6 + √2 is rational.

Then we can find coprime integers a and b (b ≠ 0) such that 6 + √2 = ab.

So, √2 = ab − 6 = a − 6bb

Since a and b are integers, a − 6bb is rational. So, √2 is rational.

But this contradicts the fact that √2 is irrational.

Hence, 6 + √2 is irrational.

2. Polynomials →