NCERT Solutions Class 10th Maths Chapter – 1 Real Numbers Exercise – 1.1

1. Express each number as a product of its prime factors:

(i) 140

Solution: 140
We use Prime factorization,
140 = 2 × 2 × 5 × 7 × 1
= 2² × 5 × 7

(ii) 156

Solution: 156
We use prime factorization,
156 = 2 × 2 × 13 × 3 × 1
= 2²× 13 × 3

(iii) 3825

Solution: 3825
We use prime factorization,
3825 = 3 × 3 × 5 × 5 × 17 × 1
= 3² × 5² ×17

(iv) 5005 

Solution: 5005
We use prime factorization,
5005 = 5 × 7 × 11 × 13 × 1
= 5 × 7 × 11 × 13

(v) 7429

Solution: 7429
We use prime factorization,
7429 = 17 × 19 × 23 × 1
= 17 × 19 × 23

2. 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

Solution: 26 and 91
We use prime factorization,
26 = 2 × 13 × 1
91 = 7 × 13 × 1
Thus, LCM (26, 91) = 2 × 7 × 13 × 1 = 182
And, HCF (26, 91) = 13

Verification
We use LCM × HCF = Product of two numbers
26 × 91 = 182 × 13
2366 = 2366
Hence, LCM and HCF equals to the product of 26 and 91.

(ii) 510 and 92

Solution: 510 and 92
We use prime factorization,
510 = 2 × 3 × 17 × 5 × 1
92 = 2² × 23 × 1
Thus, LCM (510, 92) = 2² × 3 × 5 × 17 × 23 = 23460
And, HCF (510, 92) = 2

Verification
We use LCM × HCF = Product of two numbers
510 × 92 = 23460 × 2
46920 = 46920
Hence, LCM and HCF equals to the product of 510 and 92.

(iii) 336 and 54

Solution: 336 and 54
We use prime factorization,
336 = 2⁴ × 7 × 3 × 1
54 = 2 × 3³ × 1
Therefore, LCM (336, 54) = 2⁴ × 3³ × 7 = 3024
And, HCF(336, 54) = 2 × 3 = 6

Verification
We use LCM × HCF = Product of two numbers
3024 × 6 = 3024 × 6
18,144 = 18,144
Hence, LCM and HCF equals to the product of 336 and 54.

3. Find the LCM and HCF of the following integers by applying the prime factorisation method.

(i) 12, 15 and 21

Solution: 12, 15 and 21
We use prime factorization,
12 = 2² × 3
15 = 5 × 3
21 = 7 × 3
HCF: (12, 15, 21) = 3
LCM: (12, 15, 21) = 2² × 3 × 5 × 7 = 420

(ii) 17, 23 and 29

Solution: 17, 23 and 29
We use prime factorization,
17 = 17 × 1
23 = 23 × 1
29 = 29 × 1
HCF: (17, 23, 29) = 1
LCM: (17, 23, 29) = 17 × 23 × 29 = 11339

(iii) 8, 9 and 25

Solution: 8, 9 and 25
We use prime factorization,
8 = 2³ × 1
9 = 3² × 1
25 = 5² × 1
HCF: (8,9,25) = 1
LCM: (8,9,25) = 2³ × 3² × 5² = 1800

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

Solution: We know,
First Number = 306
Second Number = 657
HCF (306, 657) = 9
LCM (306, 657) = ?
HCF × LCM = Product of the two given numbers
9 × LCM = 306 × 657
LCM = (306 × 657)/9 = 22338
Hence, LCM : (306, 657) = 22338

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

Solution: We know,
6ⁿ can end with zero, if 2 and 5 both are its prime factors.
6 = 2 × 3
Since, 5 is not a prime factor 6.
∴ 6ⁿ cannot end with zero.

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

Solution: 7 × 11 × 13 + 13

Factoring out 13, we obtain:
= 13 (7 × 11 × 1 + 1)
= 13 (77 + 1)
= 13 × 78
= 13 × 3 × 2 × 13
Hence, 7 × 11 × 13 + 13 is a composite number.

Now let’s take the other number,

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

Factoring out 5, we obtain:
= 5 (7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 (1008 + 1)
= 5 × 1009
Hence, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number.

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?

Solution: The minimum time they will meet is the LCM of 12 and 18.
12 = 2² × 3
18 = 2 × 3²
LCM (18, 12) = 2² × 3² = 36
Therefore, Sonia and Ravi meet at starting point after 36 minutes.