NCERT Solutions Class 10 Maths Chapter 1 Real Numbers Ex 1.2

NCERT Solutions Class 10 Maths
Chapter – 1 (Real Numbers)

The NCERT Solutions in English Language for Class 10 Mathematics Chapter – 1 Real Numbers Exercise 1.2 has been provided here to help the students in solving the questions from this exercise.

Chapter : 1 Real Numbers

Exercise – 1.2

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

Solutions –
(i) 140
By taking the LCM of 140, we will get the product of its prime factor.
140 = 2 × 2 × 5 × 7 × 1 = 2²× 5 × 7

(ii) 156
By Taking the LCM of 156, we will get the product of its prime factor.
156 = 2 × 2 × 13 × 3 × 1 = 2²× 13 × 3

(iii) 3825
By taking the LCM of 3825, we will get the product of its prime factor.
3825 = 3 × 3 × 5 × 5 × 17 × 1 = 3²× 5²× 17

(iv) 5005
By Taking the LCM of 5005, we will get the product of its prime factor.
5005 = 5 × 7 × 11 × 13 × 1 = 5 × 7 × 11 × 13

(v) 7429
By taking the LCM of 7429, we will get the product of its prime factor.
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
(ii) 510 and 92
(iii) 336 and 54

Solutions –

(i) 26 and 91
Expressing 26 and 91 as product of its prime factors, we get,
26 = 2 × 13 × 1
91 = 7 × 13 × 1
Therefore, LCM (26, 91) = 2 × 7 × 13 × 1 = 182
And HCF (26, 91) = 13
Verification Now, product of 26 and 91 = 26 × 91 = 2366
And product of LCM and HCF = 182 × 13 = 2366
Hence, LCM × HCF = product of the 26 and 91.

(ii) 510 and 92
Expressing 510 and 92 as product of its prime factors, we get,
510 = 2 × 3 × 17 × 5 × 1
92 = 2 × 2 × 23 × 1
Therefore, LCM(510, 92) = 2 × 2 × 3 × 5 × 17 × 23 = 23460
And HCF (510, 92) = 2
Verification
Now, product of 510 and 92 = 510 × 92 = 46920
And Product of LCM and HCF = 23460 × 2 = 46920
Hence, LCM × HCF = product of the 510 and 92.

(iii) 336 and 54
Expressing 336 and 54 as product of its prime factors, we get,
336 = 2 × 2 × 2 × 2 × 7 × 3 × 1
54 = 2 × 3 × 3 × 3 × 1
Therefore, LCM(336, 54) = 3024
And HCF(336, 54) = 2×3 = 6
Verification
Now, product of 336 and 54 = 336 × 54 = 18,144
And product of LCM and HCF = 3024 × 6 = 18,144
Hence, LCM × HCF = product of the 336 and 54.

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
Writing the product of prime factors for all the three numbers, we get,
12 = 2 × 2 × 3
15 = 5 × 3
21 = 7 × 3
Therefore,
HCF (12, 15, 21) = 3
LCM (12, 15, 21) = 2 × 2 × 3 × 5 × 7 = 420

(ii) 17, 23 and 29
Writing the product of prime factors for all the three numbers, we get,
17 = 17 × 1
23 = 23 × 1
29 = 29 × 1
Therefore,
HCF (17, 23, 29) = 1
LCM (17, 23, 29) = 17 × 23 × 29 = 11339

(iii) 8, 9 and 25
Writing the product of prime factors for all the three numbers, we get,
8 = 2 × 2 × 2 × 1
9 = 3 × 3 × 1
25 = 5 × 5 × 1
Therefore,
HCF (8, 9, 25) = 1
LCM (8, 9, 25) = 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800

4. Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution – As we know that,
HCF × LCM = Product of the two given numbers
Therefore,
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 – 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 × 5
Prime factorisation of 6ⁿ= (2 ×3)ⁿIt can be observed that 5 is not in the prime factorisation of 6ⁿ.
Hence, for any value of n, 6ⁿ will not be divisible by 5.
Therefore, 6ⁿ cannot end with the digit 0 for any natural number n.

6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution – 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 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × (77 + 1)
= 13 × 78
= 13 ×13 × 6
The given expression has 6 and 13 as its factors. Therefore, it is a composite number.
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5 × (7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 × (1008 + 1)
= 5 × 1009
1009 cannot be factorised further. Therefore, the given expression has 5 and 1009 as its factors. Hence, it 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 – Since, Both Sonia and Ravi move in the same direction and at the same time, the method to find the time when they will be meeting again at the starting point is LCM of 18 and 12.
Therefore, LCM (18, 12) = 2 × 3 × 3 × 2 × 1 = 36
Hence, Sonia and Ravi will meet again at the starting point after 36 minutes.

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NCERT Solutions Class 10 Maths Chapter 1 Real Numbers Ex 1.2