# NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2

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## 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
Solution:
(i) 140
140 = 2 x 2 x 7 x 5
140 = 22 x 7 x 5

(ii) 156 = 2 x 2 x 3 x 13 = 22 x 3 x 13

(iii) 3825 = 5 x 5 x 3 x 3 x 17
= 32 x 52 x 17

(iv) 5005 = 5 x 7 x 11 x 13

(v) 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 the two numbers.
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54
Solution:
The Prime factorisation of 26 and 91 is
26 = 2 x 13
91 = 7 x 13
Therefore, the LCM is 2 x 7 x 13 = 182.
Also HCF is 13.
By using the formula LCM x HCF = Product of two numbers
182 x 13 = 26 x 91
2366 = 2366
∴ LMC x HCF = Product of two numbers. [Verified]

(ii) 510 and 92
The prime factorisation of 510 and 92 is
510 = 2 x 3 x 5 x 17 92 = 2 x 2 x 23
Therefore, the LCM is 2 x 2 x 3 x 5 x 17 x 23 = 23,460 and HCF is 2.
Now, LCM x HCF = Product of two numbers
L.H.S. = LCM x HCF = 23640 x 2 = 46920
R.H.S. = Product of two numbers
= 510 x 92 = 46920
L.H.S. = R.H.S. [Verified]

(iii) 336 and 54
The prime factorisation of 336 and 54 is
336 = 24 x 3 x 7 54
= 2 x 3³
Therefore, the LCM is 24 x 3³ x 7 = 3024
and HCF is 2 x 3 = 6
Now, LCM x HCF = Product of two numbers
L.H.S. = LCM x HCF = 3024 x 6 = 18144
R.H.S. = Product of two numbers = 336 x 54 = 18144
Now, L.H.S.= R.H.S. [Verified]

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
Solution:
(i) 12,15 and 21
12 = 2 x 2 x 3
15 = 3 x 5
21 = 3 x 7
Therefore the LCM is 2² x 3 x 7 = 420
And the HCF is 3! = 3

(ii) 17, 23 and 29
17 = 1 x 17
23 = 1 x 23
29 = 1 x 29
Therefore tht LCM is 1 x 17 x 23 x 29 = 11339
And the HCF = 1

(iii) 8,9 and 25
8 = 2 x 2 x 2 x 1
9= 3 x 3 x 1
25 = 5 x 5 x 1
Therefore the LCM is 1 x 2³ x 3² x 5² = 1800
And HCF is = 1

Question 4.
Given that HCF (306,657) = 9, find LCM (306, 657).
Solution:
HCF (306, 657) = 9 We know that
HCF x LCM = Product of two numbers
9 x LCM = 306 x 657
LCM = $$\frac { 201042 }{ 9 }$$
LCM = 22338

Question 5.
Check wether 6n can end with the digit 0 for any natural number n.
Solution:
If the number 6n for any n, were to end with digit zero, then it will be divisible by 5. That is, the prime factorisation of 6n would contain the prime 5. This is not possible because 6n = (2 x 3)n = 2n. 3n, so the only prime in the factorisation of 6n is 2. So the uniqnessess of the fundamental theorem of arithemetic gurantees that, there are no other primes in the factorisation of 6n.

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.
Solution:
We know by fundamental theorem of arithmetic, “every composite number can be factorised as a product of primes”. So we can explain 7 x 11 x 13 + 13 and 7 x 3 x 2 x 5 x 2 x 2 x 3 x 2 x 1 + 5 are composite numbers or 7 x 11 x 13 + 13 and 7 x 32 x 24 + 5 are composite numbers.

Question 7.
There is a circular path around a sports field. Sonia takes 18 minutes to derive one round of the field, while Ravi takes 12 minutes for 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:
L.C.M (12,18)
12 = 2 x 2 x 3
18 = 2 x 3 x 3
So the L.C.M of 12 and 18 is 2 x 3 x 2 x 3 = 36 After 36 minutes they will will meet at the starting point.

We hope you found these NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2 helped you with finding the solutions easily.

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