NCERT solutions for class 10 Maths
Chapter 1
Real Numbers
Exercise 1.2
Ex 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 = 2 × 2 × 5 × 7 = 2² × 5 × 7
(ii) 156 = 2 × 2 × 3 × 13 = 2² × 3 × 13
(iii) 3825 = 3 × 3 × 5 × 5 × 17 = 3² × 5² × 17
(iv) 5005 = 5 × 7 × 11 × 13
(v) 7429 = 17 × 19 × 23
Ex 1.2 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:
(i) 26 and 91
26 = 2 ×13
91= 7 × 13
HCF = 13
LCM = 2 × 7 × 13
Product of two numbers = 26 × 91 = 2366
HCF × LCM = 13 × 182 = 2366
Hence, product of two numbers = HCF × LCM
(ii) 510 and 92
510 = 2 × 3 × 5 × 17
9 = 2 × 2 × 23
HCF = 2
LCM = 2 × 2 × 3 × 5 × 17 × 23 = 23460
Product of two numbers = 510 × 92 = 46920
HCF × LCM = 2 × 23460 = 46920
Hence, product of two numbers = HCF × LCM
(iii) 336 and 54
336 = 2 × 2 × 2 × 3 × 7 ⇒ 336 = 24 × 3 × 7
54 = 2 × 3 × 3 × 3 ⇒ 2 × 3³
HCF = 2 × 3 = 6
LCM = 24 × 3³ × 7 = 3024
Product of two numbers = 336 × 54 = 18144
HCF × LCM = 6 × 3024 = 18144
Hence, product of two numbers = HCF × LCM
Ex 1.2 Question 3.
Find the LCM and HCF of the following integers by applying the prime factorization method:
(i) 12, 15 and 21
(ii) 17, 23 and 29
(iii) 8, 9 and 25
Solution:
(i) 12, 15 and 91
12 = 2² ×3
15 = 3 × 15
91 = 3 × 7
HCF = 3
LCM = 2² × 3 × 5 × 7 = 420
(ii) 17, 23 and 29
17 = 1 × 17
23 = 1 × 23
29 = 1 × 29
HCF = 1
LCM = 17 × 23 × 29 = 11339
(iii) 8, 9 and 25
8 = 2 × 2 × 2
9 = 3 × 391= 5 × 5
25 = 5 × 5
HCF = 1
LCM = 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800
Ex 1.2 Question 4
Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution:
HCF (306, 657) = 9
Product of two numbers = HCF × LCM
∴ LCM × HCF = 306 × 657
LCM = 306 × 657 / HCF = 306 × 657 / 9
LCM = 22338
Ex 1.2 Question 5.
Check whether 6n can end with the digit 0 for any natural number n.
Solution:
If any number ends with number 0, it should be divisible by 10 0r in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5If 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 factorization of 6n = (2 ×3)n
It can be observed that 5 is not in the prime factorization 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.
Ex 1.2 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:
Ex 1.2 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?
Solution:
LCM of 12 minutes and 18 minutes.
18 = 2 × 3 × 2
12 = 2 × 2 × 3
LCM of 12 and 18 = 2 × 2 × 3 × 3 = 36
∴Ravi and Sonia will meet together at the starting point after 36 minutes.
