# AP 10th Class Maths 1st Chapter Real Numbers Exercise 1.1 Solutions

Well-designed AP Board Solutions Class 10 Maths Chapter 1 Real Numbers Exercise 1.1 offers step-by-step explanations to help students understand problem-solving strategies.

## Real Numbers Class 10 Exercise 1.1 Solutions – 10th Class Maths 1.1 Exercise Solutions

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

140 = 2 × 2 × 5 × 7
= 22 × 51 × 71

ii) 156

156 = 2 × 2 × 3 × 13
= 22 × 31 × 131

iii) 3825

3825 = 3 × 3 × 5 × 5 × 17
= 32 × 52 × 171

iv) 5005

5005 = 5 × 7 × 11 × 13

v) 7429

7429 = 17 × 19 × 23

Question 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:

HCF of 26 and 91 is 13.
LCM of 26 and 91 = 2 × 7 × 13 = 182
LCM × HCF = 182 × 13 = 2366
Product of two numbers = 26 × 91 = 2366
∴ LCM × HCF = Product of two numbers

ii) 510 and 92
Solution:

HCF of 26 and 91 is 13.
LCM of 26 and 91 = 2 × 7 × 13 = 182
LCM × HCF = 182 × 13 = 2366
Product of two numbers = 26 × 91 = 2366
∴ LCM × HCF = Product of two numbers

iii) 336 and 54

HCF of 336 and 54 = 2 × 3 = 63
LCM of 336 and 54
= 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 = 3024
LCM × HCF = 3024 × 6 = 18144
Product of two numbers
= 336 × 54 = 18144
∴ LCM HCF = Product of two numbers.

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

i) 12, 15 and 21
Solution:

HCF of 12, 15 and 21 = 3
LCM of 12, 15 and 21
= 2 × 2 × 3 × 5 × 7
= 420

ii) 17, 23 and 29
Solution:

iii) 8, 9 and 25
Solution:

Question 4.
Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution:
Given HCF = (306, 657) = 9
HCF × LCM = Product of two numbers
9 × LCM = 306 × 657
LCM = $$\frac{306 \times 657}{9}$$
∴ LCM of 306 and 657 = 22338

Question 5.
Check whether 6n can end with the digit 0 for any natural irnm ber n.
Solution:
6n = (2 × 3)n = 2n.3n
2 and 3 are primes in the factorization of 6n.
There is no other primes in the factorization of 6n = 2n.3n
5 does not occur in the prime factorization of 6n for any n.
6n does not end with the digit zero for any natural number.

Question 6.
Explain why 7 × 11 × 13 + 13 said 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution:
7 × 11 × 13 + 13 = (7 × 11 + 1) × 13 = 78 × 13 is a composite number as 13 is a factor of it and 78 is a factor.
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= (7 × 6 × 4 × 3 × 2 × 1 + 1) × 5
= 1008 × 5 is a composite number as 5 and 1008 are factors of it.

Question 7.
There is a circular path around a sports held. Sonia takes 18 minutes to drive one round of the held, 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:
Required number of minutes is the LCM of 18 and 12

= 2 × 2 × 3 × 3
= 36
Sonia and Ravi will meet again at the starting point after 36 minutes.