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

Answer:

i) 140

140 = 2 × 2 × 5 × 7

= 2^{2} × 5^{1} × 7^{1}

ii) 156

156 = 2 × 2 × 3 × 13

= 2^{2} × 3^{1} × 13^{1}

iii) 3825

3825 = 3 × 3 × 5 × 5 × 17

= 3^{2} × 5^{2} × 17^{1}

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 6^{n} can end with the digit 0 for any natural irnm ber n.

Solution:

6^{n} = (2 × 3)^{n} = 2^{n}.3^{n}

2 and 3 are primes in the factorization of 6^{n}.

There is no other primes in the factorization of 6^{n} = 2^{n}.3^{n}

5 does not occur in the prime factorization of 6^{n} for any n.

6^{n} 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.