Class 10 Maths Chapter 1 Exercise 1.1 Solutions

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Class 10 Maths Chapter 1 Exercise 1.1 Solutions

In this section of chapter 1 “Real Numbers“, you have studied the following points ⚡:

1. We have already learnt about real numbers and encountered.
2. In this class 10 maths chapter 1, we will learn The Fundamental Theorem of Arithmetic
3. The Fundamental Theorem of Arithmetic says that every composite number can be factorised as a product of primes.

In class 10 maths chapter 1 exercise 1.1 solutions, let’s try to find the answers to these questions —

EXERCISE 1.1

Question 1: Express each number as a product of its prime factors:

(i) 140

(ii) 156

(iii) 3825

(iv) 5005

(v) 7429

Answer 1: (i) 140 = 2 × 2 × 5 × 7 = 2² × 5 × 7 _Ans.

(ii) 156 = 2 × 2 × 3 × 13 = 2² × 3 × 13 _Ans.

(iii) 3825 = 3 × 3 × 5 × 5 × 17 = 3² × 5² × 17 _Ans.

(iv) 5005 = 5 × 7 × 11 × 13 _Ans.

(v) 7429 = 17 × 19 × 23 _Ans.

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

(ii) 510 and 92

(iii) 336 and 54

Answer 2: (i) 26 and 91
26 = 2 × 13
91 = 7 × 13
HCF = 13
LCM = 2 × 7 × 13 = 182
Product of the given numbers = 26 × 91 = 2366
HCF × LCM = 13 × 182 = 2366
Hence, product of two numbers = HCF × LCM

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(ii) 510 and 92
510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23
HCF = 2
LCM = 2 × 2 × 3 × 5 × 17 × 23 = 23460
Product of the given numbers = 510 × 92 = 46920
HCF × LCM = 2 × 23460 = 46920
Hence, product of two numbers = HCF × LCM

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(iii) 336 and 54
336 = 2 × 2 × 2 × 2 × 3 × 7 = 2⁴ × 3 × 7
54 = 2 × 3 × 3 × 3 = 2 × 3³
HCF = 2 × 3 = 6
LCM = 2⁴ × 3³ × 7 = 3024
Product of the given numbers = 336 × 54 = 18144
HCF × LCM = 6 × 3024 = 18144
Hence, product of two numbers = HCF × LCM

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

Answer 3: (i) 12, 15 and 21
12 = 2 × 2 × 3 = 2² × 3
15 = 3 × 5
21 = 3 × 7
HCF = 3
LCM = 2² × 3 × 5 × 7 = 420 _Ans.

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(ii) 17, 23 and 29
17 = 1 × 17
23 = 1 × 23
29 = 1 × 29
HCF = 1
LCM = 17 × 23 × 29 = 11339 _Ans.

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(iii) 8, 9 and 25
8 = 2 × 2 × 2 = 2³
9 = 3 × 3 = 3²
25 = 5 × 5 = 5²
HCF = 1
LCM = 2³ × 3² × 5² = 8 × 9 × 25 = 1800 _Ans.

Question 4: Given that HCF (306, 657) = 9, find LCM (306, 657).

Answer 4: HCF (306, 657) = 9 is given.
We know that LCM × HCF = Product of the two numbers
On applying this —
LCM × 9 = 306 × 657

$Or LCM =\frac{306 \times 657}{9}$

Or LCM = 22338 _Ans.

Question 5: Check whether 6ⁿ can end with the digit 0 for any natural number n.

Answer 5: We know that any number which ends with digit 0, is divisible by 10. The number is also divisible by 2 and 5 because 10 = 2 × 5.

Now prime factorisation of 6ⁿ = (2 × 3)ⁿ

According to the “Fundamental Theorem of Arithmetic”, the above factorisation is unique and 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.

Question 6: Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Answer 6: On taking out 13 as a common factor from the first expression —

7 × 11 × 13 + 13
= 13 (7 × 11 + 1)
= 13 (77 + 1)
= 13 × 78
= 13 × 13 × 6

This shows that the given expression has 2 factors, 6 and 13. Therefore, the first expression is a composite number.

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Now, on taking out 5 as a common factor from the second expression —

7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5 (7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5 (1008 + 1)
= 5 × 1009

5 and 1009 are prime numbers. This shows that the given expression has 2 factors, 5 and 1009. Therefore, the second expression is also a composite number.

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?

Answer 7: Sonia takes 18 mins. every time she reaches to the starting point. In other words, she takes time in multiples of 18 mins. to finish rounds of the circular path. In the same way, Ravi takes time in multiples of 12 mins. to finish rounds of the circular path.

So, the number of minutes taken by Sonia and Ravi to meet again at the starting point would be a common multiple of 18 and 12.

18 = 2 × 3 × 3
12 = 2 × 2 × 3
Hence, LCM (12, 18) = 2 × 2 × 3 × 3 = 36
Therefore, they will meet again at the starting point after 36 minutes.

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