RBSE Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 1 Real Numbers Ex 1.2 Textbook Exercise Questions and Answers.

RBSE Class 10 Maths Solutions Chapter 1 Real Numbers 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) Prime factors of 140 = 2 × 70
= 2 × 2 × 35
= 2 × 2 × 5 × 7
= 2² × 5 × 7

(ii) Prime factors of 156 = 2 × 78
= 2 × 2 × 39
= 2 × 2 × 3 × 13
= 2² × 3 × 13

(iii) Prime factors of 3825 = 3 × 1275
= 3 × 3 × 425
= 3 × 3 × 5 × 85
= 3 × 3 × 5 × 5 × 17
= 32 × 52 × 17

(iv) Prime factors of 5005 = 5 × 1001
= 5 × 7 × 143
= 5 × 7 × 11 × 13

(v) Prime factors of 7429 = 17 × 437
= 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
(ii) 510 and 92
(iii) 336 and 54
Solution:
(i) 26 and 91
Prime factors of 26 = 2 × 13
Prime factors of 91 = 7 × 13
∴ LCM of 26 and 91 = 2 × 7 × 13 = 182
and HCF of 26 and 91 = 13
Verification—HCF (26, 91) × LCM (26, 91)
= 13 × 182
= 13 × 2 × 91
= 26 × 91
= Product of given numbers

(ii) 510 and 92
Prime factors of 510 = 2 × 255
= 2 × 3 × 85
= 2 × 3 × 5 × 17 …(i)
and Prime factors of 92 = 2 × 46
= 2 × 2 × 23
= 2² × 23 ……(ii)
LCM (510, 92) = 2² × 3 × 5 × 17 × 23 = 23460
and HCF (510, 92) = 2
Verification—
HCF (510, 92) × LCM (510, 92)
= 2 × 23460
= 2 × 2² × 3 × 5 × 17 × 23
= 2 × 3 × 5 × 17 × 2² × 23
= 510 × 92
= Product of given numnbers

(iii) 336 and 54
Prime factors of 336 = 2 × 168
= 2 × 2 × 84
= 2 × 2 × 2 × 42
= 2 × 2 × 2 × 2 × 21
= 2 × 2 × 2 × 2 × 3 × 7
= 2⁴ × 3 × 7
Prime factors of 54 = 2 × 27
= 2 × 3 × 9
= 2 × 3 × 3 × 3
= 2 × 33
HCF (336, 54) = 2 × 3 = 6
LCM= 2⁴ × 3³ × 7
= 3024
Verification—
HCF (336, 54) × LCM (336, 54)
= 6 × 3024
= 2 × 3 × 2⁴ × 3³ x 7
= 2⁴ × 3 × 7 × 2 × 3³
= 336 × 54
= Product of given numbers

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
Prime factors of 12 = 2 × 2 × 3
Prime factors of 15 = 3 × 5
Prime factors of 21 = 3 × 7
∴ LCM (12, 15 and 21) = 2² × 3 × 5 × 7 = 420
and HCF (12, 15 and 21) = 3

(ii) 17, 23 and 29
Prime factors of 17 = 1 × 17
Prime factors of 23 = 1 × 23
Prime factors of 29 = 1 × 29
∴ LCM(17, 23 and 29) = 17 × 23 × 29 = 11339
and HCF (17, 23 and 29) = 1

(iii) 8, 9 and 25
Prime factors of 8 = 2 × 2 × 2 = (2)³
Prime factors of 9 = 3 × 3 = (3)²
Prime factors of 25 = 5 × 5 = (5)²
∴ LCM (8, 9 and 25) = (2)³ × (3)² × (5)²
= 8 × 9 × 25 = 1800
and HCF (8, 9 and 25) = 1

Question 4.
Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution:
According to the question, the numbers are 306 and 657.
∴ a = 306
b = 657
and HCF = 9 [Given]
We know that
L.C.M = \(\frac{a \times b}{\mathrm{H} \cdot \mathrm{C} \cdot \mathrm{F}}\)
= \(\frac{306 \times 657}{9}\)
= 34 × 657
= 22338
Hence L.C.M. (306, 657) = 22338

Question 5.
Check whether 6ⁿ can end with the digit 0 for any natural number n.
Solution:
Suppose that for any natural number n, n ∈ N, 6ⁿ ends with the digit 0. Hence 6ⁿ will be divisible by 5.
But prime factors of 6 = 2 × 3
∴ The prime factors of (6)ⁿ will be (6)ⁿ = (2 × 3)ⁿ
i.e., it is clear that there is no place of 5 in the prime factors of 6ⁿ.
By Fundamental theorem of Arithematic we know that every composite number can be factorised as a product of prime numbers and this factorisation is unique, i.e., our hypothesis assured in the beginning is wrong. Hence there is no natural number n for which 6ⁿ ends with the digit 0.

Question 6.
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1+5 are composite numbers.
Solution:
According to the question
7 × 11 × 13 + 13 = 13 (7 × 11 + 1)
Since 13 is a factor of this number obtained, therefore it is a composite number. Again according to the question
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
= 5 (7 × 6 × 4 × 3 × 2 1 1 + 1)
This obtained number is also a composite number because it has also a factor 5. Hence both the given numbers are composite numbers.

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:
Time taken by Sonia to drive one round of the field =18 minutes.
Time taken by Ravi to drive one round of the same field = 12 minutes
In order to find after how much time will they meet again at the starting point, we shall have to find out the DCM-of lS and 12. Therefore
prime factors of 18 = 2 × 9
= 2 × 3 × 3
= 2 × 3²
and prime factors of 12 = 2 × 6
= 2 × 2 × 3
= 2² × 3
Taking the product of the greatest power of each prime factor of 18 and 12
LCM (18, 12) = 2² × 3²
= 4 × 9 = 36
i.e., Sonia and Ravi will meet again of the starting point after 36 minutes.