NCERT Solutions for class 10 maths chapter 1

Real Numbers

NCERT Solutions for class 10 maths chapter 1 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

Answer 1

(i) 140
LCM of 140
140 = 2 × 2 × 5 × 7 × 1 = 2²×5×7

(ii) 156
LCM of 156
Hence, 156 = 2 × 2 × 3 × 13 × 1 = 2²× 13 × 3

(iii) 3825
LCM of 3825
3825 = 3 × 3 × 5 × 5 × 17 × 1 = 3²×5²×17

(iv) 5005
LCM of 5005
Hence, 5005 = 5 × 7 × 11 × 13 × 1 = 5 × 7 × 11 × 13

(v) 7429
LCM of 7429
Hence, 7429 = 17 × 19 × 23 × 1 = 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

Answer 2

(i) 26 and 91
26 and 91 as product of its prime factors
26 = 2 × 13 × 1
91 = 7 × 13 × 1
So, LCM (26, 91) = 2 × 7 × 13 × 1 = 182
And HCF (26, 91) = 13

Verification
LCM × HCF = product of the both number
Now, product of 26 and 91 = 26 × 91 = 2366
And Product of LCM and HCF = 182 × 13 = 2366
So,
product of 26 and 91=Product of LCM(26,91) and HCF(26,91)

(ii) 510 and 92
510 = 2 × 3 × 17 × 5 × 1
92 = 2 × 2 × 23 × 1
So, LCM(510, 92) = 2 × 2 × 3 × 5 × 17 × 23 = 23460
And HCF (510, 92) = 2

Verification
LCM × HCF = product of both number
Now, product of 510 and 92 = 510 × 92 = 46920
And Product of LCM and HCF = 23460 × 2 = 46920
Hence, LCM × HCF = product of the 510 and 92.

(iii) 336 and 54
336 = 2 × 2 × 2 × 2 × 7 × 3 × 1
54 = 2 × 3 × 3 × 3 × 1
Therefore, LCM(336, 54) =2⁴x3³x7=3024 And HCF(336, 54) = 2×3 = 6 So
Now, product of 336 and 54 = 336 × 54 = 18,144
And Product of LCM and HCF = 3024 × 6 = 18,144
Hence, LCM × HCF = product of the 336 and 54.

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
15=5×3
21=7×3
So, HCF(12,15,21) = 3
LCM(12,15,21) = 2 × 2 × 3 × 5 × 7 = 420

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

(iii) 8, 9 and 25
8=2×2×2×1 9=3×3×1
25=5×5×1
So,
HCF(8,9,25)=1
LCM(8,9,25) = 2×2×2×3×3×5×5 = 1800

Question 4

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

Answer 4

we know that, HCF×LCM=Product of the two given numbers
Therefore, 9 × LCM = 306 × 657
LCM =

(306×657) / 9

= 22338
Hence, LCM(306,657) = 22338

Question 5

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

Answer 4

If the number 6n ends with the digit zero , then it should be divisible by 5,
Prime factorization of 6n = (2×3)n
So, the prime factorization of 6n doesn’t contain prime number 5. Hence, it is clear that for any natural number n, 6n is not divisible by 5 So, 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

we know, that if a number is composite, then it means it has factors other than 1 and itself. Therefore,
7 × 11 × 13 + 13
=13(7×11×1+1) = 13(77+1) = 13×78 = 13×6×13
Hence, 7 × 11 × 13 + 13 is a composite number.
Now let’s take the other number,
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
=5(7×6×4×3×2×1+1) = 5(1008+1) = 5×1009
So, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is 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?