NCERT Solutions Maths Class 10 Chapter 1

NCERT Solutions Maths Class 10 Chapter 1

Go To:  Exercise 1.1      Exercise 1.3      Exercise 1.4 


NCERT Solutions Maths Class 10 Exercise 1.2

 

Q1. Express each number as product of its prime factors:
(i) 140                    (ii) 156                         (iii) 3825             (iv) 5005                (v) 7429

 

Solution:  (i) 140 = 2 × 2 × 5 × 7 = 22 × 5 × 7
(ii) 156 = 2 × 2 × 3 × 13 = 22 × 3 × 13
(iii) 3825 = 3 × 3 × 5 × 5 × 17 = 32 × 52 × 17
(iv) 5005 = 5 × 7 × 11 × 13
(v) 7429 = 17 × 19 × 23

 

Q2. 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 = 2 × 13
91 =7 × 13
HCF = 13
LCM =2 × 7 × 13 =182
Product of HCF and LCM = 13 × 182 = 2366
Product of two numbers = 26 × 91 = 2366
Hence, HCF × LCM = product of two numbers

(ii) 510 = 2 × 3 × 5 × 17
92 =2 × 2 × 23
HCF = 2
LCM =2 × 2 × 3 × 5 × 17 × 23 = 23460

Product of HCF and LCM = 2 × 23460 = 46920
Product of two numbers = 510 × 92 = 46920
Hence, HCF × LCM = product of two numbers

(iii) 336 = 2 × 2 × 2 × 2 × 3 × 7
54 = 2 × 3 × 3 × 3
HCF = 2 × 3 = 6
LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 7 =3024
Product of HCF and LCM = 6 × 3024 = 18144

Product of two numbers = 336 × 54 =18144
Hence, HCF × LCM = product of two numbers

Q3. Find the LCM and HCF of the following integers by applying the prime factorization method.
(i) 12, 15 and 21                   (ii) 17, 23 and 29                         (iii) 8, 9 and 25

 

Solution:  (i) 12 = 2 × 2 × 3
15 =3 × 5
21 =3 × 7
HCF = 3
LCM = 2 × 2 × 3 × 5 × 7 = 420

(ii) 17 = 1 × 17
23 = 1 × 23
29 = 1 × 29
HCF = 1
LCM = 1 × 17 × 19 × 23 = 11339

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

Q4. Given that HCF (306, 657) = 9, find LCM (306, 657).

Solution: We know that, HCF × LCM = product of number
9 × LCM = 306 × 657
Thus, LCM = (306 × 657) / 9 = 22338

 Q5. Check whether 6n can end with the digit 0 for any natural number n.

Solution:  If any digit has last digit 10 that means it is divisible by 10 and the factors of 10 = 2 × 5.
So, the value of 6n should be divisible by 2 and 5 both.

6n is divisible by 2 but not divisible by 5.

So, it can not end with 0.

 Q6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

 Solution:  7 × 11 × 13 + 13

Taking 13 common, we get
13(7 × 11 +1 )
13(77 + 1 )
13(78)
It is product of two numbers and both numbers are more than 1, so it is a composite number.

 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5

Taking 5 common, we get
5(7 × 6 × 4 × 3 × 2 × 1 + 1)
5(1008 + 1)
5(1009)
It is product of two numbers and both numbers are more than 1, so it is a composite number.

Q7. 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:  Sonia and Ravi will be meeting again after LCM of both the values at the starting point. Let us find the LCM of 18 and 12.

18 = 2 × 3 × 3
12 = 2 × 2 × 3
LCM = 2 × 2 × 3 × 3 = 36
Therefore, they will be meeting together at the starting point after 36 minutes.

 

 

 


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