NCERT Solutions For Class 10 Math Chapter – 1 Exercise – 1.2

NCERT Solutions Class 10 Math Chapter – 1 Exercise – 1.2

 

Q1. Express each number as product of its prime factors:

  • 140
  • 156
  • 3825
  • 5005
  • 7429

Solution:-

 

  • 140 = 2 x 2 x 5 x 7
  • 156 = 2 x 2 x 3 x 13
  • 3825 = 3 x 3 x 5 x 5 x 17
  • 5005 = 5 x 17 x 11 x 13
  • 7429 = 17 x 19 x 23

 

Q2. Find L.C.M and HCF of the following pairs of integers and verify that LCM x HCF = product of the two numbers.

  • 26 and 91
  • 510 and 92
  • 336 and 54

Solution:-

 

  • 26 = 2 x 13

91 =  7 x 13

HCF = 13

LCM = 2

x 7 x 13 = 182

Product of two numbers 26 x 91 = 2366

Product of HCF and LCM 13 x 182 = 2366

Hence , product of two numbers = product of HCF x LCM

 

  • 510 = 2 x 3 x 5 x 17

92 = 2 x 2 x 23

HCF = 2

LCM = 2 x 2 x 3 x 5 17 x 23 = 23460

Product of two numbers 510 x 92 = 46920

Product of HCF and LCM 2 x 23460 = 46920

Hence , product of two numbers = product of HCF X LCM

 

  • 336 = 2 x 2 x 2 x 2 x 3 x 7

54 = 2 x 3 x 3 x 3

HCF = 2 x 3 = 6

LCM = 2 x 2 x 2 x 2 x 3 x3 x 3 x7 = 3024

Product of two numbers 336 x 54 = 18144

Product of HCF and LCM 6 x 3024 = 18144

Hence  , Product of two numbers = product of HCF x LCM.

 

Q3. Find the LCM and HCF of the following integers by applying the prime factorization method.

  • 12 , 15 and 21
  • 17 , 23 and 29
  • 8 , 9 and 25

Solution:-

 

  • 12 = 2 x 2 x 3

HCF = 3

LCM = 420

 

  • 17 = 1 x 17

23 = 1 x 23

29 = 1 x 29

HCF = 1

LCM = 11339

 

  • 8 = 1 x 2 x 2 x 2

9 = 1 x 3 x 3

25 = 1 x 5 x 5

HCF = 1

LCM = 1 x 2 x 2 x 2 x 3 x 3 x 5 x 5 = 1800

 

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

Solution:-

 

We have the formula that

Product of LCM and HCF = product of number

LCM x 9 = 306 x 657

Divide both sides by 9 we get

LCM = (306 x 657) / 9 = 22338

 

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

Answer:-

 

If any digit has last digit 10 that means it is divisible by 10 and the factors of 10 = 2 x 5 So value 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 x 11 x 13 + 13 and 7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 are composite numbers.

Solution:-

 

7 x 11 x 13 + 13

Taking 13 common we get

13 (7 x 11 + 1)

13(77 + 1)

13(78)

It is product of two numbers numbers and both numbers are more than 1  so it is a composite number.

7 x 6 x 5 x 4 x 3 x 2 x 1 + 5

5(1009)

It is a product of two numbers and both the numbers are more than 1 so it is a complete number.

 

Q7. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi take 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 points ?

Solution:-

 

They will be meet again after LCM of both value at the starting point.

18 = 2 x 3 x 3

12 = 2 x 2 x 3

LCM = 2 X 2 X 3 X 3 = 36

Therefore they will meet together at the starting point after 36 minutes.

 

 

 

 

 

NCERT Solutions Class 10 Math Chapter – 1 Exercise – 1.2   Q1. Express each number as product of its prime…