X mathematics solutions chapter 1 exercise 1.2

EXERCISE 1.2

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

(i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429

(i) 140             = 2 x 2 x 5 x 7            =2^2 x 5 x 7

(ii)156             =2 x 2 x 3 x 13           =2^2 x 3 x 13

(iii)3825          =3 x 3 x 5 x 5 x 17    =3^2 x 5^2 x 17

(iv)5005          =5 x 7 x 11 x 13        =5 x 7 x 11 x 13

(v)7429           =17 x 19  x 23           =17 x 19 x 23

                                                              

                                                                                                                                           

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

(i)                     26        = 2 x 13

91        =7 x 13

HCF    = 13

LCM    =2 x 7 x 13     =182

Product of two value           26 x 91           = 2366

Product of HCF and LCM   13 x 182         = 2366

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

(ii)                    510     = 2 x 3 x 5 x 17

92        =2 x 2 x 23

HCF    =2

LCM    =2 x 2 x3 x 5 x 17 x 23         = 23460

Product of both values        510x92           = 46920

Product of HCF and LCM   2x23460         =46920

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

(iii)                   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 x 3 x 3 x 7           =3024

Product of number              336 x 54         =18144

Product of HCF and LCM   6 x 3024         = 18144

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

3. 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

(i)          12        = 2 x 2 x 3

15        =3 x 5

 21       =3 x 7

HCF    = 3

LCM = 2 x 2 x 3 x 5 x 7        = 420

(ii)        17        = 1 x 17

23        = 1 x 23

29        = 1 x 29

HCF    = 1

LCM    = 1 x 17 x 19 x 23     = 11339

(iii)

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

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

We have the formula that

Product of LCM and HCF   =          product of number

LCM x 9         =          306 x 657

Divide both side by 9 we get

LCM                =          (306 x 657)/ 9   = 22338

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

 If any digit has last digit 10 that means It is divisible by 10 And the factors of

 10 = 2x5 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.

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

      7 × 11 × 13 + 13

      Take 13 common there we get

      13  (7 x 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

      Take 5 common there we get

      5(7 x 6 x 4 x 3 x 2 x 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

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?

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

      To get the LCM we have to factorize the number.

            18        = 2 × 3 × 3

            12        = 2 × 2 × 3

            LCM    = 2 × 2 × 3 × 3 = 36 

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