Selina solutions

Selina solutions

Grade 6

Playing with Numbers | Exercise 9(C)

Question 10

Q10)Find which of the following numbers are divisible by 15 :

(i) 960

(ii) 8295

(iii) 10243

(iv) 5013

Solution :

(i) 960

Given number = 960

For a number to be divisible by 15 it should be divisible by both 3 and 5.

Sum of digits = 9+6+0 =15

Since 15 is divisible by 3 therefore the number is divisible by 3 and unit digit is 0 so it is divisible by 5.

Thus 960 is divisible by 15.

(ii) 8295

Given number = 8295

For a number to be divisible by 15 it should be divisible by both 3 and 5.

Sum of digits = 8 +2 +9+5 = 24

Since 24 is divisible by 3 therefore the number is divisible by 3 and unit digit is 5 so it is divisible by 5.

Thus 8295 is divisible by 15.

(iii) 10243

Given number = 10243

For a number to be divisible by 15 it should be divisible by both 3 and 5.

Sum of digits = 1+0+2+4+3=10

Since 10 is not divisible by 3 therefore the number is not divisible by 3 and unit digit is 3 so it is not divisible by 5.

Thus 8295 is not divisible by 15.

(iv) 5013

Given number = 5013

For a number to be divisible by 15 it should be divisible by both 3 and 5.

Sum of digits = 5+0+1+3= 9

Since 9 is divisible by 3 therefore the number is divisible by 3 and unit digit is 3 so it not is divisible by 5.

Thus, 5013 is not divisible by 15.

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subject-cta

Question 10

Q10)Find which of the following numbers are divisible by 15 :

(i) 960

(ii) 8295

(iii) 10243

(iv) 5013

Solution :

(i) 960

Given number = 960

For a number to be divisible by 15 it should be divisible by both 3 and 5.

Sum of digits = 9+6+0 =15

Since 15 is divisible by 3 therefore the number is divisible by 3 and unit digit is 0 so it is divisible by 5.

Thus 960 is divisible by 15.

(ii) 8295

Given number = 8295

For a number to be divisible by 15 it should be divisible by both 3 and 5.

Sum of digits = 8 +2 +9+5 = 24

Since 24 is divisible by 3 therefore the number is divisible by 3 and unit digit is 5 so it is divisible by 5.

Thus 8295 is divisible by 15.

(iii) 10243

Given number = 10243

For a number to be divisible by 15 it should be divisible by both 3 and 5.

Sum of digits = 1+0+2+4+3=10

Since 10 is not divisible by 3 therefore the number is not divisible by 3 and unit digit is 3 so it is not divisible by 5.

Thus 8295 is not divisible by 15.

(iv) 5013

Given number = 5013

For a number to be divisible by 15 it should be divisible by both 3 and 5.

Sum of digits = 5+0+1+3= 9

Since 9 is divisible by 3 therefore the number is divisible by 3 and unit digit is 3 so it not is divisible by 5.

Thus, 5013 is not divisible by 15.

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