Playing with Numbers NCERT Solutions - Class 8 PDF Download

Exercise 16.1 

Question 1: 

Find the values of the letters in the following and give reasons for the steps involved. 

Answer 1: 

Question 2:

Find the values of the letters in the following and give reasons for the steps involved. 

Answer 2: 

Question 3: 

Find the values of the letters in the following and give reasons for the steps involved. 

Answer 3: 

Question 4: 

Find the values of the letters in the following and give reasons for the steps involved. 

Answer 4: 

Question 5: 

Find the values of the letters in the following and give reasons for the steps involved.

Answer 5: 

Question 6: 

Find the values of the letters in the following and give reasons for the steps involved. 

Answer 6: 

Question 7: 

Find the values of the letters in the following and give reasons for the steps involved. 

Answer 7:

Here product of B and 6 must be same as ones place digit as B.

6x1 = 6, 6x2 = 12,6x3 = 18, 6x4 = 24

On putting B = 4, we get the ones digit 4 and remaining two B's value should he 44,

∴ For 6x7 = 42+ 2= 44

Hence, A = 7 and B = 4

Question 8: 

Find the values of the letters in the following and give reasons for the steps involved.

Answer 8: 

Question 9: 

Find the values of the letters in the following and give reasons for the steps involved. 

Answer 9: 

Question 10: 

Find the values of the letters in the following and give reasons for the steps involved. 

Answer 10: 

Exercise 16.2 

Question 1: 

If 21y5 is a multiple of 9, where y is a digit, what is the value of y? 

Answer 1: 

Since 2ly5 is a multiple of 9.

Therefore according to the divisibility rule of 9, the sum of all the digits should be a multiple of 9.

2+1+y+5=&+y  => 8 + y

=> 8 + y = 9  

=>  y = 1

Question 2:

If 31z5 is a multiple of 9, where z is a digit, what is the value of z? You will find that there are two answers for the last problem. Why is this so? 

Answer 2:

Since 31z5 is a multiple of 9.

Therefore according to the divisibility rule of 9, the sum of all the digits should be a multiple of 9.

3 +1 + z+ 5 = 9 + z

=>   9+z = 9

=>  z = 0

If 3+1+z+5 = 9+z

=>  9+z= 18

=>   z = 9

Hence, 0 and 9 are two possible answers.

Question 3: 

If 24x is a multiple of 3, where x is a digit, what is the value of x? 

(Since 24x is a multiple of 3, its sum of digits 6 + x is a multiple of 3; so 6 + x is one of these numbers: 0, 3, 6, 9, 12, 15, 18 ... But since x is a digit, it can only be that 6 + x = 6 or 9 or 12 or 15. Therefore, x = 0 or 3 or 6 or 9. Thus, x can have any of four different values.) 

Answer 3: 

Since 24x is a multiple of 3.

Therefore according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.

∴ 2 + 4 + x = 6+x

Since x is a digit.

=>  6 + x = 6  => x = 0

=>  6+X = 9   => x = 3

=> 6 + x = 12 => x = 6

=>  6 + x -15  => x = 9

Thus, x can have any of four different values.

Question 4:

If 31z5 is a multiple of 3, where z is a digit, what might be the values of z? 

Answer 4: 

Since 31z5 is a multiple of 3.

Therefore according to the divisibility rule of 3, the sum of all the digits should be a multiple of 3.

Since r is a digit,

3 + 1 + z + 5 = 9+z

=> 9 + z = 9  => z= 0

If   3+1+z+5 = 9+z

=> 9 + z = 12   =>  = 3

If  3 + l + z + 5 = 9 + z

=>  9+r = 15  => z = 6

If 3 + l+ z + 5 = 9 + z

=>9+r = 18  => z = 9

Hence, 0, 3,6 and 9 are four possible answers.