The average of the present ages of A, B, C and

Solution

If the present age of  D is 54 years.

The age of B is 6 years younger than D.

present age of B = 54-6 = 48 years

The average of the present ages of A, B, C and D is (y-3) years.

Sum of the present ages of A, B, C and D = 4(y-3)

present age of A + present age of B + present age of C + present age of D = 4(y-3)

present age of A+48+present age of C+54 = 4(y-3)

12 years hence, the ratio between the ages of A and C is 9:8 respectively.

Let’s assume 12 years hence, the ages of A and C are ‘9a‘ and ‘8a‘ respectively.

(9a-12)+48+(8a-12)+54 = 4(y-3)    Eq.(i)

(i) 12

Put the possible value of ‘y’ in Eq.(i).

(9a-12)+48+(8a-12)+54 = 4x(12-3)

(9a-12)+48+(8a-12)+54 = 4x9

17a+78 = 36

Here the value of ‘a’ will be negative. So this is not the possible value of ‘y’.

(ii) 24

Put the possible value of ‘y’ in Eq.(i).

(9a-12)+48+(8a-12)+54 = 4x(24-3)

17a+78 = 4x21

17a = 84-78 = 6

Here the value of ‘a’ will be in fractions. So this is not the possible value of ‘y’.

(iii) 36

Put the possible value of ‘y’ in Eq.(i).

(9a-12)+48+(8a-12)+54 = 4x(36-3)

17a+78 = 4x33

17a+78 = 132

17a = 132-78

17a = 54

Here the value of ‘a’ will be in fractions. So this is not the possible value of ‘y’.

(iv) 48

Put the possible value of ‘y’ in Eq.(i).

(9a-12)+48+(8a-12)+54 = 4x(48-3)

17a+78 = 4x45

17a+78 = 180

17a = 180-78

17a = 102

a = 6

Here the value of ‘a’ is a natural number. So this is the only possible value of ‘y’.