Question

8 years ago, a father was 4 times as old as his son was and 12 years hence, the age of the father will be twice that of his son. Find their present ages.​

Answer 1

Step-by-step explanation:

♦ Given ♦ :-

• 8 years ago, a father was 4 times as old as his son was and 12 years hence, the age of the father will be twice that of his son.

♦ To find ♦ :-

• Their present ages.

♦ Solution ♦ :-

Let assume that :

• Present age of the son = X years

• Present age of the father = Y years

8 years ago :

• The age of the son = (X-8) years

• The age of the father = (Y-8) years

According to the given problem

Father's age = 4 ×son's age

=> Y-8 = 4(X-8)

=> Y-8 = 4X-32

=> 4X-Y = 32-8

=> 4X-Y = 24 --------------(1)

After 12 years :

The age of the son = (X+12) years

The age of the father = (Y+12) years

According to the given problem:

Father's age = 2× son's age

=> Y+12 = 2(X+12)

=> Y+12 = 2X +24

=> 2X-Y = -24+12

=> 2X-Y = -12 -----------(2)

From (1)&(2)

4X-Y = 24

Y = 242X-Y = -12

(-)

_________

2X +0 = 36

_________

=> 2X = 36

=> X = 36/2

=> X = 18

Therefore,

The present age of the son = 18 years

On substituting the value of X in (2)

2(18)-Y = -12

=>36-Y = -12

=> Y = 36+12

=> Y = 48 years

The present age of the father = 48 years

♦ Answer ♦ :-

• The present ages of the son and the father are 18 years and 48 years respectively.