Question

8) Sum of the ages of father and son is 74. After 8 years
father will be twice as old as son. The present age
(in years) of the father is​

Answer 1

Answer:

52 years

Step-by-step explanation:

Let, the age of father be'x' and son be'y'.

First condition;

x + y =74

or, x = 74 - y--------(i)

After 8 years father's age = x + 8

After 8 years son's age = y + 8

Second condition;

x + 8 = 2( y + 8)

or, x + 8 = 2y + 16

or, x = 2y + 16 - 8

or, x = 2y + 8----------(ii)

Substituting the value of x from eqn. (i) to eqn. (ii), we get;

74 - y = 2y + 8

or, 74 - 8 = 2y + y

or, 66 = 3y

or,66÷3 = y

or, y = 22 years

Now,

Putting y = 22 years in eqn. (i) ;

x = 74 - 22

or, x = 52 years

So, present age of father's is 52 years.

Answer 2

Given:

Sum of ages of father and son = 74

After 8 years age of father is twice as that of son.

To find:

The present age in years of the father.

Solution:

Let the present age of father be f and that of son be s. Then:

f + s = 74 - equation 1

After 8 years:

(f+8) = 2(s+8)

f + 8 = 2s + 16

f - 8 = 2s

s = (f-8) / 2 - equation 2

Put this value in equation 1:

f + (f-8) / 2 = 74

f + f/2 - 4 = 74

3f/2 = 78

f = 78*2/3 = 52 years.

The age of father is 52 years and that of son will be 22 years.

Therefore the age of father in years will be 52 years at present.