Question

in 1980 father's age was 8 times his son's age.in 1988 father's age was 10 times what is son's age in 1980 and in 1990 father and his sons ages was?

Answer 1

Answer:

Father 42 Son 14

Step-by-step explanation:

Equ(1) F=8S

Equ(2) F+8=10S

(1) on (2)=> 8S+8=10S

S=4 and F=32

After 10 yrs F=42 and S=14

Answer 2

Given :

In 1980, Father's age was 8 times his son's age.

In 1988, father's age was 10 times what is son's age in 1980.

To find : Father and son's age in 1990=?

Solution :

Let the age of son be 'x'.

Let the age of father be '8x'.

in 1988, the age of father would be :

$8x+8=10x\\\\8=10x-8x\\\\8=2x\\\\\dfrac{8}{2}=x\\\\x=4\ years$

So, the age of father would be $8x=8\times 4=32\ years$

in 1990, age of son would be $4+10=14\ years$

The age of father would be $32+10=42\ years$

Hence, the age of his son and his father are 14 years and 42 years resp.