Question

Ten years ago father was 12 times as old as his son and 10 years hence he will be twice as old as his son. find their present ages.

Answer 1

let the age of the son be x and the father be y.
eq.1 ⇒ y-10 = 12(x-10)
⇒12x-y = 110
eq.2 ⇒ y+3 = 2(x+3)
⇒2x-y = -3
subtracting eq.2 feom eq.1
⇒10x = 113
x = 11.3 years
y = 25.6 yeas

Answer 2

The "present age" of son and father is 12 and 34 years.

Given:

x-10 = 12 (y-10)

x+10 = 2 (y+10)

To find:

The "present ages" of son and father i.e., x and y

Answer:

Assume present age of father is “x” and son is “y”

10 years ago

$x-10 = 12 (y-10)$

$x-10 = 12y-120$

$x-12y = -110 ------ (1)$

10 years from now

$x+10 = 2 (y+10)$

$x+10 = 2y+20$

$x = 10+2y ------ (2)$

By solving two equations

$10+2y-12y = -110$

$-10y = -120$

$y = \frac {120}{10}$

$y = 12 years$

$x = 10+2y$

$=10 + 2 \times 12$

$x = 34 \ years$