Question

ten years ago the age of a father was four times of his son. ten years hence the age of the father will be twice that of gis son. the present ages of the father and the son are.?

Answer 1

Let the present age of son be x

and age of father be y

10 years ago ;

age of son = x - 10

age of father = y - 10

but a/q ;

y - 10 = 4 ( x - 10 )

or, y - 10 = 4x - 40

or, 4x - y = 30 _______________1

10 years later ;

son's age = x + 10

father's age = y + 10

but a/q ;

y + 10 = 2 ( x + 10 )

or, y + 10 = 2x + 20

or, 2x - y = -10 _______________2

subtracting 2 from 1 , we get ;

2x = 40

or, x = 40/2 = 20 years (son's age)

y = 2x + 10 = 2×20 + 10 = 50 yrs (father's age)

Answer 2

Given:

Ten years ago the age of a father was four times his son

Ten years hence the age of the father will be twice that of his son

To Find:

Present age of father and son

Solution:

Let the present age of son be x years and father be y years.

Ten Years Ago

Age of son= x-10

Age of Father= y-10

According to question,

$\longrightarrow\mathrm{y-10=4(x-10)}$

$\longrightarrow\mathrm{y-10=4x-40}$

$\longrightarrow\mathrm{y=4x-40+10}$

$\longrightarrow\mathrm{y=4x-30}--------(1)$

10 years hence or 10 years later

Age of son= x+10

Age of Father= y+10

According to question,

$\longrightarrow\mathrm{y+10=2(x+10)}$

$\longrightarrow\mathrm{y+10=2x+20}$

$\longrightarrow\mathrm{y=2x+20-10}$

$\longrightarrow\mathrm{y=2x+10}-------(2)$

From (1) and (2), we get

$\longrightarrow\mathrm{4x-30=2x+10}$

$\longrightarrow\mathrm{4x-2x=30+10}$

$\longrightarrow\mathrm{2x=40}$

$\longrightarrow\mathrm{x=\dfrac{40}{2}}$

$\longrightarrow\mathrm{x=20}$

On putting value of x in (2), we get

$\longrightarrow\mathrm{y=2(20)+10}$

$\longrightarrow\mathrm{y=40+10}$

$\longrightarrow\mathrm{y=50}$

Hence, the present age of father is 50 years and present age of son is 20 years.