Math, asked by anshi77, 10 months ago

Two years ago, father was three times as old as his son and two years hence, twice his age will be equal to five times that of his son. Find their presents ages.

Answers

Answered by VIVEKPARIDA
0

Answer:

Let son's age = x

Two years ago son's age= x-2.

His Father's age at that time = 3(x-2).

Present age of Father = 3x-6+2=3x-4

Two years hence father's age=3x-4+2=3x-2.

Two years hence son's age = x+2.

Given :-5*(x+2)=2*(3x-2)

5x + 10= 6x - 4

10+4=6x-5x

14=x

SON'S PRESENT AGE : 14 years.

FATHER'S PTESENT AGE: 38 years

please,plz,plz,plz Mark me as brainliest

if you do then i will follow you

Answered by Anonymous
0

Let son's age = x

Two years ago son's age= x-2.

His Father's age at that time = 3(x-2).

Present age of Father = 3x-6+2=3x-4

Two years hence father's age=3x-4+2=3x-2.

Two years hence son's age = x+2.

Given :-5*(x+2)=2*(3x-2)

5x + 10= 6x - 4

10+4=6x-5x

14=x

SON'S PRESENT AGE : 14 years.

FATHER'S PTESENT AGE: 38 years.

hope it helps u ✌

Similar questions