Math, asked by yuvrajratna502, 19 days 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 present ages. ​

Answers

Answered by himab8420
2

Answer:

Let son's age = x

Two years ago son's 1ge = x - 2 . His Father's age at that time = 3(x - 2) . Given :-5*(x+2)=2*(3x-2) 5x + 10 = 6x - 4

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.

10 + 4 = 6x - 5x

14 = x SON'S PRESENT AGE: 14 years. FATHER'S PTESENT AGE: 38 years.

Answered by yadavanushka611
0

Answer:

The present ages of the father and son are 38 years and 14 years respectively.

Similar questions