Question

Two years ago, Sahil 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. Check your solution.​

Answer 1

Answer:

Step-by-step explanation:

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.

Answer 2

Answer:

Let present age of son = x

present age of Sahil = y

y - 2 = 3 ( x-2)  .....................(1)

2(y + 2) = 5 ( x + 2)...................(2)

Solving (1) and (2),

x = 14 , y = 38

Hence , presently Sahil is 38 years old and his son is 14 years old.

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