Question

Two years ago, dilip was 3 times as old as his son and two years hence twice his age will be equal to 5 times that of his son's age. Find their present ages.​

Answer 1

Answer:

sons present age = 14

fathers present age = 38(3×14) -4

Step-by-step explanation:

sons age = x

two years ago =x-2

present age of father =3x-6+2 (3x-4)

two years ago= 3(x-2)

two years hence fathers age =3x-4+2=3x-2

two years hense sons age=x+2

5×(x+2)= 2×(3x-2)

5x+10=6x-4

x=14

Answer 2

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.