Question

Two years agi a father was 5 times as odd as his sin 2years later his age will be 8 more than3 times the age of the son. Find the present age of father and son

Answer 1

Hiii friend,


Let the age of father and his son be X and Y year's.


Two years ago age of father = (X-2) years

Two years ago age of his son = (Y-2) years


According to question,

(X-2) = 5(Y-2)

X -2 = 5Y -10

X - 5Y = -10+2

X - 5Y = -8--------------(1)

Two years later age of father = (X+2) years.

Two years later age of son = (Y+2) years.

According to question,

(X+2) = 3(Y+2) + 8

X + 2 = 3Y + 6 +8

X - 3Y = 14-2

X - 3Y = 12-------------(2)


From equation (1) we get,


X - 5Y = -8

X = -8+5Y-----------(3)

Putting the value of X in equation (2)

X - 3Y = 12

-8+5Y -3Y = 12

2Y = 12+8

Y = 20/2 = 10 years.


Putting the value of Y in equation (3)

X = -8 + 5Y => -8 + 5 × 10 = -8 + 50

X = 42


Age of father = X = 42 years.


AND,

Age of son = Y = 10 years.



HOPE IT WILL HELP YOU..... :-)