Question
Four years back in time, a father was 3 times as old as his son then was. 8 years later father will be 2 times as old as his son will then be. Find the ages of son and father.
Hint:
let present ages of son is x years and present age of father is y years
Find both the ages 4 years back and apply the condition.
Find both ages after 8 yrs and apply the condition.
Solve the two equation to get the age of son and father.
The correct answer is: 40 years .
Ans :- Age of Son is 16 years and Age of father is 40 years .
Explanation :-
Step 1:- frame the equations from the given set of conditions .
let present ages of son is x years and present age of father is y years
Fours years back, age of son = x-4
Fours years back, age of father = y-4
Four years back in time, a father was 3 times as old as his son then was
—Eq1
After 8 years, age of son = x+8
After 8 years, age of father = y+8
8 years later father will be 2 times as old as his son will then be
— Eq2
Step 2:- find x by eliminating y
Doing Eq1-Eq2 to eliminate x
∴ x= 16
Step 3:- substitute x=16 in Eq2.
∴ y = 40
∴ The present age of son =x=16 years and present age of father = 40 years
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