Maths-
General
Easy

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
    left parenthesis y minus 4 right parenthesis equals 3 space cross times space left parenthesis x minus 4 right parenthesis
    not stretchy rightwards double arrow y equals 3 x minus 8 —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
    not stretchy rightwards double arrow left parenthesis y plus 8 right parenthesis equals 2 left parenthesis x plus 8 right parenthesis not stretchy rightwards double arrow y plus 8 equals 2 x plus 16
    not stretchy rightwards double arrow y equals 2 x plus 8 — Eq2
    Step 2:- find x by eliminating y
    Doing Eq1-Eq2 to eliminate x
    left parenthesis 3 x minus 8 right parenthesis minus left parenthesis 2 x plus 8 right parenthesis equals y minus y not stretchy rightwards double arrow x minus 16 equals 0
    ∴ x= 16
    Step 3:- substitute x=16 in Eq2.
    y equals 2 left parenthesis 16 right parenthesis plus 8 not stretchy rightwards double arrow y equals 32 plus 8
    ∴ y = 40
    ∴ The present age of son =x=16 years and present age of father = 40 years

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