Question

Six years ago a father was seven times as old as his son. After six years he will be three times as old as his son. Determine their present ages.

Answer 1

Let the father's age and the son's age be x and y respectively

Given

Six years ago a father was seven times as old as his son.

$\implies x-6=7(y-6)\\\implies x-6=7y-42\\\implies x-7y=-42+6\\\implies x-7y=-36--(1)$

After six years he will be three times as old as his son.

$\implies x+6=3(y+6)\\\implies x+6=3y+18\\\implies x-3y=18-6\\\implies x-3y=12--(2)$

Solving (1) and (2), we get,

$x-7y=-36\\\underline{x-3y=12}\\\underline{\underline{-4y=-48}}\\\implies y = 12$

Substituting the value of y in the second equation, we get,

$\implies x-3(12)=12\\\implies x-36=12\\\implies x = 12+36\\\implies x = 48$

$\boxed{\boxed{\bold{Father's \ age = 48 \ years}}}\\\boxed{\boxed{\bold{Son's \ age = 12 \ years}}}$

                                                                             

Answer 2

GIVEN :-

•Six years ago a father was seven times as old as his son.

•After six years he will be three times as old as his son.

TO FIND :-

• Father and son's present age.

SOLUTION :-

Let, father age be R and son's age be T.

Case (I)

•Six years ago a father was seven times as old as his son.

=> R - 6 = 7(T - 6)

=> R - 6 = 7T - 42

=> R - 7T = - 36 -----------(1)

Case (II)

•After six years he will be three times as old as his son.

=> R + 6 = 3(T + 6)

=> R + 6 = 3T + 18

=> R - 3T = 12 --------------(2)

Subtracting eqn. (1) and (2) ,

R - 7T = -36

R - 3T = 12

- + +

-----------------

4T = 48

.°. T = 48/4

.°. T = 12 .

Put T=12 in eqn. (2) we get,

=> R - 3(12) = 12

=> R - 36 = 12

=> R = 36 + 12

=> R = 48 .

Hence,

• The present age of Father = 48 years.

• The present age of son = 12 years.