Question

Two years ago, a father was three times as old as his son and two years hence, twice his age will be equal to five times that of his son. Find their present ages.

Answer 1

Hint: We may consider the present age of father and son to be x and y respectively and make equations according to the given conditions and we can find the values of x and y using these equations.

Complete step by step answer:
Let us consider the present age of the father be = x
And present age of son be = y
Now we will form equations involving x and y according to the given conditions.
Since, two years ago, the age of a father was three times that of his son and we have their present ages x years and y years respectively. So, their ages two years ago would have been (x-2) years and (y-2) years respectively.

Therefore, according to the given condition we have following equation:

3(y-2) = x- 2

3y-6=x-2

3y-x-4=0……….(1)

Now, the second condition given in the question is that two years hence, twice the age of father will be equal to 5 times the age of his son.

The age of father after two years will be = (x+2) years

The age of son after two years = (y+2) years

Therefore, according to the given condition we have following equation:

2(x+2) = 5(y+2)

2x + 4 = 5y +10

2x – 5y -6 = 0 …………(2)

On multiplying equation (1) by 2 and adding it to the equation (2) we get:

2(3y-x-4) +2x -5y -6 = 0

6y -2x -8 +2x -5y – 6 =0

y -14 =0

y = 14

On substituting the value y = 14 in equation (1) we get:

$\begin{align}

  & 3\times 14-x-4=0 \\

 & 42-x-4=0 \\

 & 38-x=0 \\

 & x=38 \\

\end{align}$

So, the values of x and y are 38 and 14 respectively.

Hence, the present age of the father is 38 years and the present age of son is 14 years.


Note: Students should note here that while finding the ages of father and son before and after 2 years, we have to subtract and add 2 respectively to their present ages and chances of mistakes are there while forming equations , so care should be taken.