Question

Two years ago, Dilip 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. Then the present age of Dilip is:
(a) 14 years
(b) 38 years
(c) 32 years
(d) 34 years

Answer 1

Hint: Let us assume the present age of Dilip be x years and his son’s age be y years. Now, two years ago, the age of Dilip was $\left( x-2 \right)$ years and the age of his son was $\left( y-2 \right)$ years and the relation between their age was $\left( x-2 \right)=3\left( y-2 \right)$. Now, after two years from the present age of Dilip and his son, their ages will become $\left( x+2 \right)\And \left( y+2 \right)$ respectively and the relation between their ages is given as: .$2\left( x+2 \right)=5\left( y+2 \right)$. Now, solve these two equations in x and y and find the values of x and y. And the value of x will give you the present age of Dilip.

Complete step by step solution:
In the above problem, we are asked to find the present age of Dilip. Let us assume the present age of Dilip is x years and the present age of his son be y years.
Now, two years ago from the present age the ages of Dilip and his son is as follows:
Age of Dilip was $\left( x-2 \right)$ years.
And age of his son was $\left( y-2 \right)$ years.
It is given that the relation between their ages was Dilip was three times as old as his son so multiplying Dilip’s son age by 3 and equating this multiplication to Dilip’s age we get,
$\left( x-2 \right)=3\left( y-2 \right).....(1)$
Now, two years after present age, the age of Dilip and his son becomes:
Age of Dilip was $\left( x+2 \right)$ years.
And age of his son was $\left( y+2 \right)$ years.
It is given that the relation between their ages is that twice the age of Dilip will be five times as old as his son so multiplying Dilip’s son age by 5 and equating this multiplication to Dilip’s age we get,
$2\left( x+2 \right)=5\left( y+2 \right).....(2)$
Simplifying eq. (1) and eq. (2) we get,
$\left( x-2 \right)=3\left( y-2 \right).....(1)$
Multiplying 3 inside the bracket of the above equation we get,
$\begin{align}
  & x-2=3y-6 \\
 & \Rightarrow x-3y+4=0.....(3) \\
\end{align}$
Simplifying eq. (2) we get,
$2\left( x+2 \right)=5\left( y+2 \right).....(2)$
Multiplying 5 and 2 inside the bracket in the R.H.S and L.H.S of the above equation we get,
$\begin{align}
  & 2x+4=5y+10 \\
 & \Rightarrow 2x-5y-6=0......(4) \\
\end{align}$
Now, multiplying eq. (3) by 2 and then subtracting eq. (4) from eq. (3) we get,
Multiplying eq. (3) by 2 we get,
$\begin{align}
  & 2\left( x-3y+4=0 \right) \\
 & =2x-6y+8=0.....(5) \\
\end{align}$
Now, subtracting eq. (5) from eq. (4) we get,
$\begin{align}
  & 2x-5y-6=0 \\
 & 2x-6y+8=0 \\
 & \underline{\begin{matrix}
   - & + & - & - \\
\end{matrix}} \\
 & y-14=0 \\
\end{align}$
The result of the subtraction of two equations is equal to:
$y-14=0$
Adding 14 on both the sides of the above equation we get,
$y=14$
Substituting the value of y in eq. (3) we get,
$\begin{align}
  & x-3\left( 14 \right)+4=0 \\
 & \Rightarrow x-42+4=0 \\
 & \Rightarrow x-38=0 \\
 & \therefore x=38 \\
\end{align}$
From the above, we got the value of x as 38 years which is the present age of Dilip.

So, the correct answer is “Option b”.

Note: The mistake that could happen in the above problem is that in the haste of solving the problem in the examination, you might tick the option (a) which is the age of Dilip’s son so make sure you won’t make this mistake in the examination.