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# The age of a father is twice that of the elder son. Ten years later the age of the father will be three times that of the younger son. If the difference of ages of the two sons is 15 years. Find the age of the father.A. 100B. 70C.60D. 50

Last updated date: 29th May 2024
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Hint: We first assume the age of the father. Then using the given relations, we find the age of his two sons. We form the algebraic equation as it is given that the difference of ages of the two sons is 15 years. We solve that to find the age of the father.

Let us assume the age of the father is $2x$ . The age of a father is twice that of the elder son which means the age of the elder son is $\dfrac{2x}{2}=x$ .
10 years later the age of the father will be $2x+10$ and that will be 3 times that of the younger son.
So, the age of the younger son 10 years later is $\dfrac{2x+10}{3}$ . Present age is $\dfrac{2x+10}{3}-10$
This gives the equation of $x-\left( \dfrac{2x+10}{3}-10 \right)=15$ which gives $x-\dfrac{2x+10}{3}=5$
\begin{align} & 3x-\left( 2x+10 \right)=15 \\ & \Rightarrow 3x-2x-10=15 \\ & \Rightarrow x=15+10=25 \\ \end{align}
The age of the father is $25\times 2=50$ years. The correct option is D.