The age of a father is twice that of his elde | vedclass

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(D) Let the ages of the elder and younger sons be $x_{1}$ and $x_{2}$ respectively.$	herefore$ Age of father $= 2x_{1}$.Ten years hence:Age of father

Vedclass · Accepted Answer

$50$

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(D) Let the ages of the elder and younger sons be $x_{1}$ and $x_{2}$ respectively.$	herefore$ Age of father $= 2x_{1}$.Ten years hence:Age of father $= 2x_{1} + 10$.Age of younger son $= x_{2} + 10$.According to the problem,$2x_{1} + 10 = 3(x_{2} + 10)$ .... $(1)$.Given that the difference between the ages of the two sons is $15 	ext{ years}$,so $x_{1} - x_{2} = 15 \implies x_{2} = x_{1} - 15$.Substitute $x_{2}$ into equation $(1)$:$2x_{1} + 10 = 3(x_{1} - 15 + 10)$.$2x_{1} + 10 = 3(x_{1} - 5)$.$2x_{1} + 10 = 3x_{1} - 15$.$x_{1} = 25$.Therefore,the age of the elder son is $25 	ext{ years}$.Age of the father $= 2x_{1} = 2 	imes 25 = 50 	ext{ years}$.

The age of a father is twice that of his elder son. Ten years hence,the age of the father will be $3$ times that of the younger son. If the difference of ages of two sons is $15 \text{ years}$,the age of the father is (in $\text{years}$)

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