Five years hence the age of father will be thrice the age of his son. Two years ago father's age was four times his son's age. Find the sum of their present ages (in years).

74

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58

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50

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48

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Solution

The correct option is A $74$
Let the present age of father be $f$ and present age of son be $s$
Given,
$f + 5 = 3 (s + 5)$
$=> f + 5 = 3 s + 15$
$=> f - 3 s = 10$ --- $(1)$
Also
$f - 2 = 4 (s - 2)$
$=> f - 2 = 4 s - 8$
$=> f - 4 s = - 6$ --- $(2)$
Subtracting eqn $(2)$ from $(1)$ , we get
$s = 16$
From eqn $(1), f - 3 (16) = 10 => f = 58$
And their sum of present ages is $58 + 16 = 74$ years

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