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Four years ago, the father's age was three times the age of his son. The total of the ages of the father and the son after four years will be 64 years. What is the father's age at present?

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$\begin{align}\text{At present, father's age = x years}\\ \text{& son's age = y years} \end{align}$

$\begin{align}\text{∴ 4 years ago, father's age would be (x-4) years} \\ \text{ & son's age would be (y-4) years} \end{align}$

Now, given that,

Four years ago, the father's age was three times the age of his son.

$∴ (x-4) = 3(y-4) ---------- 1)$

Also given that,

The total of the ages of the father and the son $\underline{\text{after four years}}$ will be 64 years.

$∴ ( x+4 )+ (y+4 ) = 64 \\Or, x = 64 - 8 - y \\ Or, x = 56 - y ------ 2)$

Now, putting the $x's$ value in the equation $1)$

$ (x-4) = 3(y-4) \\ Or, 56-y-4 = 3y - 12 \\ Or, 52-y = 3y-12 \\ Or, 52+12 = 3y+y \\ Or, 4y = 64 \\ Or, y = \dfrac{64}{4} = 16$

$\color{maroon}{\text{∴ Son's present age is 16 years}}$

Now, putting the value of $y = 16$ in equation $2)$, we get -

$x = 56 - y \\ = 56 - 16 \\ = 40$

$\color{green}{\text{∴ Father's present age is 40 years. }}$

+1 vote

Assum father's present age to be $x$ $years$ and son's present age to $y$ $years$.

Four years ago father's age is 3 times the age of his son means:

$x - 4 = 3(y-4)$

The total of the ages of the father and the son after four years** will be** 64 years. **Will be means that the question is talking about future, and therefore the correct second equation is:**

$x + 4 + y + 4 = 64$

Solving the two equations we get $x = 40$, father is presently $40$ **years old** and $y = 16$, son is presently $16$ **years old.**

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