Find Marvin's Age

Let the following be true:

$x = \text{Marvin's current age}$

$y = \text{Marvin's age 3 years ago}$

$z = \text{Marvin's age 2 years from now}$

If that's true, we're solving for $x$ , so:

$x = x$

$y = x - 3$

$z = x + 2$

So, according to the problem, here's our equation:

$\frac{1}{2}z + \frac{1}{3}y = 20$

Once the above equation is simplified, we have:

$3z + 2y = 120$

Once we put both $z$ and $y$ in terms of $x$ , our equation becomes:

$3(x+2) + 2(x-3) = 120$

Which becomes:

$3x + 6 + 2x - 6 = 120$

Simplifying the above equation gives:

$5x = 120$

Therefore, $x=24$ , meaning the answer is that Marvin's current age is 24.

• Let the present age of Marvin be $x$ .
• Two years from now, Marvin will be $x+2$ years old and one-half of that is $\dfrac {x+2}2$ .
• Three year ago, Marvin was $x-3$ years old and one-third of that is $\dfrac {x-3}3$ .

Then we have:

$\begin{aligned} \frac {x+2}2 + \frac {x-3}3 & = 20 \\ \frac x2 + 1 + \frac x3 - 1 & = 20 \\ \frac {3x+2x}6 & = 20 \\ 5x & = 120 \\ \implies x & = \frac {120}5 = \boxed {24} \end{aligned}$

Marvin's age is $\dfrac{20}{\frac{1}{2}+\frac{1}{3}}=\boxed {24}$ years. Correct? :)

Generalization :

The sum of ( $\frac{1}{x}$ )'th of someone's age $x$ years from now and ( $\frac{1}{y}$ )'th of his age $y$ years ago is $\dfrac{37(x+y)}{xy}$ years. How old is he now?

Let M = present age of Marvin

M + 2 = Marvin’s age two years from now

M – 3 = Marvin’s age three years ago

(1/2)(M + 2) + (1/3)(M - 3) = 20

M/2 + 1 + M/3 – 1 = 20

M/2 + M/3 = 20

6(M/2 + M/3 = 20)

3M + 2M = 120

5M = 120

M = 24