# A tricky logical (age) problem

It's an old question, may be from 7th grade, but I am really looking for a good explanation for this question:

A says to B, "I am three times as old as you were, when I was as old as you are". If the sum of their present ages is 64, find the ages of A and B respectively.

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Let $x$ be the age of A, $y$ the age of B. We know that $x+y=64$, and the sentence "I (A) am three times as old as you (B) were, when I was as old as you are" can be translated as: $3(y - (x - y)) = x$. Thus $y = \frac{2}{3}x$ and we get $x = \frac{3\cdot64}{5} = 38.4$ and $y=\frac{128}{5} = 25.6$.

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In that kind of riddle, the answer lives surely in $\mathbb N$. – Damien L Feb 4 '13 at 11:29
Hey your method is correct! I have made a change in the question by reducing the sum of the ages to be 60. – Hemant Feb 4 '13 at 11:57
@DamienL: Yes, I know, I was kind of unsure, too. But the method is correct and the answer is right, so... – Daniel Robert-Nicoud Feb 4 '13 at 21:30

Let $x$ be the current age of A and $y$ the current age of B. When A was as old as B now, that was $x - y$ years ago and B's age that time is $y - (x - y)$, which can be simplified as $2y -x$. Now, the statement "I am three times as old as you were, when I was as old as you are" can now be represented by the following expression:

$$x = 3(2y - x)$$

$$x = 6y - 3x$$

$$4x = 6y$$

Since $x + y = 60$, then $y = 60 - x$. So now we have,

$$4x = 6(60 - x)$$

$$4x = 360 - 6x$$

$$10x = 360$$

$$x = 36$$

$$y = 60 - 36 = 24$$

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The current time will be denoted as $t$. The age of $A$ at time $t$ will be $A(t)$ and the age of $B$ at the same time is $B(t)$. There exists $n$ such that $A(t) = B(t) + n$, $n$ is just the age difference.

Now for any past time $t_0$, we still have $A(t_0) = B(t_0) + n$.

The sentence says that $A(t) = 3 B(t_0)$ where $t_0$ is chosen so that $A(t_0) = B(t)$.

After so equations, you end up with $B(t) = \frac{n}{2}$ and $A(t) = n + \frac{n}{2}$. Knowing the sum is equal to 64, you find that

$$A(t) = 48 \text{ and } B(t) = 16$$

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You seem to have a sign issue here - when $A$ was $16$, $B$ was $-16$, thus $A$ should now be $-48$. – Matthew Pressland Feb 4 '13 at 11:55
@ Damien L: This is not the correct answer. I Have made a change in the question, please try with the new age. – Hemant Feb 4 '13 at 11:58

The present ages of A and B are $36$ and $24$ respectively. and the number of years ago in the question is $12$ years (value of $n$)

Steps:

Let the present ages of A and B be $x$ and $y$.

$n$ years back the ages of A and B are $x-n$ and $y-n$

Given $x = 3(y-n)$ --> "I am three times as old as you were"

$x-n = y$ --> "when I was as old as you are"

Also given $x+y = 60$

Solving these equations will give the above results.

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May I know if my answer is correct? – Santosh Rao Oct 21 '13 at 5:06