# How long would a track be?

A and B are swimming in lanes right next to each other, but in different directions. They both start at the same time, and they pass each other after person A has swam 84 feet. When they reach the end, they turn around and swim back, meeting again 36 feet away from person A's starting point. How long is the pool?

I would assume the two swimmers are swimming at different rates, but I don't know how I would go about this problem. I would assume trying to find ratios would be the best, but I don't know what to do.

Thank you!

• Do they start at the same place? Sep 20, 2017 at 3:03
• They start at opposite ends Sep 20, 2017 at 3:06
• @MathLover: we only care about the ratio of the swimming speeds, not the individual values. That decreases the number of variables by $1$, making the problem soluable. Sep 20, 2017 at 3:08

I recall a similar problem from the Moscow Puzzles. A clever way of solving with minimal algebra is as follows:

When they pass each other first, they have together crossed $1$ length of the pool. When they cross again, they have crossed $3$ lengths of the pool, and since their speeds are constant, we know that it took $3$ times as long.

So in the whole time elapsed, swimmer A traveled $84*3=252$ feet and met $36$ feet from his starting point. Thus $2L=252+36$ and $L=144$.

• I was thinking of that too! Hey! Sep 20, 2017 at 3:18
• Yes, it is a very nice way of solving the problem. Had you heard the problem before as well @neonpokharkar? Sep 20, 2017 at 3:21
• Well, as I'm giving IIT advanced (most honourable and toughest logical exams in India). So I'm in par with such easy ways to solve problems. And no I hadn't heard of this problem. Sep 20, 2017 at 3:34
• Wow. I'm gonna look that up now. Sep 20, 2017 at 3:38

Let the length of the pool be $L$. When they first meet, $A$ has swum $84$ feet and $B$ has swum $L-84$. How far have they each swum the next time they meet? Now require that the speeds be constant, so the ratios of distances are the same at each meeting. That gives an equation in $L$.

• I understand... but I also don't..? How does that give an equation in $L$? Sep 20, 2017 at 3:08
• Wouldn't that give an equation for their speeds? Sep 20, 2017 at 3:09
• The ratio at the first meeting is then $\frac {84}{L-84}$. Equate that to the ratio at the second meeting. See the $L$ there? Sep 20, 2017 at 3:09
• Ah. So the second would be $\frac{2L-36}{L+36}$! Thank you so much! (I'll accept the answer in 3 minutes because that's when I can accept) Sep 20, 2017 at 3:10

Let's say, that the pool's length is $X$ metres ($X>0$)

1. When they pass each other first time, A has swam $84$ metres and B has swum $X-84$ metres.
2. When they pass each other second time, A has swam $2X-36$ metres and B has swum $X+36$ metres.
3. Let's say, that A swims $a$ times faster than B. Thus: $$a=\frac{84}{X-84}=\frac{2X-36}{X+36}$$ Now let's solve the equation we've just obtained: $$\frac{84}{X-84}=\frac{2X-36}{X+36}\\ X^2 - 144X = 0\\ X(X-114)=0\\ X=0 \vee X=144$$ It was said, that $X>0$, thus the length of the pool is $144$ metres.