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Two swimmers start swimming in a swimming pool from opposite ends. They met first at a distance of $50\,\text m$ from east and return back. They met again for a second time at $20\,\text m$ from west. Find the length of pool. Assume their speeds are constant.

My Approach:

Let there be 2 swimmers $E$ and $W$ swimming from east and west direction respectively and the speeds be $S_e$ and $S_w$.

Case 1:

$$\begin{align} \text{Time taken by } E &= \frac{50}{S_e}\\[5pt] \text{Time taken by } W &= \frac{d-50}{S_w} \end{align}$$

Clearly, $$\frac{50}{S_e}=\frac{d-50}{S_w}\implies\frac{S_w}{S_e}=\frac{d-50}{50}\tag1$$

I couldn't form the second equation for the second meeting.

Can anyone guide me how to solve the rest of the problem?

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$W - - - - - - - ->\bullet<= = 50m = = E$

$W <= = = = = = = = = = \bullet - - - - - - >E$

$W = = 20m = =>\bullet<- - - - - - - - E$

It can be solved creatively with a minimum of algebra.
From the diagram, it should be clear that when they first meet, together they have covered the length of the pool, and then twice the length the next time they meet.

Since their respective speeds are constant,
each has travelled twice the distance on the 2nd leg compared to the first leg.

Let the length of the pool be L m, and we will track the $= = = =>$ swimmer.

In the first leg she swam $50m$, so in the 2nd leg she must have swum $100m$

In the 2nd leg, she swam $L-50+20,$ so equating

$L - 50 + 20 = 100$, which yields $L = 130m$

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  • $\begingroup$ Why 2*50 Sir I have not understood? $\endgroup$ – justin takro Oct 20 '15 at 17:23
  • $\begingroup$ On first leg, the swimmer from east travels 50m, so on the 2nd leg, she must have travelled double the distance, 100m. On the 2nd leg, she travelled L-50 + 20, so equate it to 100 $\endgroup$ – true blue anil Oct 20 '15 at 17:46
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For second equation,
Time taken by E = $\frac{d+20}{S_E}$
Time taken by W = $\frac{2d-20}{S_W}$
Hence we have $$\frac{d+20}{S_E}=\frac{2d-20}{S_W}$$ and for your work we have $$\frac{50}{S_E}=\frac{d-50}{S_W}$$ Therefore $$\frac{d+20}{50}=\frac{2d-20}{d-50}$$ $$d^2-30d-1000=100d-1000$$ $$d^2-130d=0$$ $$d(d-130)=0$$

Answer is $d=130\,\mathrm{m}$

EDIT: E covers the entire length and further $20\,\mathrm{m}$. And W covers the entire length and further $(d-20)\,\mathrm{m}$. Hence the second equation. Actually earlier I had assumed that the swimmers return back after they meet for the first time i.e. they meet and return. Then I saw the other answer where it was assumed that after meeting the swimmers go on to complete the entire distance and then only return. However your question did not specify that.

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  • $\begingroup$ Ans is $130$.Check your equations again.Isn't the second equation be 2D-20/Se=?/Sw.In place of ? its D+20 but i don't know how they got it. $\endgroup$ – justin takro Oct 20 '15 at 17:18
  • $\begingroup$ I have updated my answer. See if it fits in. $\endgroup$ – SchrodingersCat Oct 21 '15 at 5:59

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