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I was asked this problem during an interview recently: A pedestrian starts walking from town A to town B. At the same time, another pedestrian starts walking from town B to town A. They pass each other at noon and continue on their paths. One of them arrives at 4 PM, the other at 9 PM. How many hours had each walked before passing each other?

My logic so far is that: B is slower than A, since it took him more time. From noon onward, the 4 hours it took A to finish was what B had walked until now. The 9 hours it took B to finish is was A walked until now. For some reason I cannot find how to finish it with the S = V * T formula...

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  • $\begingroup$ Hint: draw a graphic of their movement, you'll have similar triangles. $\endgroup$ Sep 6, 2016 at 11:05

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Let the distance between $A$ and $B$ be $x$, the faster pedestrian's speed $s$ and the slower pedestrian's speed $t$. From their meeting point $C$ at $12$ pm, the distance to $A$ is $4s$ and the distance to $B$ is $9t$. Therefore, we get that $4s+9t=x$. Hence, the distance walked by the faster pedestrian before meeting the other one is $9t$ and he does this in the same time that the other pedestrian does the distance $4s$. So, $\frac{4s}{t} = \frac{9t}{s}$, so $4s^2=9t^2$. Now, on simplifying, $s=\frac{3t}{2}$, hence we can find the original time now, by doing distance by time:$\frac{9t}{\frac{3t}{2}} = 6$. Hence, each walked for six hours before the meeting.

Verifying the answer:

Both start out at six am. One walks for ten hours, one for fifteen hours, so the speeds are $\frac{x}{10}$ and $\frac{x}{15}$. Hence, the meeting point should be six hours afterwards for both, which is true because $6\frac{x}{10} + 6\frac{x}{15} =1$. Hence, both walk for $6$ hours is correct.

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We have: $$A-------------M--------B$$

Let $|MB|=x$ and $|AM|=kx$.

Pedestrian1 covers $x$ in $4$hrs, Pedestrian2 covers $kx$ in $9$hrs.

Pedestrian1 covers $kx$ in $4k$hr, Pedestrian2 covers $x$ in $\dfrac9k$ hrs.

So $4k=\dfrac9k\to k^2=\dfrac94\to k=\dfrac32$.

Which gives $6$ hours as the answer.

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