Hardy and Andy start a two-length swimming race at the same moment but

Let Length of Pool be L, speed of Hardy be H and that of Andy be A. Given H>A

They 1st pass at a point 18.5m from the deep end

(L - 18.5)/H = 18.5/A --- eqn1

1/A = (L - 18.5)/18.5 H ------eqn2

They cross for 2nd time at a point 10.5m from the shallow end

(L - 10.5)/ H = 10.5/A ----eqn3

now total time spent by both from 1st end to 2nd end ,plus wait period and crossing for second time will be equal i.e.

L/H + 45 + (L - 10.5)/ H = L/A +45 + 10.5/A

on solving this we get
(2L - 10.5)/H = (L+10.5)/A----eqn 4

substituting the value of 1/A from eqn2 in eqn 4 we get

(2L - 10.5)/H = (L+10.5) * (L - 18.5)/18.5 H, on solving this we get L = 45

How have you got 3x. I don't get it. explain please.

Posted from my mobile device

Hi Naeem,

x=length of pool
Andy's distance to second meeting=
18.5+(x-18.5)+10.5=x+10.5 m
Hardy's distance to second meeting=
(x-18.5)+18.5+(x-10.5)=2x-10.5 m
combined distance to second meeting=
(x+10.5)+(2x-10.5)=3x m

I hope this helps.
gracie

Hi Gracie

Although I do understand (at least i think i do) this part: if andy swims 18.5 m in x, then he will swim 3*18.5=55.5 m in 3x
Could you please explain that mathematically?
Is it because the distance covered by each party till the point of meeting is in proportion to their speed?

Thanks
Saumya

Hi Saumya,

yes, because time of each to meeting is equal, distance of each is proportional to their speeds
combined distance to first meeting is x meters
andy swims 18.5 of those x meters
thus, the ratio of andy's distance to combined distance is 18.5/x meters
assuming consistent speeds and multiplying by 3,
the same ratio will apply to second meeting: 55.5/3x meters

I hope this helps.
gracie

How do you say that the 45 seconds wait time does not matter?
And can something of this sort be expected on the GMAT?

The wait time doesn't matter since after their first meet, they both rest for 45 secs and then meet again.
Say, they meet again in 2 mins. But actually both swam for only 1 min 15 secs. Doesn't matter to us either way. Time for which they swam is still the same. So ratio of distance covered will still be the same as the ratio of their speeds.

I am stilling struggling to understand the question since I'm not english native speaker. Could anyone help explain the meaning of "deep end" and "shallow end" mentioned in the question? Without understanding correctly, I cannot depict the question....

A swimming pool is typically constructed in a rectangular fashion (L X B). It doesn't really matter what the deep and shallow ends actually are, other than the fact that they are opposite ends of a swimming pool.

Shallow end (4ft depth) Deep end (17ft)
_____________________________________________
|-------------------------------------------------------------|
|-------------------------------------------------------------|
|-------------------------------------------------------------|
|-------------------------------------------------------------|
|-------------------------------------------------------------|
<---------------Length of the pool---------------------->

The question indicates the race length as 2-length, which means the swimmer:

- goes from one end to and another
- and returns to that end to finish the race

Hope this helps.

The key to solving is getting the right distance and the distance travelled by other contestants which is
let the distancee of the pool be x then the total distance covered = 3x
and if amit ravelled x+10.5 which is equal to 3*18.55
=>x+10.5 =55.5
=>x=45
Therefore IMO B

­Thanks, this was a very good approach at the problem.