Forum Home
PC World Chat
 
Thread ID: 95701 2008-12-14 23:48:00 Maths Problem Dally (6292) PC World Chat
Post ID Timestamp Content User
729106 2008-12-14 23:48:00 Having read the missing dollar post here is a proper problem
M Phelps (Olympic champion) and J Smith (Good swimmer)start off from opposite ends of a swimming pool simultaneously. Phelps from end A and Smith from end B. They swim at a constant speed Phelps being the quicker. They are allowed exactly one minute rest at the turns. They pass 35m from end B and on turning they pass 15m from end A
How long is the swimming pool from end A to end B 		Dally (6292)
729107 2008-12-14 23:51:00 I've never been too good at these things, but wouldn't we need to know how fast they swim (or how much faster mike is) to work this out? wratterus (105)
729108 2008-12-15 00:32:00 The minute rest is irrelevant because it happens on both sides of the equation so it cancels itself out.

It needs to be longer than 70m, as M is faster than J.

I don't know the answer, but I'm thinking that I need some more info before I can say. 		Thebananamonkey (7741)
729109 2008-12-15 00:43:00 Sorry - No more information is required Dally (6292)
729110 2008-12-15 00:54:00 Can I just make a guess and say 100m? Thebananamonkey (7741)
729111 2008-12-15 01:40:00 I thought 100m would be the guess but it is wrong Dally (6292)
729112 2008-12-15 02:20:00 90m MushHead (10626)
729113 2008-12-15 02:41:00 85m? --Wolf-- (128)
729114 2008-12-15 03:16:00 OK then, to clarify:

let X be our pool size, Vp & Vs the swim speeds of Phelps & Smith respectively.

We have 2 points in time that we know - when the swimmers cross each other.


t1 = (X - 35)/Vp = 35/Vs

and at the other end (I'm assuming Smith has swum a length, rested 60s, then started back, rather than Phelps overtaking him on Smith's first length)


t2 = (2X - 15)/Vp + 60 = (X + 15)/Vs + 60

from these 2 relationships we can say


Vp = (X - 35)Vs/35 (from t1)

We substitute this into the t2 equation (after cancelling out the 60s on each side):


(2X - 15) = (X + 15)
((X-35)Vs/35) Vs

Now we can cancel out Vs & reorganise to get:


35(2X - 15) = (X + 15)(X - 35)

or


70X - 525 = X^2 - 20X - 525

this reduces down to


90X = X^2

so X=90. 		MushHead (10626)
729115 2008-12-15 04:05:00 Very good Mushead - There is also a beautiful intuitive solution

When they first cross their combined distance equals one length. When they cross again their combined distance equals three lengths therefore JS has swum three times 35m which is 105m but he has swum one length plus 15m - Length of pool 90m 		Dally (6292)
1 2