Need answer to this puzzle??

[FoMoCoDave]

The Harvard Bridge and the Longfellow Bridge are one mile apart. The MIT crew team starts rowing upstream at the Longfellow. As the crew passes under the Harvard, the coxswain’s neoprene visor falls into the river. Exactly 10 minutes later, the coxswain notices, turns the boat around instantaneously and has the crew row back to get it, rowing at the same constant rate as they had rowed previously (but this time with the current rather than against it). By the time the team reaches the visor, they are back at the Longfellow Bridge. How fast is the river flowing?

Anyone??

 

[SHOE]

:idea: What color was the visor? :-D

 

[nj7703]

Since the crew's speed is measured relative to the river and the visor is floating on the river, the time it takes the crew to get back to the visor is also 10 minutes. So, in 20 minutes the visor travelled 1 mile down the river. The river's speed is 1 mile/20 minutes or 3 miles per hour.

 

[Canoodler]

Originally posted by nj7703
Since the crew's speed is measured relative to the river and the visor is floating on the river, the time it takes the crew to get back to the visor is also 10 minutes. So, in 20 minutes the visor travelled 1 mile down the river. The river's speed is 1 mile/20 minutes or 3 miles per hour.

No, because they turn around 10 minutes later to go after it, and then it takes them an unspecified amount of time to reach it after they realized it was missing.

 

[03cobrablack]

I think a better question is why would anyone want to be called a coxswain?????

 

[FoMoCoDave]

Originally posted by nj7703
Since the crew's speed is measured relative to the river and the visor is floating on the river, the time it takes the crew to get back to the visor is also 10 minutes. So, in 20 minutes the visor travelled 1 mile down the river. The river's speed is 1 mile/20 minutes or 3 miles per hour.

nj7703 MENSA??

 

[SHOE]

At 3 mph, the visor would have floated 1/2 mile in 10 minutes (3 miles X 10/60 or .166667). In another 10 minutes it would have floated 1 mile. So I think he's right. The question is, how fast was the crew rowing?

 

[nj7703]

He is right.. i looked it up, i just copied and pasted the answer from an MIT test, which is where this originated. I'll post the link after dinner :-D

 

[esqeddy]

nj7703 is correct. This is really a classic trick question. The real puzzler is to ask someone: "...and what was the speed of the boat?"

Here is the proof:

First, this is a simple problem in linear algebra. Normally, to lock any variable down you must have at least as many independant relationship (equations) as you have unknowns (variables). Otherwise the answer would not be a specific value but another equation(s).

let's look at what we have:

Knowns and Variables:
1 mile = distance between bridges.
R = rate the river flows in mph.
S = the speed the boat goes relative to the water in mph.
1/6 hr = time before turning around.
t = hours it takes the boat to catch up to the visor after turning around.

While traveling up stream the boat is traveling at its speed S minus the speed of the river R flowing against it (S-R). So, At the moment the boat turns around it is 1/6(S-R) beyond the second bridge, and the visors have floated a distance of 1/6R back toward the first bridge. In the time t that it takes the boat to catch up to the visor, the visor will have floated a total of 1 mile:

1/6R + Rt = 1 <----- Our first equation.

The boat having traveled past the second bridge has more ground to cover. On the way back it is traveling its speed plus the speed of the river (S+R) so it will travel a total distance of (S+R)t to reach the visor. We know this distance minus the distance it traveled past the second bridge, which is 1/6(S-R), is one mile.

(S+R)t - 1/6(S-R) = 1 <----- Our second equation.

But wait. We have 3 variables and only two equations. So we have to get a bit tricky. Maybe we can pin some variables down anyway. Let's try substitution.

From the first equation we can derive that:

t = 1/R - 1/6

substitute this into the second equation:

(S+R)(1/R-1/6) - 1/6(S-R) = 1

S/R +R/R -S/6 - R/6 -S/6 +R/6 = 1

S/R +1 -2S/6 = 1

S/R -2S/6 = 0

S/R = 2S/6

1/R = 2/6

R = 3 miles per hour.

Of course, from this we can determine that t = 1/3 - 1/6 = 1/6 hrs.

Now you may ask what is the speed. Well remember that there were only 2 equations? Guess what, go back now and try and isolate S and you will find that you can't. The S term always falls out of the second equation. So perhaps our problem allows for more than one speed for the boat to still satisfy the conditions.

Consider. S must be greater than R otherwise the boat would have never made it up stream to begin with!

So.... S > 3. But is there an upper limit? No. OK OK....don't get cute. We all know that the speed of light is the ultimate speed limit, and that there is a limit of speed dependant on home much river there is beyond the bridge. So, what this means, is that it doesn't matter how fast the boat was going relative to the water in this problem. You will still get the same answer.

QED

 

[KevinB120]

All I know is that it makes no difference in a tripple-502ci Fountain 47' go-fast MOVE OUT OF THE WAY ROW BOYS....EAT WAKE!!! VAROOOOOOOOOOOOOOOMMMMMMM:burnout:

 

[Canoodler]

Meh :shrug:

 

[ZincDude]

Dude just get a new frinkin cap!!

 

[99cobrablack]

Originally posted by KevinB120
All I know is that it makes no difference in a tripple-502ci Fountain 47' go-fast MOVE OUT OF THE WAY ROW BOYS....EAT WAKE!!! VAROOOOOOOOOOOOOOOMMMMMMM:burnout:

I was thinking why the hell would someone want to row when we have things like this^^

 

[FuryShift]

I say, who the hell chases down a damn visor for ten minutes? And how the hell does it take some dumbass another 10 minutes to realize it's gone, I wouldn't let him be MY coxswain.. :nonono:

 

[nj7703]

http://www.plainsandpeaks.us.mensa.org/2003/oct03.htm

The answer is about 1/2 way down.