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Upstream-downstream question

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by r2kins » Sat Apr 24, 2010 8:23 am
Can somebody pls suggest a short method for solving this question:

Q - A crew can row a certain course upstream in 84 min; they can row the same course downstream in 9 minutes less than they can row it in still water. How long will they take to row down with the stream?

A -

(a)45 or 23 minutes
(b)63 or 12 minutes
(c)60 minutes
(d)19 minutes
(e)25 minutes
Last edited by r2kins on Sat Apr 24, 2010 11:20 am, edited 1 time in total.
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by iamseer » Sat Apr 24, 2010 9:24 am
Is that all the information.
You sure nothing is missing in the question? I think the speed of the stream is missing.

And once we know that the answer is 9/(speed of stream)
"Choose to chance the rapids and dance the tides"
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by r2kins » Sat Apr 24, 2010 11:18 am
iamseer wrote:Is that all the information.
You sure nothing is missing in the question? I think the speed of the stream is missing.

And once we know that the answer is 9/(speed of stream)
Well I checked to see but this is all thats given. The answer choices, which might be helpful, in case you are plugging in are

(a)45 or 23 minutes
(b)63 or 12 minutes
(c)60 minutes
(d)19 minutes
(e)25 minutes

Any ideas now ? Can any of the gmat experts here help?
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by Rahul@gurome » Sat Apr 24, 2010 8:08 pm
Solution
According to question
d/(x - y) = 84.
And d/x - d/(x + y) = 9 or dy/{x*(x + y)} = 9.
Dividing second by first we get that {y*(x - y)}/{x*(x + y)} = 9/84 = 3/28.
Divide both numerator and denominator by x^2.
Let y/x = k
So we get that {k*(1 - k)}/(1 + k) = 3/28.
From above we get a quadratic equation 28k^2 - 25k + 3 = 0.
Or k = 1/7 or k = ¾.
What the question wants is d/(x + y) or (d/x)/(1 + y/x) = (d/x)/(1 + k)
Now (d/x)/(1 - k) = 84.
So (d/x)/(1-1/7) = 84, if we take the first value of k = 1/7.
Or d/x = 84 * 6/7 = 72.
So d/(x+y) = 72 - 9 = 63
If k = ¾, then (d/x)/(1 - ¾) = 84 or d/x = 21.
So d/(x+y) = 21 - 9 = 12.
So d/(x+y) is either 63 or 12 minutes.
The correct answer is (b).
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by Rahul@gurome » Sat Apr 24, 2010 8:10 pm
Let the distance the crew rows be d.
x - speed of crew in still water.
y - speed of the stream.

Solution
According to question
d/(x - y) = 84.
And d/x - d/(x + y) = 9 or dy/{x*(x + y)} = 9.
Dividing second by first we get that {y*(x - y)}/{x*(x + y)} = 9/84 = 3/28.
Divide both numerator and denominator by x^2.
Let y/x = k
So we get that {k*(1 - k)}/(1 + k) = 3/28.
From above we get a quadratic equation 28k^2 - 25k + 3 = 0.
Or k = 1/7 or k = ¾.
What the question wants is d/(x + y) or (d/x)/(1 + y/x) = (d/x)/(1 + k)
Now (d/x)/(1 - k) = 84.
So (d/x)/(1-1/7) = 84, if we take the first value of k = 1/7.
Or d/x = 84 * 6/7 = 72.
So d/(x+y) = 72 - 9 = 63
If k = ¾, then (d/x)/(1 - ¾) = 84 or d/x = 21.
So d/(x+y) = 21 - 9 = 12.
So d/(x+y) is either 63 or 12 minutes.
The correct answer is (b).
Rahul Lakhani
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Gurome, Inc.
https://www.GuroMe.com
On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits
1-800-566-4043 (USA)
+91-99201 32411 (India)
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by frank1 » Sat Apr 24, 2010 8:26 pm
Thanks
but to be frank,
The question itself was confusing for me (that means 30 seconds for reading question )
and all other in 1.30 minutes?
seems to be frightening for me....lol
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by r2kins » Sat Apr 24, 2010 9:10 pm
Rahul@gurome wrote:Let the distance the crew rows be d.
x - speed of crew in still water.
y - speed of the stream.

Solution
According to question
d/(x - y) = 84.
And d/x - d/(x + y) = 9 or dy/{x*(x + y)} = 9.
Dividing second by first we get that {y*(x - y)}/{x*(x + y)} = 9/84 = 3/28.
Divide both numerator and denominator by x^2.
Let y/x = k
So we get that {k*(1 - k)}/(1 + k) = 3/28.
From above we get a quadratic equation 28k^2 - 25k + 3 = 0.
Or k = 1/7 or k = ¾.
What the question wants is d/(x + y) or (d/x)/(1 + y/x) = (d/x)/(1 + k)
Now (d/x)/(1 - k) = 84.
So (d/x)/(1-1/7) = 84, if we take the first value of k = 1/7.
Or d/x = 84 * 6/7 = 72.
So d/(x+y) = 72 - 9 = 63
If k = ¾, then (d/x)/(1 - ¾) = 84 or d/x = 21.
So d/(x+y) = 21 - 9 = 12.
So d/(x+y) is either 63 or 12 minutes.
The correct answer is (b).
Thanks a bunch Rahul! This really helped!
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by iamseer » Mon Apr 26, 2010 10:13 am
iamseer wrote:Is that all the information.
You sure nothing is missing in the question? I think the speed of the stream is missing.

And once we know that the answer is 9/(speed of stream)
My mistake: Nothing missing. the answer is 9*(speed of boat in still water)/(speed of stream)... and indeed very nicely solved by Rahul.

Thanks.
"Choose to chance the rapids and dance the tides"
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by SANTOSH MOHANTY » Sun Sep 17, 2017 10:12 am
PLEASE read the question carefully ..it's very easy.
Get to question
Remember we need to find the time taken by boat to travel X distance in DOWNSTREAM.

distance travelled by boat is same for ( upstream=downstream=stillwater )

1-> The speed of the boat in still water = x / t
t=time travelled to take x distance in still water


2-> A crew can row a certain course UPSTREAM in 84 min means :
travelling 84min in X distance i.e ( speed of upstream = X / 84 )

3-> They can row the same course DOWNSTREAM in 9 minutes less than they can row it in STILL WATER.

TIME TAKEN IN DOWNSTREAM IS ............. ( T = t -9 )

T= time taken to go X distance in DOWNSTREAM
t= time taken to X distance in STILL WATER

Downstream speed = X / (t-9)

we know the formula for finding the Speed of boat in still water is equal to :

FORMULA=
The speed of boat = (a+b )/2= (upstream speed + downstream speed ) / 2

now put the value in terms of distance/ time ....you will get the answer:

X/t = ( X/84) + (X/ t-9) / 2

solve this you will get the answer
t = 84*2*(t-9) / 84+ (t-9 )

t - 9 = 63
don't forget
t is time take by boat in still water
t-9 is time taken by boat in downstream
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