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
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Upstream-downstream question
This topic has expert replies
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)
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"
Well I checked to see but this is all thats given. The answer choices, which might be helpful, in case you are plugging in areiamseer 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)
(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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Rahul@gurome
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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).
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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Quant Expert
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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Rahul@gurome
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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).
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
Quant Expert
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)
Quant Expert
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)
Thanks a bunch Rahul! This really helped!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).
My mistake: Nothing missing. the answer is 9*(speed of boat in still water)/(speed of stream)... and indeed very nicely solved by Rahul.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)
Thanks.
"Choose to chance the rapids and dance the tides"
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SANTOSH MOHANTY
- Newbie | Next Rank: 10 Posts
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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
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
