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M06 #19

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M06 #19 [#permalink] New post 24 Sep 2008, 20:16
A swimmer practiced in a still water inlet a river made in its bank, and achieved a speed of 4 km/h. When she was swimming upstream in the river, however, she could only make 3 km/h. One day the swimmer swam into the river from the inlet, then doubled the covered distance going downstream, and finally came back. She averaged 3.8 km/h that day. What was her downstream speed?

To solve the problem, pick a number for the distance. Take 6 km for each leg up and down the stream, which leaves 3 km for the still water. { Can someone please tell me here how we know that she stopped in between and swam in still water? Is it because question states that she "finally came back". Also, how do we know that she swam 3 KM in still water? Thank you. I answered this question correctly but using a different approach, which now I'm not so sure is valid: Her speed upstream at 3m/hr should equal her still water rate of 4km/hr minus the speed of the current - so 1km/hr. If the speed of the current is 1 km/hr then downstream she should swim 4km/hr + 1 km/hr with the stream helping her, resulting in 5 km/hr. Is that incorrect? Thank you :shock: .} The total swim was then 6+6+3 km and took 257 minutes.


Swimming in still water took hours 3/4 = 45 minutes, and upstream 6/3 hours = 120 minutes. This leaves 6 km and minutes = 1.2 hours for downstream. We can find a downstream speed by constructing an equation ( denotes the downstream speed):




The correct answer is B.
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Re: M06 #19 [#permalink] New post 09 Oct 2008, 02:27
This question is up for rewording. We see that it needs some corrections and disambiguations.

In the OE we only suppose that the distances covered are 3 km in still water and 6 km downstream and 6 km upstream.

"Finally came back" refers to swimming upstream after the downstream swim. Let's put it this way: the swimmer swims in a still water (the inlet), then swims down the river and then swims back (covering the same distance as the downstream swim, which is twice the distance in a still water).

Sorry for some ambiguities.
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Re: M06 #19 [#permalink] New post 02 Apr 2009, 22:48
Anyone solved this question using algebra?
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Re: M06 #19 [#permalink] New post 22 May 2009, 17:42
I tried to solve this question with algebra while doing the test. It has since been reworded:

A boat crossed a lake from North to South at the speed of 4 km/h, entered a river and covered twice as much distance going upstream at 3 km/h. It then turned around and stopped at the south shore of the lake. If it averaged 3.8 km/h that day, what was its approximate downstream speed?


The answer choices and numbers remain the same.

So d1 (distance crossing the lake) = 4t1 (where t1 = time spent cross the lake)

d2 (going upstream) = 3 * t2 = 2 * d1 = 8 * t1
d3 = d2 = s * t3 (where s is the downstream speed) = 2 * d1 = 8 * t1

So average = 3.8 = t1(4 + 8 + 8)
---------------
t1 + 8/3*t1 + t3


Solve and you get something like
18.2 * t1/3 = 3.8 * t3

t3 = (8 * t1)/s

substitute and you can get s = (3.8 * 8 * 3) / 18.2
~~ 5.01

IMO numbers were hard to manipulate. Thats why I got it wrong :)
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Re: M06 #19 [#permalink] New post 17 Sep 2009, 02:50
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I dont think too much is reqd in this question. Speed in lake (still water) 4 kph, speed upstream (ie relative speed of boat and upstream river) is 3 kmph. This means the speed of river is 4-3 kph ie 1 kmph. so, downstream speed will be 4 + 1 = 5 kph.
Is this simplistic approach OK
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Re: M06 #19 [#permalink] New post 22 Sep 2009, 11:45
Any approach is good provided you get the right answer with it :). The faster you can get to the right answer, the better for you.
tejal777 wrote:
I dont think too much is reqd in this question. Speed in lake (still water) 4 kph, speed upstream (ie relative speed of boat and upstream river) is 3 kmph. This means the speed of river is 4-3 kph ie 1 kmph. so, downstream speed will be 4 + 1 = 5 kph.
Is this simplistic approach OK

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Re: M06 #19 [#permalink] New post 28 May 2010, 05:38
Do you mean this question needs corrections?

If yes, then please. I was baffled by the statements and took 4+ min to guess :(


dzyubam wrote:
This question is up for rewording. We see that it needs some corrections and disambiguations.

In the OE we only suppose that the distances covered are 3 km in still water and 6 km downstream and 6 km upstream.

"Finally came back" refers to swimming upstream after the downstream swim. Let's put it this way: the swimmer swims in a still water (the inlet), then swims down the river and then swims back (covering the same distance as the downstream swim, which is twice the distance in a still water).

Sorry for some ambiguities.

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Re: M06 #19   [#permalink] 28 May 2010, 05:38
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