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# M10 #04

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Manager
Joined: 14 Oct 2008
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05 Nov 2008, 13:13
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This topic is locked. If you want to discuss this question please re-post it in the respective forum.

The steamer going upstream would cover the distance between town A and town B in 4 hours and 30 minutes. The same steamer going downstream would cover the distance between the towns in 3 hours. How long would it take a raft moving at the speed of the current to float from town B to town A?

(A) 10 hours
(B) 12 hours
(C) 15 hours
(D) 18 hours
(E) 20 hours

[Reveal] Spoiler: OA
D

Source: GMAT Club Tests - hardest GMAT questions

I didn't quite understand the explanation by gmat club for this Qs. Any help is appreciated.
Thanks.

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Manager
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05 Nov 2008, 13:41
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Let the speed of steamer=s
speed of current=c
Total distance covered=d
So the equation 1 is d/(s+c)=3
Equation 2 is d/(s-c)=9/2
(when the steamer and current in same direction then the resultant speed is added and when the steamer direction and current direction in opposite direction the resultant speed is deducted.)

Solving both equation you get s=5c
and applying the value of s in equation 1 you get the d=18c
So it will take 18 hr.
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04 Dec 2009, 08:36
Say the distance between A and B is 4.5 miles.

Moving from A to B and 1 mile/hr (4.5 hours expended).
Move from B to A at 1.5 mile/hr (3 hours expended).

So from B to A, the downstream acceleration provided by water is 0.5 miles/hour. So if the engines were shut-off the boat would float down 4.5 miles at 0.5 m/h - so it would take 9 hours.

9 is not a choice in the answers provided, why is my reasoning above wrong?

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Manager
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04 Dec 2009, 09:34
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The quickest and most accurate way to solve these problems is through the method of tabular representation.

Note for table : The values in black are those that have been given and the values in blue are those that have been calculated.
Attachment:

P1.png [ 21.2 KiB | Viewed 7379 times ]

Since we want to find out the time taken by the raft to go from Town A to Town B, let us assign it as variable 'x'.

Now let us assume speed of boat in calm water to be 'b' and speed of current to be 's'. This gives us the expressions we require for speed in all three cases.

Note : The first thing that should strike us in this problem is that the distances are all the same. This implies that the solution will in all probability lie in equating the expression for the distances.

In row 1, we are given the time and we know that since the boat is traveling upstream, it's speed must be 'b - s'. Thus we can form an expression for the distance.

Distance(1) = Speed(1) * Time(1) = (b - s)*4.5

In row 2, we are given the time and we know that since the boat is traveling downstream, it's speed must be 'b + s'. Thus we can form an expression for the distance.

Distance(2) = Speed(2) * Time(2) = (b + s)*3

Now let us use our knowledge of the distances being equal in order to get an expression for 'b' in terms of 's'. This can be done by equating Distance(1) and Distance(2) since they are the same.

(b - s)*4.5 = (b + s)*3 ---> b = 5s

Now lets move on to row 3. We have assumed the time taken to be 'x' and we know the speed is that of the current 's'. Thus we can obtain an expression in for the distance in terms of 's' and 'x'.

Distance(3) = Speed(3) * Time(3) = s*x

Again, we know that this must be equal to both Distance(1) and Distance(2). So let us equate it with any one of them to obtain an expression for 'x' in terms of 'b' and 's'.

Equating it to Distance(2) we get :

(b + s)*3 = s*x ---> 3b + 3s = s*x ---> Substituting b = 5s ---> 18s = s*x ---> x = 18 hours.

OR

In case we would have equated it to Distance(1) we would still have got the same result :

(b - s)*4.5 = s*x ---> 4.5b - 4.5s = s*x ---> Substituting b = 5s ---> 22.5s - 4.5s = s*x ---> x = 18 hours.

junker wrote:
Say the distance between A and B is 4.5 miles.

Moving from A to B and 1 mile/hr (4.5 hours expended).
Move from B to A at 1.5 mile/hr (3 hours expended).

So from B to A, the downstream acceleration provided by water is 0.5 miles/hour. So if the engines were shut-off the boat would float down 4.5 miles at 0.5 m/h - so it would take 9 hours.

9 is not a choice in the answers provided, why is my reasoning above wrong?

Moving from A to B and 1 mile/hr (4.5 hours expended)
Move from B to A at 1.5 mile/hr (3 hours expended)

Your reasoning is correct till this point. However, you cannot subtract one speed from the other to get the speed of the current. In one case it is the speed of the boat - speed of current while in the other case it is the speed of boat + speed of current.

Thus you will have the following two expressions :

b + s = 1.5

b - s = 1

Solving them you will get b = 1.25 mph and s = 0.25 mph

Then you can calculate the time taken by raft to be = 4.5/0.25 = 18 hours.

Although this method is not wrong, there is a lot of scope for silly mistakes (as you would have realized) if your concepts are not one hundred percent clear.

Personally, I believe that the most fool proof way to solve these problems is through tabular representation. (At least until your concepts become strong and maybe even then).

You can check out my post on these types of word problems (you'll find the link below). You might find it helpful.

Cheers.
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Click below to check out some great tips and tricks to help you deal with problems on Remainders!
http://gmatclub.com/forum/compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

1) Translating the English to Math : http://gmatclub.com/forum/word-problems-made-easy-87346.html

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Manager
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12 Feb 2010, 23:03
Great explanation sriharimurthy! Kudos!

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29 Sep 2010, 05:03
I am wondering why does the substitution not work for this problem?
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Manager
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06 Dec 2010, 09:24
D.

let the speed of steamer be s and current be c
Respective speeds =>
upstream = s - c
downstream = s + c

distances covered being equal ->
(s - c) *270 = (s+c)*180
(s-c)/(s+c) = 2/3 => s = 5c

s + c = 5c + c = 6c
3 * 6c = x * c
=> x = 18 hrs

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Manager
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11 Dec 2011, 17:13
This question seems pretty confusing. Anyone know where this one would rank?

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Manager
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12 Dec 2011, 07:59
Good explanation Sriharimurthy. One needs to understand the concept of upstream and downstream to convert it into equations.

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10 Dec 2012, 09:02
Let the speed of the boat be B and that of the stream be S. An ideal raft will travel at the speed of the current so time = distance/S
Upstream - the effective velocity of the boat = velocity of boat in still water - velocity of stream = B-S
time = D/(B-S) or D = 4.5(B-S) ......I
Downstream - the effective velocity of the boat = velocity of boat in still water + velocity of stream = B+S
time = D/(B+S) or D = 3(B+S) .....II

equating both values of D we get B = 5S
substitute the value of B in any of the equation to get D= 18S

Therefore, time taken by ideal raft = D/S = 18S/S = 18 hours.

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Current Student
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20 Dec 2012, 15:11
I did this as follows -

s=> speed f boat and r=> speed of river

s-r = 4.5
s+r = 3

solving we get s=7.5/2
r=1.5/2

D= 4.5 * (s-r) = 13.5

Now my issue is understanding the problem it says -

How long would it take a raft moving at the speed of the current to float from town B to town A

How would you know the time is d/r and not d/2*r
I thought since the speed of the raft = current the time taken will be half..again if this is the case the answer would be straight forward (i know) .. but at first glance that is what I understood.. can anyone please me understand the wordings

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Re: M10 #04   [#permalink] 20 Dec 2012, 15:11
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# M10 #04

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