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 Post subject: M10 #04 [#permalink]
PostPosted: Wed Nov 05, 2008 12:13 pm 
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Kudos (?): 11 (0), given: 0

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?

10 hours
12 hours
15 hours
18 hours
20 hours

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


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 Post subject: Re: m10 -#4 [#permalink]
PostPosted: Wed Nov 05, 2008 12:41 pm 
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Kudos (?): 10 (0), given: 1

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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 Post subject: Re: M10 #04 [#permalink]
PostPosted: Fri Dec 04, 2009 7:36 am 
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Kudos (?): 2 (0), given: 2

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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 Post subject: Re: M10 #04 [#permalink]
PostPosted: Fri Dec 04, 2009 8:34 am 
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Kudos (?): 96 (2), given: 18

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
P1.png [ 21.2 KiB | Viewed 298 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.


Answer : 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!
compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

Word Problems Made Easy!
1) Translating the English to Math : word-problems-made-easy-87346.html
2) 'Work' Problems Made Easy : work-word-problems-made-easy-87357.html
3) 'Distance/Speed/Time' Word Problems Made Easy : distance-speed-time-word-problems-made-easy-87481.html


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 Post subject: Re: M10 #04 [#permalink]
PostPosted: Fri Feb 12, 2010 10:03 pm 
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Kudos (?): 7 (0), given: 19

Great explanation sriharimurthy! Kudos!


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