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Distance Problem - Upstream/Downstream/Current Speed

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Distance Problem - Upstream/Downstream/Current Speed

by ogbeni » Fri Sep 25, 2009 6:50 am
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

OA is D. I was way off on this one :(

I think I may have bombed it coz I'm not too clear what direction the rafter would float (upstream or downstream). Anyway you bright folks should chime in.
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Re: Distance Problem - Upstream/Downstream/Current Speed

by Brent@GMATPrepNow » Fri Sep 25, 2009 7:06 am
ogbeni wrote: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

OA is D. I was way off on this one :(

I think I may have bombed it coz I'm not too clear what direction the rafter would float (upstream or downstream). Anyway you bright folks should chime in.
Let B = cruising speed of the boat (steamer) in calm water
Let C = speed of current.

So, traveling upstream, the boat's net (overall) speed is B-C
Traveling downstream, the boat's net (overall) speed is B+C

Time = distance/speed

We'll let d = the distance from town A to town B

We get two equations: d/(B-C) = 4.5 and d/(B+C)=3
When we solve this system of equations, we get B=5C

Now we want to know the time it takes a raft to travel the same distance (d). Since the raft has no power, the raft will move along at the speed of the current (C)

So, the time for the raft to get from A to B equals d/C. What is the value of d/C?

Since we know that B=5C, we can take one of our earlier equations, d/(B+C)=3, and replace B with 5C to get: d/(5C+C)=3
simpplify --> d/6C=3
Multiply both sides by 6 to get d/C=18

So our answer is D.
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by ssmiles08 » Fri Sep 25, 2009 7:11 am
Let A to B = upstream. Which means B to A is downstream.

The Streamer has its own rate with which it moves (+ or - the rate of the current)

Let d = distance.

Upstream: (r-c)*(9/2) = d; it is r - c b/c the streamer is flowing against the current and therefore the total rate would be the rate of the streamer - rate of the current.

Downstream: (r+c)(3) = d: it is r + c b/c the streamer is moving alongside with the current; which makes sense b/c it covers the same distance d in a shorter period of time.

set two equations equal to each other: (r-c)*(9/2) = (r+c)(3); r = 5c.

The question is asking how many hours it would take to get from B to A (downstream) only with the rate of the current.

(c)(t) = d

we know that from above: (5c+c)*3 = d; 18c = d

so (c)(t) = 18c; The c's cancel out. t = 18 hours. (D)
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by ogbeni » Fri Sep 25, 2009 7:18 am
Bret,

Wow. Here's how I approached it and let me know where my assumption was faulty.

I chose a distance that would be easy to work with D=18 (one way).

1. Going upstream with a time of 4.5 hrs, speed will be 18/4.5=4
2. Going downstream with a time of 3 hrs, speed will be 18/3 = 6.

Therefore, speed of currrent will be 6-4=2.

That said, if the rafter is floating downstream at the speed of the current it will take 18/2 = 9 hours to complete the journey. What's wrong with this approach? I see that 9 is half of 18 but I'm not connecting the dots :(
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by Brent@GMATPrepNow » Fri Sep 25, 2009 8:19 am
ogbeni wrote:Bret,

Wow. Here's how I approached it and let me know where my assumption was faulty.

I chose a distance that would be easy to work with D=18 (one way).

1. Going upstream with a time of 4.5 hrs, speed will be 18/4.5=4
2. Going downstream with a time of 3 hrs, speed will be 18/3 = 6.

Therefore, speed of currrent will be 6-4=2.

That said, if the rafter is floating downstream at the speed of the current it will take 18/2 = 9 hours to complete the journey. What's wrong with this approach? I see that 9 is half of 18 but I'm not connecting the dots :(
This is a great (better) approach.
The only problem is how you set it up.
You got two speeds of 4 and 6 (perfect)
Using my earlier definitions of C and B, we get the two equations:
B+C = 6
B-C = 4
Subtract bottom equation from top to get 2C=2
So, C=1.
Since the speed of the current is 1 and the distance is 18, the time will be 18/1

Great approach - I should have caught that :)
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by ogbeni » Fri Sep 25, 2009 9:36 am
Bret,

Again wow! You know math is truly amazing and I wish I had my present attitude when I was a kid. I'd just look at problems and say 'it's hard'.

So my mistake was an incorrect/simplistic assumption of the speed of the current. Looking at the 2 equations you set up:

B+C=6
B-C=4

It becomes obvious that I 'over-attributed' to the speed of the current when I just subtracted 4 from 6 to get 2.

How do you math mavens just get to think so properly mathematically? I wanna get there!!! ha ha ha ha :)

And ssmiles08, lest you think I ignored you. Thanks for your contribution!!
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