Distance Problem - Upstream/Downstream/Current Speed

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.

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