Thread: Is this Problem solvable as written?

I'm normally not bad at word problems. This problem comes from a take home test my instructor gave us Tuesday.

The problem reads:

A boat takes five hours to move upstream against a current of 3mi/hour. The return trip takes only 2.5 hours. What is the relative speed of the boat?

I know the total time of the trip is 7.5 hours and the current is +/- 3 of relative speed.

So.. I started with this equation...

7.5 = (Distance/(r+3)) + (Distance/(r-3))

In every example our textbook did of these types of problems, a distance was always given. I know you can manipulate the equation to d = rt, but I can't get d if I don't know r.

The instructor is saying this is solvable. I don't know mind scratching my head to figure it out as long as someone here can confirm it is indeed solvable. I don't think it is without a distance being given.

let R = the speed of the boat in still water.Write two equations for the distance and solve for R

Thanks for the reply! I ended up with 9mph for relative speed which seems to check out correctly in the original problem!

d upstream = d downstream

d(up)=5(v1-3)
d(dn)=2.5(v2+3)
d(up)+d(dn)=7.5

Solve for v1 & v2:

v1=3.75
v2=-1.5

relative v(up)=v1-3=.75
relative v(dn)=-1.5+3=1.5

Here's how I would have done the problem: Let v be the speed of the boat relative to the water and let d be the length of the trip. Upstream, the speed of the boat relative to the bank is v- 3. Since the trip took 5 hours, d= 5(v- 3). Downstream, the speed of the boat is v+ 3. Since the trip took 2.5 hours, d= 2.5(v+3). So we have d= 5(v- 3)= 2.5(v+ 3). Solve that for v.

That is, of course, assuming that the speed, relative to the water, is the same both up and downstream. MaxJasper does not assume that. I would be interested in knowing how he solved the single equation, , for both v1 and v2.