I need help w/ a difficult word problem?

Question

A sailboat travels 20 mi. downstream in 3 hrs. It returns in 4 hrs. Find the speed of the sailboat in still water AND the rate of the current.

This is how I set it up:

rate x time= distance
Downstream: r+c 3 20
Upstream: r-c 4 20



What do I do now?

Answer 1

speed of sailboat = r
rate of current = c

For downstream, sailboat travel 20 mi in 3 hours.
r + c = 20 / 3 ---(1)
For upstream, sailboat travel 20 mi in 4 hours.
r - c = 20 / 4 ---(2)

(1) + (2)
2r = 20/3 + 20/4
2r = (4x20 + 3x20)/12
2r = 140/12
r = 35/6 = 5.83

from (1)
c = 20/3 - 35/6 = 5/6 = 0.83

speed of sailboat in still water = 5.83 mi/hr
rate of current = 0.83 mi/hr

Answer 2

You started off correctly. Assuming that you are using r for the sailboat speed in still water (why not use s) and c for speed of current you got (r + c)*3 = 20 and (r - c)*4 = 20. You now need to solve these simultaneous equations. In this particular example the easiest thing to do is divide the first equation by 3 (leaving the right hand side as 20/3). Also divide the second equation by 4. When you add the resulting equations one letter disappears and gives you an easy equation to solve for the other. Then you can put that result back into either of the original equations to find the value of the other letter. If you do it correctly both speeds turn out to be integer/6 with one 7 times the other.

Answer 3

I'll set it up, and you punch the numbers, OK?

The sailboat is **drifting** downstream. The speed of the current is 20 miles / 3 hours.

Going upstream, the sailboat is being pushed by the wind **against** the current. 20 miles / 4 hours tells you the difference between the wind speed and the current.

Since the sailboat is actually making progress upstream, that means the wind is **faster** than the current. The wind speed is (20 miles / 3 hours) PLUS (20 miles / 4 hours).

The wind speed is also the sailboat's speed in still water. Good luck with the problem!

Answer 4

You should solve it like this:

Consider Vb=boat speed in still water
Vc=current rate

you have these equations:
in going: Vb+Vc=20/3 (mi/hr)
in returning Vb-Vc=20/4 (mi/hr)

so you have two equations with two parameters,
then Vb=5.83 (mi/hr)
and Vc=0.83 (mi/hr)