Answer:
speed in still water is 15 miles per hour; the speed of the current is 9 miles per hour
Explanation:
This is a distance = rate times time problem.
We have to set up a table with the info we have and then take it from there.
d = r x t
downstream (w/ current)
upstream (against current)
We know the distance is 72 miles each way, so filling that in (I am also abbreviating downstream to d.s. and upstream to u.s.)
d = r x t
d.s 72
u.s 72
We also know that the trip with the current took 3 hours, and the trip back took 12. Filling that in:
d = r x t
d.s 72 = x 3
u.s 72 = x 12
Now we just need the rate values. But we don't have anything solid to put in there. We only know that WITH the current, the rate of the boat is faster; we also know that AGAINST the current, the rate of the boat is slower. So the rate with the current is whatever the rate in still water is + the rate of the current or r + c. The rate against the current is whatever the rate is in still water - the rate of the current or r - c. Those values will fit into the rate column:
d = r x t
d.s 72 = (r + c) x 3
u.s. 72 = (r - c) x 12
Since distance = rate * time, we set that equation up for each part of the trip. For the first part:
72 = 3(r + c) and
72 = 3r + 3c
For the second part:
72 = 12(r - c) and
72 = 12r - 12c
Since the rate of the boat in still water is going to be the same whether you are being pushed along or being pushed against by the current, I solved the first equation for r:
72 = 3r + 3c and
3r = 72 - 3c so
r = 24 - c
and subbed that in for r in the second equation to solve for the rate of the current:
72 = 12(24 - c) - 12c and
72 = 288 - 12c - 12c
Combining like terms and we have
-216 = -24c so
c = 9
Now we can go back up the rate in terms of current, r = 24 - c, and plug in 9 for c to solve for r:
r = 24 - 9 so
r = 15