SOLUTION: A motorboat left a harbor and traveled to an island at an average rate of 15 knots. The average speed on the return trip was 10 knots. If the total trip took 7.5 hours, how far is

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Question 161670: A motorboat left a harbor and traveled to an island at an average rate of 15 knots. The average speed on the return trip was 10 knots. If the total trip took 7.5 hours, how far is the harbor from the island?
Found 2 solutions by checkley77, josmiceli:
Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
D=RT
T=D/R
7.5=D/15+D/10
7.5=(10D+15D)/150
7.5=25D/150
25D=7.5*150
25D=1125
D=1125/25
D=45 MILES FROM THE HARBOR TO THE ISLAND.
PROOF:
7.5=45/15+45/10
7.5=3+4.5
7.5=7.5

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
For both trips,
d+=+r%2At is the general formula
For harbor to island trip:
d%5B1%5D+=+r%5B1%5D%2At%5B1%5D
For the island to harbor trip:
d%5B2%5D+=+r%5B2%5D%2At%5B2%5D
--------------------------------
r%5B1%5D+=+15knots
r%5B2%5D+=+10knots
t%5B1%5D+%2B+t%5B2%5D+=+7.5hrs
t%5B2%5D+=+7.5+-+t%5B1%5Dhrs
d%5B1%5D+=+d%5B2%5D (both trips are the same distance)
r%5B1%5D%2At%5B1%5D+=+r%5B2%5D%2At%5B2%5D
15%2At%5B1%5D+=+10%2A%287.5+-+t%5B1%5D%29
15%2At%5B1%5D+=+75+-+10%2At%5B1%5D
25%2At%5B1%5D+=+75
t%5B1%5D+=+3hrs
t%5B2%5D+=+7.5+-+3
t%5B2%5D+=+4.5hrs
-------------------------------
Going back to original equations,
d%5B1%5D+=+r%5B1%5D%2At%5B1%5D
d%5B2%5D+=+r%5B2%5D%2At%5B2%5D
I looked up a conversion table for knots to mi/hr
10 knots = 11.5 mi/hr
15 knots = 17.3 mi/hr
d%5B1%5D+=+17.3%2A3
d%5B1%5D+=+51.9mi
d%5B2%5D+=+11.5%2A4.5
d%5B2%5D+=+51.75
d%5B1%5D should equal d%5B2%5D, but they don't quite agree so,
the answer is somewhere in between 51.75 and 51.9 mi.