SOLUTION: A cabin cruiser travels 20 miles in the same time that a power boat travels 40 miles. The cruiser travels 5 mph slower than the power boat. Find the speed of each boat.

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Question 182580: A cabin cruiser travels 20 miles in the same time that a power boat travels 40 miles. The cruiser travels 5 mph slower than the power boat. Find the speed of each boat.
Found 2 solutions by stanbon, josmiceli:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A cabin cruiser travels 20 miles in the same time that a power boat travels 40 miles. The cruiser travels 5 mph slower than the power boat. Find the speed of each boat.
---------------
Cruiser DATA:
distance = 20 miles ; rate = x-5 mph ; time = 20/(x-5) hrs
------------------------
Power boat DATA:
distance = 40 miles ; rate = x mph ; time = 40/x hrs
------------------------
Equation:
time = time
20/(x-5)= 40/x
20x = 40(x-5)
20x = 40x - 200
-20x = -200
x = 10 mph (speed of the power boat)
x-5 = 5 mph (speed of the cruiser)
======================================
Cheers,
Stan H.

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
For both boats, the time travelling is the same
For cabin cruiser:
(1) d%5B1%5D+=+r%5B1%5D%2At
For power boat:
(2) d%5B2%5D+=+r%5B2%5D%2At
given:
d%5B1%5D+=+20
d%5B2%5D+=+40
r%5B1%5D+=+r%5B2%5D+-+5
Rewriting (1) and (2):
(1) 20+=+%28r%5B2%5D+-+5%29%2At
(2) 40+=+r%5B2%5D%2At
This is 2 equations and 2 unknowns, so it's solvable
(1) 20+=+r%5B2%5D%2At+-+5t
(2) 40+=+r%5B2%5D%2At
Subtract (1) from (2)
(2) 40+=+r%5B2%5D%2At
(1) -20+=+-r%5B2%5D%2At+%2B+5t
20+=+5t
t+=+4 hrs
------------------
(1) 20+=+r%5B1%5D%2At
(1) 20+=+r%5B1%5D%2A4
r%5B1%5D+=+5 mi/hr
--------------------
(2) 40+=+r%5B2%5D%2At
(2) 40+=+r%5B2%5D%2A4
r%5B2%5D+=+10 mi/hr
--------------------
The cabin cruiser goes 5 mi/hr and
the power boat goes 10 mi/hr