SOLUTION: a private plane traveled from seattle to a rugged wilderness at an average of 132 mph. On the return trip, the average speed was 231 mph. If the total traveling time was 7 hours,

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Question 737518:
a private plane traveled from seattle to a rugged wilderness at an average of 132 mph. On the return trip, the average speed was 231 mph. If the total traveling time was 7 hours, how far is Seattle from the wideness?
Found 2 solutions by josmiceli, checkley79:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +d+ = the one-way distance the plane flies
Let +t+ = the time in hours for the 1st trip
+7+-+t+ = the time for the return trip
-------------
1st trip:
+d+=+132t+
Return trip:
+d+=+231%2A%28+7-t+%29+
By substitution:
+132t+=+231%2A%28+7-t+%29+
+132t+=+1617+-+231t+
+363t+=+1617+
+t+=+1617+%2F+363+
and
+d+=+132t+
+d+=+132%2A%28+1617+%2F+363+%29+
+d+=+588+
The distance from Seattle to wilderness is 588 mi
check:
+d+=+231%2A%28+7-t+%29+
+588+=+231%2A%28+7-t+%29+
+2.54545+=+7+-+4.45454+
+2.54545+=+2.54545+
OK

Answer by checkley79(3341) About Me  (Show Source):
You can put this solution on YOUR website!
D=RT
D=(132*T)
D=231(7-T)
132T=231(7-T)
132T=1,617-231T
132T+231T=1,617
363T=1,617
T=1,617/363
T=4.45 OR 4 HOURS 27 MINUTES FOR THE 132 MPH. TRIP.
7-4.45=2.55 OR 2 HOURS 33 MINUTES FOR THE 231 MPH. TRIP.
PROOF:
D=132*4.45=587.4 MILES.
D=231*2.55=589 MILES. (DIFFERENCE DUE TO ROUNDING ERROR.)