SOLUTION: A pilot flew a jet from city A to city B, a distance of 2200 mi. On the return trip, the average speed was 20% faster than the outbound speed. The round-trip took 9 h 10 min. What

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Question 1020280: A pilot flew a jet from city A to city B, a distance of 2200 mi. On the return trip, the average speed was 20% faster than the outbound speed. The round-trip took 9 h 10 min. What was the speed from city A to city B?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +s+ = plane's speed for A to B
+1.2s+ = plane's speed from B to A
Let +t+ = plane's time in hrs going from A to B
+9.1667+-+t+ = plane's time in hrs going from B to A
( I converted 10 min to .1667 hrs )
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Trip from A to B:
(1) +2200+=+s%2At+
Trip from B to A:
(2) +2200+=+1.2s%2A%28+9.1667+-+t+%29+
-----------------------------
(1) +s+=+2200%2Ft+
Plug (1) into (2)
(2) +2200+=+1.2%2A%28+2200%2Ft+%29%2A%28+9.1667+-+t+%29+
(2) +2200%2F1.2+=+2200%2A%28+1%2Ft+%29%2A%28+9.1667+-+t+%29+
(2) +t%2F1.2+=+9.1667+-+t+
(2) +t%2A%28+1+%2B+1%2F1.2+%29+=+9.1667+
(2) +1.8333t+=+9.1667+
(2) +t+=+5+
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(1) +s+=+2200%2Ft+
(1) +s+=+2200%2F5+
(1) +s+=+440+ mi/hr
From A to B, the speed was 440 mi/hr
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check:
(2) +2200+=+1.2s%2A%28+9.1667+-+t+%29+
(2) +2200+=+1.2%2A440%2A%28+9.1667+-+t+%29+
(2) +2200+=+528%2A9.1667+-+528t+
(2) +2200+=+4840+-+528t+
(2) +528t+=+2640+
(2) +t+=+5+
OK