SOLUTION: A pilot flew a jet from City A to City B, a distance of 1500 mi. On the return trip, the average speed was 20% faster than the outbound speed. The round-trip took 7 h 20 min. What

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Question 1200263: A pilot flew a jet from City A to City B, a distance of 1500 mi. On the return trip, the average speed was 20% faster than the outbound speed. The round-trip took 7 h 20 min. What was the speed from City A to City B?
Found 3 solutions by josgarithmetic, ikleyn, greenestamps:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
                      SPEED       TIME            DIST.
GOING                   r         1500/r          1500
RETURN                 1.2r       1500/(1.2r)     1500
TOTAL                             7&1/3

Solve for r:
1500%2Fr%2B1500%2F%281.2r%29=7%261%2F3
.
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Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.
A pilot flew a jet from City A to City B, a distance of 1500 mi.
On the return trip, the average speed was 20% faster than the outbound speed.
The round-trip took 7 h 20 min. What was the speed from City A to City B?
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Let x be the speed from A to B, in miles per hour.

Then the speed from B to A is 1.2 x, according to the problem.


The flight time from A to B is 1500%2Fx  hours.

The flight time from B to A is 1500%2F%281.2x%29 = 1250%2Fx hours.


Total flight time for the round trip is  1500%2Fx + 1250%2Fx = 2750%2Fx.


We are given that

    2750%2Fx = 71%2F3 hours = 22%2F3 hours.


It gives us x = 2750%2F%28%2822%2F3%29%29 = %282750%2A3%29%2F22 = 375.


ANSWER.  The speed from A to B was 375 miles per hour.

Solved.



Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Here is an alternative to the standard algebraic solution method shown by the other tutors. For a particular similar problem, this alternative method might (or might not!) make the solution easier.

The distances both directions are of course the same; and the ratio of the two speeds is 1:1.2, or 5:6. That means the ratio of times at the two speeds was 6:5.

The total time for the trip was 7 hours 20 minutes, or 440 minutes. Dividing the 440 minutes in the ratio 6:5 gives 240 minutes for the flight at the lower speed from A to B and 200 minutes for the flight at the higher speed from B to A.

The question asks for the speed on the flight from A to B, which was 1500 miles in 240 minutes, or 4 hours; the speed was 1500/4 = 375 mph.

ANSWER: 375 mph