SOLUTION: sam flew a distance of 500 miles at a normal speed. on the trip back (also 500 miles) sam doubled the speed of the plane. how fast did sam fly on the trip back if the total travell

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Question 1147317: sam flew a distance of 500 miles at a normal speed. on the trip back (also 500 miles) sam doubled the speed of the plane. how fast did sam fly on the trip back if the total travelling time was 5 hours?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


The two distances are the same, and the speed returning was double the speed going.

That means the time returning was half the time going.

That means 1/3 of the total time was returning and 2/3 of the total time was going.

1/3 of the total 5 hours is 5/3 hours.

500 miles in 5/3 hours is (500/(5/3)) = 300 mph.

ANSWER: 300 mph on the return trip

CHECK:
going: 500 miles at 150 mph = 500/150 = 10/3 hours
returning: 500 miles at 300 mph = 500/300 = 5/3 hours
total time: 10/3 hours + 5/3 hours = 15/3 hours = 5 hours
Answer by ikleyn(52800)   (Show Source): You can put this solution on YOUR website!
.

            Algebra solution is  THIS : 


Let "x" be the speed flying 500 miles "to there", in miles per hour.


Then the speed flying back was  2x  miles per hour.


Then the "time" equation is


     +  = 5  hours


saying that the total time flying "to there" and back was 5 hours.


To solve the "time" equation, multiply both sides by 2x.  You will get


    1000 + 500 = 10x

    1500       = 10x

      x        =  = 150 miles per hour.


Thus the average speed flying "to there" was 150 mph;  hence, the average speed flying back was twice of it, i.e. 300 mph.    ANSWER


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