Question 15678: A consultant traveled 3 hours to attend a meeting. The return trip took only 2 hours because the speed was 8 miles per hour faster. The total trip was 291 miles. What was the consultant's speed each way?
Answer by rapaljer(4671)   (Show Source):
You can put this solution on YOUR website! Let x = rate going (for 3 hours)
x+8 = rate returning (faster) (for 2 hours)
D=RT
(RT going) + (RT returning) = Total Distance
3x + 2(x+8) = 291
3x + 2x + 16 = 291
5x = 275
x= 55 mph going
x+8 = 63 mph returning
Check: See if total distance = 291 miles
3(55) + 2(63) = 291
165 + 126 = 291
However, there is a problem here. Notice that the distance going is 165 miles and the return trip was only 126 miles. What happened?? Did the consultant find a shortcut to come home?? We would normally assume that the distance going and coming would be the same. However, that would give a completely different problem in which
D going = D returning
RT (going) = RT returning
3x = 2(x+8)
3x = 2x + 16
x=16 mph going
x+8 = 24 mph returning
However this does NOT give you a total trip of 291 miles. I think there is a problem (an oversight!) in this problem.
R^2 at SCC
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