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17 December, 18:05

An aircraft carrier made a trip. The trip there took 5 hours and the trip back took 6 hours. It averaged 3 mph faster on the trip there then on the return trip. Find the aircraft carriers speed on the outbound trip.

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  1. 17 December, 18:12
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    Answer: The outbound trip is 18 miles per hour

    Step-by-step explanation:

    Let x represent the speed of the plane.

    An aircraft carrier made a trip. Let us assume that this trip was outbound. The trip there took 5 hours.

    Distance = speed * time. Therefore

    Distance = 5x

    The trip back took 6 hours. It averaged 3 mph faster on the trip there then on the return trip. This means that the speed on the trip back is x - 3 mph.

    Distance = 6 (x-3) = 6x - 18

    Since the distance is the same,

    5x = 6x - 18

    6x - 5x = 18

    x = 18

    The speed on return or inbound trip would be 18 - 3 = 15 mph
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