##### i dont understand this

 Mathematics Tutor: None Selected Time limit: 1 Day

a container ship made a trip to a navigational buoy and back. the trip there took four hours and the trip back took five hours. it averaged 5 km/h faster on the trip there than on the return trip. what was the container ships average speed on the out

Aug 29th, 2015

Thank you for the opportunity to help you with your question!

let the speed of the return trip be x
So, the trip to there would have taken = x + 5 km/h

so, distance traveled while going = speed x time = (x + 5) * 4 = 4x + 20 km
and distance traveled while returning = speed x time = x*5 = 5x

Since the distance traveled while going and returning remains the same.
So,
4x + 20 = 5x
20 = 5x - 4x
x = 20

so, the average speed of trip there = x +5 = 20+ 5 = 25 km/h
and average speed of return trip = x = 20 km

Let me know incase you need any further help ! Thanks :)
Aug 29th, 2015

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Aug 29th, 2015
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Aug 29th, 2015
Dec 10th, 2016
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