SOLUTION: A cabin cruiser leaves a harbour and travels south at 15km/h. One hour later a speed boat leaves the same harbour and travels on the same course at 25 km/h. How far from the harbou
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Question 1173675: A cabin cruiser leaves a harbour and travels south at 15km/h. One hour later a speed boat leaves the same harbour and travels on the same course at 25 km/h. How far from the harbour will the speed boat pass the cruiser? Found 3 solutions by Boreal, MathTherapy, greenestamps:Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! One way is the cabin cruiser has a 15 km head start and is being caught at 10 km/h, so it is 1.5 hours chase and 2.5 hours after leaving the harbor, or 37.5 km.
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another way is cruiser has gone 15km/h*x h km.
the speed boat has gone 25 km/h (x-1) h km.
Set those equal to each other'
15x=25x-25
10x=25
x=2.5 hours, time the cabin cruiser has gone 15*2.5 or 37.5 km and the speed boat 25*1.5 or 37.5 km.
You can put this solution on YOUR website!
A cabin cruiser leaves a harbour and travels south at 15km/h. One hour later a speed boat leaves the same harbour and travels on the same course at 25 km/h. How far from the harbour will the speed boat pass the cruiser?
Let distance from the harbour, be D
Then time the slower boat took to get to that point =
And, time the faster took to get to that point =
We then get the following TIME equation:
5D = 3D + 75 ------ Multiplying by LCD, 75
5D - 3D = 75
2D = 75
Distance from the harbour, or
