SOLUTION: Could you please help me with this problem: The excursion boat "Holiday" travels 35 km upstream and then back again in 4hours and 48 minutes. If the speed on the "Holiday" in still

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: Could you please help me with this problem: The excursion boat "Holiday" travels 35 km upstream and then back again in 4hours and 48 minutes. If the speed on the "Holiday" in still      Log On

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Question 26014: Could you please help me with this problem: The excursion boat "Holiday" travels 35 km upstream and then back again in 4hours and 48 minutes. If the speed on the "Holiday" in still water is 15 km/h, what is the speed of the current?
Thank you.
Answer by Paul(988) About Me  (Show Source):
You can put this solution on YOUR website!
4 hours and 48 minutes = 4.8hours
Still water speed = 15
Speed with the current = 15+x
Speed against the current = 15-x
total distance == 35
Equation:
35%2F15%2Bx+-+35%2F15-x=4.8
35[(15+x)+(15-x)]=4.8[(15+x)(15-x)] SIMPLIFY
35%2830%29=4.8%28-x%5E2%2B225%29
1050=-4.8x%5E2%2B1080
4.8x%5E2=1080-1050
4.8x%5E2=30
4.8x%5E2%2F4.8=30%2F4.8
x%5E2=6.25
x=sqrt%286.25%29%29
x=2.5

Hence, the speed of the current is 2.5km/h.
Paul.