An algebra problem by Nishita Tripathy
Algebra
Level
2
Two boats are crossing the river from opposite sides. When they first meet, they are 720 feet from the near shore. When they reach the opposite shores, they stop for 10 minutes and cross the river again, but this time they meet 400 ft from the far shore. How wide is the river?
1761
1750
1759.9
1760
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1 solution
Let the width of the river be w. Let the speed of the first boat be f. Let the speed of the second boat be s.
Time of first meeting is 720/f = (w - 720)/s f reaches far shore at w/f waits 10 minutes and heads back s reaches near shore at w/s waits 10 minutes and heads back time of second meeting is w/f + 10 + 400/f = w/s + 10 + (w-400)/s
The two 10 minute delays cancel each other out.
720s = fw - 720f f(w-720) = 720 s f/s = 720/(w-720)
w/f + 10 + 400/f = w/s + 10 + (w-400)/s ws + 10fs + 400s = wf + 10fs + wf - 400f s (400 + w) = f (2w - 400) f/s = (400 + w)/(2w - 400)
720/(w-720) = (400 + w)/(2w - 400) 720 (2w - 400) = (400 + w) (w - 720) 1440 w - 288000 = 400w - 720w + w^2 - 288000 w^2 - 1760w = 0 w = 1760