A boater travels 21 miles upstream against a 2 mi/h current, then returns downstream to the starting point. If the entire trip took 10 hours, what is the rate of the boat in still water?
A. 4 mi/h
B. 7 mi/h
C. 5 mi/h
D. 6 mi/h
E. 3 mi/h
Boater
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Let speed of the boat in still water = x
Upstream
Since, the boat is going upstream, more speed is required. The speed would be x+2
D/S = T
T= 21/(x+2)
Downstream
Since, the boat is going downstream, less speed is required. The speed would be x-2
D/S = T
T= 21/(x-2)
Total time = 21/(x+2) + 21/(x-2) = 10
Solve the equation, you'll get
5x^2 -21x -20 = 0
(5x+4)(x-5) = 0
discard 5x+4, coz speed can never be in negative
x-5 = 0
x = 5 mph
Hence C.
Hope this helps.
Upstream
Since, the boat is going upstream, more speed is required. The speed would be x+2
D/S = T
T= 21/(x+2)
Downstream
Since, the boat is going downstream, less speed is required. The speed would be x-2
D/S = T
T= 21/(x-2)
Total time = 21/(x+2) + 21/(x-2) = 10
Solve the equation, you'll get
5x^2 -21x -20 = 0
(5x+4)(x-5) = 0
discard 5x+4, coz speed can never be in negative
x-5 = 0
x = 5 mph
Hence C.
Hope this helps.
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An easy solution would be to plug in the answer choices. Pick C
U = 3
D = 7
downstream travel time - 21 = T * 7 -> T = 3
upstream travel time - 21 = 3 (10 - T)
21 = 3 * (10-3) -> 21 = 21
Hence answer C.
U = 3
D = 7
downstream travel time - 21 = T * 7 -> T = 3
upstream travel time - 21 = 3 (10 - T)
21 = 3 * (10-3) -> 21 = 21
Hence answer C.