Bunuel wrote:
A boat travels from point A to point B upstream and returns from point B to point A downstream. If the round trip takes the boat 5 hours and the distance between point A and point B is 120 kms and the speed of the stream is 10 km/hr, how long did the upstream journey take?
A. 2.5 hrs
B. 3 hrs
C. 3.5 hrs
D. 4 hrs
E. 4.5 hrs
Solution:We let r = the speed of the boat in still water. Thus, the speed of the boat when it is going downstream is (r + 10), and its upstream speed is (r - 10). Since distance / rate = time, we can create the equation for the sum of the combined time of the downstream and upstream legs of the journey as:
120 / (r + 10) + 120 / (r - 10) = 5
Multiplying the equation by (r + 10)(r - 10), we have:
120(r - 10) + 120(r + 10) = 5(r + 10)(r - 10)
120r - 120 + 120r + 120 = 5(r^2 - 100)
240r = 5r^2 - 500
r^2 - 48r - 100 = 0
(r - 50)(r + 2) = 0
r = 50 or r = -2
Since r can’t be negative, r = 50 km/hr. Since the speed of the boat going upstream is r - 10 = 50 - 10 = 40 km/hr, the time it took to go upstream is 120/40 = 3 hours.
Alternate Solution:Let t be the time, in hours, the boat spends traveling upstream. Notice that if r is the speed of the boat in still water, the speed of the boat traveling upstream is r - 10 and the speed of the boat traveling downstream is r + 10; hence, the speed of the boat traveling downstream is r + 10 - (r - 10) = 20 km/h faster than the speed of the boat traveling upstream. In terms of t, the speed of the boat traveling downstream is 120/(5 - t), and the speed of the boat traveling upstream is 120/t; hence we can create the following equation:
120 / (5 - t) - 120 / t = 20
Multiplying the equation by t(5 - t), we have:
120t - 120(5 - t) = 20t(5 - t)
120t - 600 + 120t = 100t - 20t^2
20t^2 + 140t - 600 = 0
Dividing each side by 20, we obtain:
t^2 + 7t - 30 = 0
(t - 3)(t + 10) = 0
t = 3 or t = -10
Since t cannot be negative, the boat spends t = 3 hours traveling upstream.
Answer: B _________________
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