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Difficulty: Very hardRate, Time, and Distance Problems

A motorboat travels downstream along a straight river from Port Alpha to Port Beta, a distance of 120120 miles, completing the downstream leg in exactly 33 hours. At the moment the motorboat departs from Port Alpha, an unpowered raft is released from Port Alpha and drifts downstream driven solely by the river's constant current. Upon reaching Port Beta, the motorboat immediately turns around and travels upstream toward Port Alpha. If the motorboat meets the drifting raft at a point 3030 miles downstream from Port Alpha, and the motorboat's speed relative to the water remains constant throughout the journey, what is the motorboat's speed in still water, in miles per hour?

  1. A
    20
  2. B
    25
  3. C
    32
  4. 35Answer
  5. E
    45

Answer

35 miles per hour
The correct answer is 35 miles per hour. Since the motorboat covers the 120-mile downstream distance in 3 hours, its downstream rate is 40 mph, which means the boat's speed in still water vv plus the current speed cc equals 40. The raft drifts 30 miles at speed cc, requiring 30c\frac{30}{c} hours. In that same total time, the boat spends 3 hours going downstream and then travels 90 miles upstream at speed v−c=40−2cv - c = 40 - 2c. Equating the two time expressions 30c=3+9040−2c\frac{30}{c} = 3 + \frac{90}{40-2c} produces the quadratic equation c2−45c+200=0c^2 - 45c + 200 = 0. The valid root is c=5c = 5 mph, yielding v=40−5=35v = 40 - 5 = 35 mph.

Step-by-Step Solution

1
Determine the downstream speed of the motorboat and express the still-water speed vv in terms of the current cc.
The downstream speed is 120 miles3 hours=40\frac{120 \text{ miles}}{3 \text{ hours}} = 40 mph. Therefore, v+c=40v + c = 40, which implies v=40−cv = 40 - c mph and upstream speed v−c=40−2cv - c = 40 - 2c mph.
Downstream speed combines the motorboat's still-water speed and the river's current speed.
2
Formulate expressions for the time elapsed for both the raft and the motorboat until they meet.
The raft drifts 3030 miles at speed cc, so raft time T=30cT = \frac{30}{c} hours. The motorboat travels 120120 miles downstream in 33 hours, and then 120−30=90120 - 30 = 90 miles upstream at speed (40−2c)(40 - 2c) mph. So motorboat time T=3+9040−2cT = 3 + \frac{90}{40 - 2c} hours.
Both vessels start at the exact same moment from Port Alpha, so their total travel times until they meet are equal.
3
Equate total times and solve the resulting algebraic equation for cc.
Setting 30c=3+9040−2c\frac{30}{c} = 3 + \frac{90}{40 - 2c} and dividing by 33 yields 10c=1+1520−c=35−c20−c\frac{10}{c} = 1 + \frac{15}{20 - c} = \frac{35 - c}{20 - c}. Cross-multiplying gives 10(20−c)=c(35−c)  ⟹  c2−45c+200=010(20 - c) = c(35 - c) \implies c^2 - 45c + 200 = 0. Factoring gives (c−5)(c−40)=0(c - 5)(c - 40) = 0. Since c<40c < 40, c=5c = 5 mph.
Solving the quadratic equation isolates the valid physical value for the current speed.
4
Calculate the motorboat's speed in still water, vv.
v=40−5=35v = 40 - 5 = 35 mph.
Subtracting the current speed from the downstream speed yields the speed in still water.

Key Concept

Relative speed in current and multi-leg journey time equilibrium
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Topics

Rate, Time, and Distance Problems

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