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Difficulty: Very hardAlgebraic Word Problems and Modeling

An express train travels along a straight track between Station A and Station B. For the first 40%40\% of the total distance, the train travels at a constant speed of vv miles per hour. For the next 50%50\% of the remaining distance, due to track maintenance, it travels at a constant speed that is 25%25\% slower than vv. For the final leg of the journey, the train increases its speed to a constant rate that is 20%20\% faster than vv. If the average speed for the entire journey from Station A to Station B is 6060 miles per hour, what is the value of vv, in miles per hour?

  1. A
    57.557.5
  2. B
    60.060.0
  3. 63.063.0Answer
  4. D
    68.068.0
  5. E
    69.069.0

Answer

The base speed vv is 63.063.0 miles per hour.
The correct answer is derived by setting up a model for distance, speed, and time across all three segments. The first leg covers 0.4D0.4D at speed vv (time =0.4D/v= 0.4D/v). The second leg covers 50%50\% of the remaining 0.6D0.6D, which is 0.3D0.3D, at speed 0.75v0.75v (time =0.3D/(0.75v)=0.4D/v= 0.3D/(0.75v) = 0.4D/v). The third leg covers the remaining 0.3D0.3D at speed 1.2v1.2v (time =0.3D/(1.2v)=0.25D/v= 0.3D/(1.2v) = 0.25D/v). Summing these gives total time T=1.05D/vT = 1.05D/v. The average speed is D/(1.05D/v)=v/1.05=20v/21D / (1.05D/v) = v / 1.05 = 20v/21. Setting 20v/21=6020v/21 = 60 yields v=63v = 63.

Step-by-Step Solution

1
Define distances for each leg of the trip in terms of total distance DD.
Leg 1 distance d1=0.40Dd_1 = 0.40D. Remaining distance is D−0.40D=0.60DD - 0.40D = 0.60D. Leg 2 distance d2=0.50×0.60D=0.30Dd_2 = 0.50 \times 0.60D = 0.30D. Leg 3 distance d3=0.60D−0.30D=0.30Dd_3 = 0.60D - 0.30D = 0.30D.
The problem specifies percentages of remaining distance, requiring step-by-step subtraction of completed distance.
2
Determine the speed for each leg in terms of vv.
Leg 1 speed v1=vv_1 = v. Leg 2 speed v2=v(1−0.25)=0.75vv_2 = v(1 - 0.25) = 0.75v. Leg 3 speed v3=v(1+0.20)=1.20vv_3 = v(1 + 0.20) = 1.20v.
Percentage increases and decreases are applied to the base rate vv.
3
Calculate time spent on each leg (t=dvt = \frac{d}{v}) and sum for total time TT.
t1=0.40Dvt_1 = \frac{0.40D}{v}, t2=0.30D0.75v=0.40Dvt_2 = \frac{0.30D}{0.75v} = \frac{0.40D}{v}, t3=0.30D1.20v=0.25Dvt_3 = \frac{0.30D}{1.20v} = \frac{0.25D}{v}. Total time T=0.40D+0.40D+0.25Dv=1.05Dv=21D20vT = \frac{0.40D + 0.40D + 0.25D}{v} = \frac{1.05D}{v} = \frac{21D}{20v}.
Average speed requires total distance divided by total time.
4
Set up the average speed equation and solve for vv.
Average speed =DT=D21D20v=20v21=60  ⟹  20v=1260  ⟹  v=63= \frac{D}{T} = \frac{D}{\frac{21D}{20v}} = \frac{20v}{21} = 60 \implies 20v = 1260 \implies v = 63.
Equating the algebraic average speed expression to the given value of 60 mph yields the value of vv.

Key Concept

Weighted Average Speed in Multi-Phase Motion Problems
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Topics

Algebraic Word Problems and Modeling

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