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Difficulty: MediumAlgebraic Word Problems and Modeling

A delivery truck traveled from Warehouse X to Warehouse Y at a constant speed of 5050 miles per hour. On the return trip along the exact same route, heavy traffic reduced the truck's constant speed to 3030 miles per hour. If the total driving time for the entire round trip was 88 hours, what was the distance, in miles, between Warehouse X and Warehouse Y?

  1. A
    120120
  2. 150150Answer
  3. C
    160160
  4. D
    250250
  5. E
    300300

Answer

The distance between Warehouse X and Warehouse Y is 150150 miles.
The correct distance between the two warehouses is 150150 miles. Since time equals distance divided by rate, the time taken going to Warehouse Y is d50\frac{d}{50} hours and returning is d30\frac{d}{30} hours. Summing these expressions to equal the total 88 hours gives d50+d30=8\frac{d}{50} + \frac{d}{30} = 8. Solving for dd yields 8d150=8\frac{8d}{150} = 8, which simplifies to d=150d = 150 miles.

Step-by-Step Solution

1
Define variables and write time expressions for each leg of the trip
Let dd be the distance in miles between Warehouse X and Warehouse Y. Time outbound is t1=d50t_1 = \frac{d}{50} hours, and time inbound is t2=d30t_2 = \frac{d}{30} hours.
Distance divided by rate gives the time taken for each individual leg of the journey.
2
Set up the equation using total elapsed time
d50+d30=8\frac{d}{50} + \frac{d}{30} = 8
The total driving time for both legs combined is given as 88 hours.
3
Solve the algebraic equation for dd
\frac{3d + 5d}{150} = 8 \implies \frac{8d}{150} = 8 \implies 8d = 1200 \implies d = 150
Finding a common denominator of 150150 allows combining the fractions to solve for the unknown distance dd.

Key Concept

Distance-Rate-Time Relationship & Harmonic Mean Rate Modeling
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Topics

Algebraic Word Problems and Modeling

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