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Difficulty: Very hardAlgebraic Word Problems and Equation Modeling

A delivery truck completes an outbound trip along Route X of length dd miles at a constant average speed of vv miles per hour. On the return trip along the exact same route, heavy traffic reduces the truck's average speed by 2020 miles per hour, which increases the travel time for the return trip by 22 hours compared to the outbound trip. Later, the truck completes a trip along Route Y, which has the same distance dd as Route X, at an average speed of (v+60)(v + 60) miles per hour in exactly 22 hours. What was the truck's average speed vv, in miles per hour, on the outbound trip along Route X?

  1. A
    4040
  2. B
    5050
  3. 6060Answer
  4. D
    8080
  5. E
    150150

Answer

The truck's average outbound speed vv on Route X was 6060 miles per hour.
By setting up time equations for the two legs of Route X, the distance dd is expressed as d=v(v−20)10d = \frac{v(v-20)}{10}. Combining this with the relation for Route Y (d=2(v+60)d = 2(v+60)) yields the quadratic equation v2−40v−1200=0v^2 - 40v - 1200 = 0. Factoring gives (v−60)(v+20)=0(v - 60)(v + 20) = 0, yielding the valid positive speed v=60v = 60 miles per hour.

Step-by-Step Solution

1
Express the distance dd in terms of vv using the time difference on Route X.
d=v(v−20)10d = \frac{v(v - 20)}{10}
The outbound time is t1=dvt_1 = \frac{d}{v} and the return time is t2=dv−20t_2 = \frac{d}{v - 20}. Given t2−t1=2t_2 - t_1 = 2, we have dv−20−dv=2  ⟹  20dv(v−20)=2  ⟹  d=v(v−20)10\frac{d}{v - 20} - \frac{d}{v} = 2 \implies \frac{20d}{v(v - 20)} = 2 \implies d = \frac{v(v - 20)}{10}.
2
Express the distance dd in terms of vv using the conditions for Route Y.
d=2(v+60)d = 2(v + 60)
Route Y has distance dd, speed (v+60)(v + 60) mph, and takes 22 hours, so d=rate×time=2(v+60)d = \text{rate} \times \text{time} = 2(v + 60).
3
Equate the two expressions for dd to form a quadratic equation in terms of vv.
v2−40v−1200=0v^2 - 40v - 1200 = 0
Setting v(v−20)10=2(v+60)\frac{v(v - 20)}{10} = 2(v + 60) yields v2−20v=20v+1200v^2 - 20v = 20v + 1200, which simplifies to v2−40v−1200=0v^2 - 40v - 1200 = 0.
4
Solve the quadratic equation for vv and verify physical feasibility.
v=60v = 60
Factoring (v−60)(v+20)=0(v - 60)(v + 20) = 0 gives solutions v=60v = 60 or v=−20v = -20. Since speed must be positive and greater than 2020 mph (to avoid a zero or negative return speed), v=60v = 60 mph.

Key Concept

Algebraic Modeling of Multi-Leg Motion using Systems of Rational and Quadratic Equations
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Topics

Algebraic Word Problems and Equation Modeling

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