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Difficulty: Very hardBasic Algebraic Expressions and One-Step Equations

A commuter train travels at a constant speed to complete a trip of DD miles in tt hours. Due to track maintenance, the train's speed is decreased by 15 miles per hour for a 20-minute portion of the trip. Which of the following expressions represents the distance, in miles, the train traveled during this 20-minute portion of the trip?

  1. A
    Dt−5\frac{D}{t} - 5
  2. B
    D3t−15\frac{D}{3t} - 15
  3. D3t−5\frac{D}{3t} - 5Answer
  4. D
    D3t−45\frac{D}{3t} - 45
  5. E
    3Dt−5\frac{3D}{t} - 5

Answer

The expression D3t−5\frac{D}{3t} - 5 represents the distance traveled during the 20-minute portion of the trip.
The train's original speed is Dt\frac{D}{t} miles per hour. During the maintenance segment, its speed is reduced by 15 miles per hour, giving a speed of Dt−15\frac{D}{t} - 15 miles per hour. To find the distance traveled over a 20-minute interval (which represents 2060=13\frac{20}{60} = \frac{1}{3} of an hour), the reduced speed is multiplied by the elapsed time. Applying the distributive property to (Dt−15)×13\left(\frac{D}{t} - 15\right) \times \frac{1}{3} results in D3t−5\frac{D}{3t} - 5.

Step-by-Step Solution

1
Determine the train's original speed.
Original speed is Dt\frac{D}{t} miles per hour.
Speed is defined as distance divided by time.
2
Express the reduced speed during the track maintenance portion.
Reduced speed is Dt−15\frac{D}{t} - 15 miles per hour.
The speed was decreased by 15 miles per hour from the original speed.
3
Convert the duration of the maintenance portion from minutes to hours.
20 minutes is equal to 2060=13\frac{20}{60} = \frac{1}{3} hour.
Since the speed is in miles per hour, the time must be converted to hours to ensure unit consistency.
4
Calculate the distance traveled during the 20-minute portion.
Distance is (Dt−15)×13=D3t−5\left(\frac{D}{t} - 15\right) \times \frac{1}{3} = \frac{D}{3t} - 5 miles.
Distance is equal to the product of speed and time. The distributive property must be applied to both terms in the speed expression.

Key Concept

Translating a verbal rate and time relationship into a simplified algebraic expression using the distributive property and unit conversion.

Alternative Method

Alternatively, you can assign plug-in values to make the problem concrete. Let the total distance D=120D = 120 miles and the total time t=2t = 2 hours. This gives an original speed of 6060 miles per hour. The reduced speed is 60−15=4560 - 15 = 45 miles per hour. Traveling at 4545 miles per hour for 2020 minutes (13\frac{1}{3} hour) results in a distance of 45×13=1545 \times \frac{1}{3} = 15 miles. Substituting D=120D = 120 and t=2t = 2 into the correct expression gives 1203(2)−5=1206−5=20−5=15\frac{120}{3(2)} - 5 = \frac{120}{6} - 5 = 20 - 5 = 15 miles, which matches our result.
Estimated Time:1m 30s
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Topics

Basic Algebraic Expressions and One-Step Equations

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