An express train travels a distance of 180 miles from Station A to Station B at a constant… — GMAT

An express train travels a distance of $180$ miles from Station A to Station B at a constant speed of $v$ miles per hour. On the return trip from Station B to Station A, the train travels along the same route at a constant speed that is $15$ miles per hour slower. If the return trip takes $1$ hour longer than the outbound trip, what is the value of $v$ ?

A
$45$
B
$50$
$60$ Answer
D
$75$
E
$90$

Answer

60

The correct answer is

60

. The outbound time is

\frac{180}{v}

hours and the return time is

\frac{180}{v - 15}

hours. Setting their difference equal to

1

gives

\frac{180}{v - 15} - \frac{180}{v} = 1

. Solving the resulting quadratic equation

v^2 - 15v - 2700 = 0

yields

v = 60

miles per hour for the positive root.

Step-by-Step Solution

Express the time taken for each leg of the journey in terms of

v

Outbound time

t_1 = \frac{180}{v}

hours, return time

t_2 = \frac{180}{v - 15}

hours.

Time is equal to distance divided by speed.

Set up the equation based on the given time difference of

1

hour.

\frac{180}{v - 15} - \frac{180}{v} = 1

The return trip takes

1

hour longer than the outbound trip.

Clear the denominators by multiplying the equation by

v(v - 15)

180v - 180(v - 15) = v(v - 15) \implies 2700 = v^2 - 15v

Simplifying rational expressions into standard quadratic form.

Rearrange into

v^2 - 15v - 2700 = 0

and factor to solve for

v

(v - 60)(v + 45) = 0 \implies v = 60 \text{ or } v = -45

Factoring the quadratic equation.

Select the valid positive speed value.

Since speed must be positive,

v = 60

miles per hour.

Negative values for speed are not physically valid in this context.

Key Concept

Algebraic Word Problems and Equation Modeling

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