2014 AMC 8 Problems/Problem 17

Problem

George walks $1$ mile to school. He leaves home at the same time each day, walks at a steady speed of $3$ miles per hour, and arrives just as school begins. Today he was distracted by the pleasant weather and walked the first $\frac{1}{2}$ mile at a speed of only $2$ miles per hour. At how many miles per hour must George run the last $\frac{1}{2}$ mile in order to arrive just as school begins today?

$\textbf{(A) }4\qquad\textbf{(B) }6\qquad\textbf{(C) }8\qquad\textbf{(D) }10\qquad \textbf{(E) }12$

Solution

Note that on a normal day, it takes him $1/3$ hour to get to school. However, today it took $\frac{1/2 \text{ mile}}{2 \text{ mph}}=1/4$ hour to walk the first $1/2$ mile. That means that he has $1/3 -1/4 = 1/12$ hours left to get to school, and $1/2$ mile left to go. Therefore, his speed must be $\frac{1/2 \text{ mile}}{1/12 \text { hour}}=\boxed{6 \text{ mph}}$ , so $\text{(B)}$ is the answer.

Solution 1

Using the harmonic mean formula, and making the speed he needs to take to get to school for the last half $x$ , we can make the expression:

$\frac{2 * 2x}{2x}$

Which simplifies to:

$\frac{4x}{2x}$

Then, because we know that since he goes at $3$ mph on a normal day, we can say that it is the harmonic mean of the two rates he goes at today, so we add that to our expression and turn it into something familiar:

$\frac{4x}{2x} = 3$

Solving that equation gives us:

$\boxed{\text {(B)} 6}$

2014 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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2014 AMC 8 Problems/Problem 17

Contents

Problem

Solution

Solution 1

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