# 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:

$\frac{(2 * 2x)/(2x)}$ (Error compiling LaTeX. ! Missing } inserted.)

Which simplifies to:

$\frac{(4x)/(2x)}$ (Error compiling LaTeX. ! Missing } inserted.)

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 out expression and turn it into something familiar: $\frac{(4x)/(2x)} = 3$

Solving that equation gives us: $\boxed{\text { (B)} 6}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 