# 2002 AMC 12A Problems/Problem 11

The following problem is from both the 2002 AMC 12A #11 and 2002 AMC 10A #12, so both problems redirect to this page.

## Problem

Mr. Earl E. Bird gets up every day at 8:00 AM to go to work. If he drives at an average speed of 40 miles per hour, he will be late by 3 minutes. If he drives at an average speed of 60 miles per hour, he will be early by 3 minutes. How many miles per hour does Mr. Bird need to drive to get to work exactly on time? $\text{(A)}\ 45 \qquad \text{(B)}\ 48 \qquad \text{(C)}\ 50 \qquad \text{(D)}\ 55 \qquad \text{(E)} 58$

## Solution

### Solution 1

Let the time he needs to get there in be t and the distance he travels be d. From the given equations, we know that $d=\left(t+\frac{1}{20}\right)40$ and $d=\left(t-\frac{1}{20}\right)60$. Setting the two equal, we have $40t+2=60t-3$ and we find $t=\frac{1}{4}$ of an hour. Substituting t back in, we find $d=12$. From $d=rt$, we find that r, and our answer, is $\boxed{\text{(B)}\ 48 }$.

### Solution 2

Since either time he arrives at is 3 minutes from the desired time, the answer is merely the harmonic mean of 40 and 60. The harmonic mean of a and b is $\frac{2}{\frac{1}{a}+\frac{1}{b}}=\frac{2ab}{a+b}$. In this case, a and b are 40 and 60, so our answer is $\frac{4800}{100}=48$, so $\boxed{\text{(B)}\ 48}$.

### Solution 3

A more general form of the argument in Solution 2, with proof:

Let $d$ be the distance to work, and let $s$ be the correct average speed. Then the time needed to get to work is $t=\frac ds$.

We know that $t+\frac 3{60} = \frac d{40}$ and $t-\frac 3{60} = \frac d{60}$. Summing these two equations, we get: $2t = \frac d{40} + \frac d{60}$.

Substituting $t=\frac ds$ and dividing both sides by $d$, we get $\frac 2s = \frac 1{40} + \frac 1{60}$, hence $s=\boxed{48}$.

(Note that this approach would work even if the time by which he is late was different from the time by which he is early in the other case - we would simply take a weighted sum in step two, and hence obtain a weighted harmonic mean in step three.)

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