# 2003 AMC 12A Problems/Problem 4

The following problem is from both the 2003 AMC 12A #4 and 2003 AMC 10A #4, so both problems redirect to this page.

## Problem

It takes Anna $30$ minutes to walk uphill $1$ km from her home to school, but it takes her only $10$ minutes to walk from school to her home along the same route. What is her average speed, in km/hr, for the round trip? $\mathrm{(A) \ } 3\qquad \mathrm{(B) \ } 3.125\qquad \mathrm{(C) \ } 3.5\qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ } 4.5$

## Solution 1

Since she walked $1$ km to school and $1$ km back home, her total distance is $1+1=2$ km.

Since she spent $30$ minutes walking to school and $10$ minutes walking back home, her total time is $30+10=40$ minutes = $\frac{40}{60}=\frac{2}{3}$ hours.

Therefore her average speed in km/hr is $\frac{2}{\frac{2}{3}}=\boxed{\mathrm{(A)}\ 3}$.

## Solution 2

The average speed of two speeds that travel the same distance is the harmonic mean of the speeds, or $\dfrac{2}{\dfrac{1}{x}+\dfrac{1}{y}}=\dfrac{2xy}{x+y}$ (for speeds $x$ and $y$). Mary's speed going to school is $2\,\text{km/hr}$, and her speed coming back is $6\,\text{km/hr}$. Plugging the numbers in, we get that the average speed is $\dfrac{2\times 6\times 2}{6+2}=\dfrac{24}{8}=\boxed{\mathrm{(A)}\ 3}$.

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