2003 AMC 10A Problem 4

Average speed always means total distance divided by total time, so find those two totals.

1. Distance: $1$ km to school and $1$ km back, so $2$ km in all.
2. Time: $30$ minutes up plus $10$ minutes back, so $40$ minutes.
3. Change minutes to hours, because the answer is in km per hour: $40$ minutes is $\frac{40}{60} = \frac{2}{3}$ of an hour.
4. Average speed:
$$\frac{2 \text{ km}}{\frac{2}{3} \text{ hour}} = 2 \times \frac{3}{2} = 3 \text{ km per hour}.$$

Another way to see step 4: $2$ km in $40$ minutes is $1$ km every $20$ minutes, and there are three $20$-minute blocks in an hour, so she goes $3$ km per hour.

The answer is $\boxed{\textbf{(A)}\ 3}$.

Anna walks uphill at $2$ km per hour and downhill at $6$ km per hour, but she spends three times as long going uphill. The slow part counts for more of the trip, so the average sits closer to $2$ than to $6$. That is why averaging the two speeds (to get $4$) is wrong; always go back to total distance over total time.

There is a shortcut for going the same distance at two speeds $u$ and $v$: the average speed is $\frac{2uv}{u+v}$. Uphill $1$ km in half an hour is $u = 2$ km per hour; downhill $1$ km in $10$ minutes is $v = 6$ km per hour. Then $\frac{2 \cdot 2 \cdot 6}{2 + 6} = \frac{24}{8} = 3$.