2002 AMC 10B Problem 21

The problem only says how the lawns and mowers compare, never their actual sizes. So choose easy numbers that fit the comparisons.

1. Lawn sizes. Andy's lawn is $2$ times Beth's and $3$ times Carlos'. A number that splits nicely into halves and thirds is $6$. So:
- Andy: $6$ units, Beth: $3$ units, Carlos: $2$ units.
2. Mower speeds. Carlos' mower is the slowest: half of Beth's and a third of Andy's. Give Carlos a speed of $1$ unit per minute. Then:
- Beth: $2$ units per minute, Andy: $3$ units per minute, Carlos: $1$ unit per minute.
3. Time = lawn size ÷ speed.

1. Beth takes the least time. Andy and Carlos tie for second.

The answer is $\boxed{\textbf{(B)}\ \text{Beth}}$.

When a problem gives only comparisons like "twice as much" or "half as fast," the answer cannot depend on the real sizes, so any numbers that fit the comparisons will give the right answer. Pick numbers that avoid fractions. Then one simple rule, time equals amount of work divided by speed, settles it.

Use letters instead of numbers. Say Andy's lawn is $A$ and his speed is $s$. Then Beth's lawn is $\frac{A}{2}$ with speed $\frac{2s}{3}$, and Carlos' lawn is $\frac{A}{3}$ with speed $\frac{s}{3}$. Their times are $\frac{A}{s}$ for Andy, $\frac{A}{2} \div \frac{2s}{3} = \frac{3A}{4s}$ for Beth, and $\frac{A}{3} \div \frac{s}{3} = \frac{A}{s}$ for Carlos. Beth's time is three quarters of the others', so she finishes first.