AMC 10 · 2002 · #17

Grade 6 geometry-2d
rateratio-proportion dimensional-analysis ↑ Prerequisites: rateratio-proportion
📏 Medium solution 💡 2 insights
Problem
Three friends mow their own lawns at the same start time. Andy's lawn is twice Beth's and three times Carlos'. Carlos' mower cuts half as fast as Beth's and one third as fast as Andy's. Decide who finishes mowing first.

Pick an answer.

(A)
$\ \text{Andy}$
(B)
$\ \text{Beth}$
(C)
$\ \text{Carlos}$
(D)
$\ \text{Andy\ and \ Carlos\ tie\ for\ first.}$
(E)
$\ \text{All\ three\ tie.}$
How to solve
Strategy Analyze the Units

The question hides a rate problem. Tool #8 (Analyze the Units) supplies the one relationship that settles everything: a mowing speed is area cut per unit time, so time to finish = lawn area ÷ mowing speed. Bigger lawn means more time; faster mower means less time. Because only ratios are given, Tool #4 (Introduce a Variable) lets us pin down convenient stand-in numbers that respect every ratio — pick Andy's area as a multiple of both 2 and 3, and pick Carlos' speed as the base 1. Tool #13 (Convert to Algebra) then turns each 'finishes when?' into a single division, and comparing the three quotients names the winner. Choosing numbers is legitimate here precisely because the answer depends only on ratios, not on the raw sizes.

1STEP 1

Turn the area ratios into numbers

Pick Andy's area as 6 so the ratios give whole numbers: areas 6, 3, 2.

Area: A = 6, B = 3, C = 2
2STEP 2

Turn the speed ratios into numbers

Taking the slowest mower as 1 gives speeds 3, 2, 1.

Speed: A = 3, B = 2, C = 1
3STEP 3

Compute each finishing time

Time is area divided by speed, giving 2, 1.5, 2.

t_A = 6/3 = 2, t_B = 3/2 = 1.5, t_C = 2/1 = 2
4STEP 4

Compare the times

The other two tie at 2, so the smallest time belongs to Beth, choice (B).

1.5 < 2 → Beth first (B)
Answer
Beth
The winner must not depend on the numbers chosen, so test a different set that keeps the ratios. Let Andy's area be 12: then Beth's is 6 and Carlos' is 4. Let Carlos' speed be 2: then Beth's is 4 and Andy's is 6. Times are t_A = 12 ÷ 6 = 2, t_B = 6 ÷ 4 = 1.5, t_C = 4 ÷ 2 = 2. The same trio of times appears and Beth is again first, confirming choice (B) is stable under any legal scaling.
💡Key takeaway

When something happens at a rate, time equals the amount of work divided by the speed — so a smaller job with a fast-enough tool can beat a bigger job every time.

  • Turn the area ratios into numbers
  • Turn the speed ratios into numbers
  • Compute each finishing time
  • Compare the times