AMC 10 · 2002 · #21
Grade 6 rate-ratioAndy's lawn has twice as much area as Beth's lawn and three times as much area as Carlos' lawn. Carlos' lawn mower cuts half as fast as Beth's mower and one third as fast as Andy's mower. If they all start to mow their lawns at the same time, who will finish first?
Pick an answer.
AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: Andy's lawn area is twice Beth's lawn area; Andy's lawn area is three times Carlos' lawn area; Carlos' mowing speed is half of Beth's mowing speed; Carlos' mowing speed is one third of Andy's mowing speed; All three begin mowing at the same moment; Answer choices: (A) Andy, (B) Beth, (C) Carlos, (D) Andy and Carlos tie for first, (E) All three tie
Unknowns: Which person finishes mowing their lawn first
Understand
Restated: 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.
Givens: Andy's lawn area is twice Beth's lawn area; Andy's lawn area is three times Carlos' lawn area; Carlos' mowing speed is half of Beth's mowing speed; Carlos' mowing speed is one third of Andy's mowing speed; All three begin mowing at the same moment; Answer choices: (A) Andy, (B) Beth, (C) Carlos, (D) Andy and Carlos tie for first, (E) All three tie
Plan
Primary tool: #8 Analyze the Units
Secondary: #4 Introduce a Variable, #13 Convert to Algebra
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.
Execute — Answer: B
6.RP.A.3 Step 1 Turn the area ratios into numbers
- Andy's lawn is twice Beth's and three times Carlos', so Andy's area is a common multiple of $2$ and $3$.
- Pick Andy's area to be $6$ so the others come out whole.
- Then Beth's area is half of Andy's, $6 \div 2 = 3$, and Carlos' area is a third of Andy's, $6 \div 3 = 2$.
💡 Choosing the biggest lawn as a multiple of both ratio numbers keeps every area a whole number.
6.RP.A.3 Step 2 Turn the speed ratios into numbers
- Carlos is the slowest mower, so use him as the base: let Carlos' speed be $1$.
- Carlos cuts half as fast as Beth, so Beth cuts twice as fast, giving Beth's speed $2$.
- Carlos cuts one third as fast as Andy, so Andy cuts three times as fast, giving Andy's speed $3$.
💡 Anchoring the slowest mower at $1$ makes the faster mowers simple whole-number multiples.
6.RP.A.2 Step 3 Compute each finishing time
- Mowing speed is area cut per unit of time, so the time to finish a lawn is its area divided by the mower's speed: $\text{time} = \text{area} \div \text{speed}$.
- Andy finishes in $6 \div 3 = 2$; Beth finishes in $3 \div 2 = 1.5$; Carlos finishes in $2 \div 1 = 2$.
💡 Dividing how much work by how fast you work tells you how long the job takes.
6.RP.A.3 Step 4 Compare the times
- The smallest time finishes first.
- Andy and Carlos each take $2$ units of time, while Beth takes only $1.5$.
- Beth's time is the least, so she finishes before both of the others — note that Andy and Carlos merely tie each other for second, which is the trap in choice (D).
- The one who finishes first is Beth, choice (B).
💡 Least time wins, and Beth's mid-size lawn with a fast-enough mower beats both bigger and slower matchups.
6.RP.A.3 Andy's lawn is twice Beth's and three times Carlos', so Andy's area is a common 6.RP.A.3 Carlos is the slowest mower, so use him as the base: let Carlos' speed be $1$. C 6.RP.A.2 Mowing speed is area cut per unit of time, so the time to finish a lawn is its a 6.RP.A.3 The smallest time finishes first. Andy and Carlos each take $2$ units of time, w Review
Reasonableness: 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 $t_A = 12 \div 6 = 2$, $t_B = 6 \div 4 = 1.5$, $t_C = 4 \div 2 = 2$. The same trio of times appears and Beth is again first, confirming choice (B) is stable under any legal scaling.
Alternative: Skip numbers and reason with ratios directly. Relative to Carlos, Andy mows a lawn $3$ times as big at $3$ times the speed, so their finishing times are equal — Andy and Carlos always tie. Beth mows a lawn $\tfrac{3}{2}$ times Carlos' area (since Carlos is $2$ and Beth is $3$ when Andy is $6$) at $2$ times Carlos' speed, so her time is $\tfrac{3/2}{2} = \tfrac{3}{4}$ of Carlos' time. Three-quarters of Carlos' time is less than Carlos' (and Andy's) time, so Beth finishes first.
CCSS standards used (min grade 6)
6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Translating the 'twice as much', 'half as fast' comparisons into a consistent set of areas and speeds, and comparing the resulting times to name the winner.)6.RP.A.2Understand the concept of a unit rate and use rate language (Reading mowing speed as area per unit time so that finishing time equals area divided by speed.)
⭐ 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.
⭐ 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.
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