AMC 10 · 2002 · #21
Grade 6 rate-ratioPick an answer.
AMC 10 2002 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
Turn the area ratios into numbers
Only ratios matter, so set Andy's area to 6; then Beth's is 6 ÷ 2 = 3 and Carlos' is 6 ÷ 3 = 2.
Choosing the biggest lawn as a multiple of both ratio numbers keeps every area a whole number.
6.RP.A.3Introduce A VariableTurn the speed ratios into numbers
Carlos is slowest, so let his speed be 1; then Beth cuts twice as fast at 2 and Andy three times as fast at 3.
Anchoring the slowest mower at 1 makes the faster mowers simple whole-number multiples.
6.RP.A.3Introduce A VariableCompute each finishing time
Speed is area per unit time, so time = area ÷ speed: Andy 6 ÷ 3 = 2, Beth 3 ÷ 2 = 1.5, Carlos 2 ÷ 1 = 2.
Dividing how much work by how fast you work tells you how long the job takes.
Dividing how much work there is by how fast you work tells you how long the job takes.
▸ Why?
At a steady rate the work done is the rate multiplied by the time, so dividing runs that backwards.
▸ Why?
With the rate fixed, a bigger job takes proportionally longer, so the comparison is a pure ratio.
Compare the times
Least time wins: 1.5 beats 2, so Beth is first and Andy and Carlos merely tie for second — 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.3Convert To AlgebraWhen 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