Competition · AMC preparation · step 4 of 4
AMC 10 · 2021B · #6
Grade 6 rate-ratioPick an answer.
AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Tool #9 (Easier Related Problem): replace the unspecified class sizes with the smallest whole numbers in ratio 3:4 — pick 3 morning students and 4 afternoon students. The combined mean does not depend on the absolute counts, so this concrete case gives the same answer. Tool #8 (Units / total-over-count) computes the mean as total points ÷ total students. Tool #3 (Eliminate) checks: since the afternoon group is larger, the combined mean must sit below the midpoint (70 + 84)/2 = 77, eliminating (D) and (E) before any arithmetic.
Pick easy class sizes
Reduce the ratio: let the morning class be 3 students, the afternoon class 4 — smallest counts in 3 : 4; the combined mean is unchanged.
Grade 6 ratios: 3 : 4 is the same whether the counts are 3, 4 or 30, 40 or 300, 400.
6.RP.A.1Solve An Easier Related ProblemFind each class total
Each class's points = mean × size: morning 252 = 84 · 3, afternoon 280 = 70 · 4.
Grade 6 measure of center: mean × count = sum of all scores.
Multiplying each class's mean by its size rebuilds that class's total.
▸ Why?
An average is a total shared over a count, so multiplying by the count undoes the sharing.
▸ Why?
A ratio fixes only relative sizes, so any matching pair of class sizes gives the same final answer.
Add the two totals
Add across classes: grand total 532 = 252 + 280 points over 7 = 3 + 4 students.
Grade 3 addition within 1000: just stack the partial sums.
3.NBT.A.2Analyze The UnitsDivide for the combined mean
Combined mean = grand total ÷ students = 532 ÷ 7 = 76.
Grade 4 division: 7 · 76 = 532, so the quotient is exactly 76.
4.NBT.B.6Analyze The UnitsSanity-check the range
Eliminate: the mean sits between 70 and 84, pulled below the midpoint 77 by the bigger class — only 76 fits, choice (C).
Grade 4 comparing numbers: the bigger group pulls the mean toward its own mean.
4.NBT.A.2Eliminate PossibilitiesThis AMC 10 problem only needs Grade 6 ratio thinking you already know — pretend there are just 3 morning students and 4 afternoon students (the ratio is the same), add up 3 · 84 + 4 · 70 = 532 points across 7 students, and divide: 532 ÷ 7 = 76, choice (C).
- Pick easy class sizes
- Find each class total
- Add the two totals
- Divide for the combined mean
- Sanity-check the range
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