AMC 10 · 2021 · #4
Grade 6 rate-ratioPick an answer.
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.
Turn the ratio into class sizes
Write the ratio as actual counts.
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's total
Mean times size gives the total.
Grade 6 measure of center: mean × count = sum of all scores.
A mean multiplied by its count gives back the sum of all the scores.
▸ Why?
An average is a total shared out over a count, so multiplying it back returns the total.
▸ Why?
The combined group is exactly the two classes put together, so their totals and counts add.
Combine everything
Add the totals and the sizes.
Grade 3 addition within 1000: just stack the partial sums.
3.NBT.A.2Analyze The UnitsDivide
Unequal sizes make it a weighted mean.
Grade 4 division: 7 · 76 = 532, so the quotient is exactly 76.
4.NBT.B.6Analyze The UnitsMatch the choice
The mean is 76.
Grade 4 comparing numbers: the bigger group pulls the mean toward its own mean.
4.NBT.A.2Eliminate PossibilitiesThis AMC 12 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).