AMC 10 · 2021 · #6
Grade 6 rate-ratioProblem
Ms. Blackwell gives an exam to two classes. The mean of the scores of the students in the morning class is 84, and the afternoon class's mean score is 70. The ratio of the number of students in the morning class to the number of students in the afternoon class is 43. What is the mean of the scores of all the students?
Pick an answer.
AMC 10 2021 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: Ms. Blackwell gave one exam to two classes. The morning class averaged $84$ and the afternoon class averaged $70$. The morning class is smaller than the afternoon class — their student counts are in the ratio $3 : 4$. Find the overall (combined) mean of all students' scores.
Givens: Morning class mean $= 84$; Afternoon class mean $= 70$; $\dfrac{\#\text{morning}}{\#\text{afternoon}} = \dfrac{3}{4}$; Answer choices: (A) $74$, (B) $75$, (C) $76$, (D) $77$, (E) $78$
Unknowns: The overall mean across all students
Understand
Restated: Ms. Blackwell gave one exam to two classes. The morning class averaged $84$ and the afternoon class averaged $70$. The morning class is smaller than the afternoon class — their student counts are in the ratio $3 : 4$. Find the overall (combined) mean of all students' scores.
Givens: Morning class mean $= 84$; Afternoon class mean $= 70$; $\dfrac{\#\text{morning}}{\#\text{afternoon}} = \dfrac{3}{4}$; Answer choices: (A) $74$, (B) $75$, (C) $76$, (D) $77$, (E) $78$
Plan
Primary tool: #9 Easier Related Problem
Secondary: #8 Analyze the Units, #3 Eliminate Possibilities
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 $\text{total points} \div \text{total students}$. 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.
Execute — Answer: C
6.RP.A.1 Step 1 - Use the easier problem: let the morning class have $3$ students and the afternoon class have $4$ students (smallest whole numbers in the ratio $3 : 4$).
- The combined mean is the same for any other pair in this ratio.
💡 Grade 6 ratios: $3 : 4$ is the same whether the counts are $3, 4$ or $30, 40$ or $300, 400$.
6.SP.A.3 Step 2 - Total points from each class $=$ (class mean) $\times$ (class size).
- Morning total $= 84 \cdot 3 = 252$.
- Afternoon total $= 70 \cdot 4 = 280$.
💡 Grade 6 measure of center: mean $\times$ count $=$ sum of all scores.
3.NBT.A.2 Step 3 - Add the two totals to get the grand total of all students' scores: $252 + 280 = 532$ points.
- Add the two class sizes: $3 + 4 = 7$ students.
💡 Grade 3 addition within $1000$: just stack the partial sums.
4.NBT.B.6 Step 4 Combined mean $=$ grand total $\div$ total students $= 532 \div 7$.
💡 Grade 4 division: $7 \cdot 76 = 532$, so the quotient is exactly $76$.
4.NBT.A.2 Step 5 - Sanity-check by Tool #3 (Eliminate): the combined mean must be between $70$ and $84$, and because the larger ($4$-part) group has mean $70$, the answer must lean toward $70$ — below the midpoint $77$.
- Only (A) $74$, (B) $75$, (C) $76$ survive.
- Our value $76$ matches (C).
💡 Grade 4 comparing numbers: the bigger group pulls the mean toward its own mean.
6.RP.A.1 Use the easier problem: let the morning class have $3$ students and the afternoo 6.SP.A.3 Total points from each class $=$ (class mean) $\times$ (class size). Morning tot 3.NBT.A.2 Add the two totals to get the grand total of all students' scores: $252 + 280 = 4.NBT.B.6 Combined mean $=$ grand total $\div$ total students $= 532 \div 7$. 4.NBT.A.2 Sanity-check by Tool #3 (Eliminate): the combined mean must be between $70$ and Review
Reasonableness: Sanity: $76$ lies between $70$ and $84$ (good) and is closer to $70$ than to $84$ ($76 - 70 = 6$ vs $84 - 76 = 8$) — consistent with the afternoon class being the bigger group. Also, scaling the class sizes by $10$ ($30$ morning, $40$ afternoon) gives total $84 \cdot 30 + 70 \cdot 40 = 2520 + 2800 = 5320$, divided by $70$ students $= 76$ — same answer, confirming the ratio invariance.
Alternative: Tool #5 (Look for a Pattern) via the weighted-mean shortcut: split the $3 : 4$ split as fractions of the whole, $\dfrac{3}{7}$ and $\dfrac{4}{7}$. Combined mean $= \dfrac{3}{7} \cdot 84 + \dfrac{4}{7} \cdot 70 = \dfrac{252 + 280}{7} = \dfrac{532}{7} = 76$. Same answer (C), no need to choose concrete counts.
CCSS standards used (min grade 6)
6.RP.A.1Understand the concept of a ratio and use ratio language (Replacing the unknown class sizes by $3$ and $4$ — the smallest pair in ratio $3 : 4$.)6.SP.A.3Recognize that a measure of center summarizes all its values with a single number (Using mean $\times$ count $=$ total to recover each class's score total from its mean.)3.NBT.A.2Fluently add and subtract within 1000 (Adding $252 + 280 = 532$ and $3 + 4 = 7$.)4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Dividing $532 \div 7 = 76$ to get the combined mean.)4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Eliminating choices above the midpoint $77$ and matching $76$ to choice (C).)
⭐ This 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 $3 \cdot 84 + 4 \cdot 70 = 532$ points across $7$ students, and divide: $532 \div 7 = 76$, choice (C).
⭐ This 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 $3 \cdot 84 + 4 \cdot 70 = 532$ points across $7$ students, and divide: $532 \div 7 = 76$, choice (C).
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