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 , and the afternoon class's mean score is . The ratio of the number of students in the morning class to the number of students in the afternoon class is . What is the mean of the scores of all the students?
Pick an answer.
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).