AMC 10 · 2007 · #16
Grade 6 arithmeticA teacher gave a test to a class in which 10% of the students are juniors and 90% are seniors. The average score on the test was 84. The juniors all received the same score, and the average score of the seniors was 83. What score did each of the juniors receive on the test?
Pick an answer.
AMC 10 2007 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: In a class, $10\%$ of the students are juniors and $90\%$ are seniors. The whole class averaged $84$ on a test. Every junior got the exact same score, and the seniors averaged $83$. Find the score each junior received.
Givens: Juniors are $10\%$ of the class; seniors are $90\%$; The class average is $84$; The seniors' average is $83$; All juniors scored identically; Answer choices: (A) 85, (B) 88, (C) 93, (D) 94, (E) 98
Unknowns: The single score that each junior received
Understand
Restated: In a class, $10\%$ of the students are juniors and $90\%$ are seniors. The whole class averaged $84$ on a test. Every junior got the exact same score, and the seniors averaged $83$. Find the score each junior received.
Givens: Juniors are $10\%$ of the class; seniors are $90\%$; The class average is $84$; The seniors' average is $83$; All juniors scored identically; Answer choices: (A) 85, (B) 88, (C) 93, (D) 94, (E) 98
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #7 Identify Subproblems, #11 Work Backwards
Percentages with no actual head count feel abstract, so replace the class with a convenient concrete size (Tool #9): pick $10$ students, which turns $10\%$ and $90\%$ into a clean $1$ junior and $9$ seniors. An average is just a total divided by a count, so turn each average into a total number of points (Tool #7). The class total minus the seniors' total is exactly the one junior's score, recovered by working backwards (Tool #11).
Execute — Answer: C
6.RP.A.3 Step 1 Pick a convenient class size
- Nothing fixes how many students there are, so choose the size that makes the percentages whole.
- With $10$ students, $10\%$ is $1$ junior and $90\%$ is $9$ seniors.
- Any multiple of $10$ works and gives the same answer, so take the smallest one.
💡 Swapping vague percentages for a real head count of $10$ makes every quantity something you can just add up.
6.SP.B.5 Step 2 Turn averages into total points
- An average is the total of all scores divided by how many scores there are, so total $=$ average $\times$ count.
- The whole class of $10$ averaged $84$, giving $84 \times 10 = 840$ points in all.
- The $9$ seniors averaged $83$, giving $83 \times 9 = 747$ points among the seniors.
💡 Turning each average back into a raw total lets you add and subtract groups of scores like ordinary numbers.
4.NBT.B.4 Step 3 Peel off the seniors to find the junior
- The $840$ class points are split between the seniors and the one junior.
- The seniors account for $747$ of them, so the junior holds whatever is left: $840 - 747 = 93$.
- Since there is a single junior, that leftover is exactly the score each junior received.
- This is choice $\textbf{(C)}$.
💡 The class total is the seniors' points plus the junior's, so subtracting the seniors leaves the junior's score alone.
6.RP.A.3 Nothing fixes how many students there are, so choose the size that makes the per 6.SP.B.5 An average is the total of all scores divided by how many scores there are, so t 4.NBT.B.4 The $840$ class points are split between the seniors and the one junior. The sen Review
Reasonableness: Check the weighted average: $10\%$ of $93$ plus $90\%$ of $83$ is $0.1\times 93 + 0.9\times 83 = 9.3 + 74.7 = 84$, exactly the class average, so $93$ is consistent. It also makes sense that the juniors' score ($93$) sits above the class average ($84$) while the seniors sit just below ($83$): since juniors are only a tenth of the class, their high score barely nudges the class mean up by $1$ point, which is why the answer is well above $84$ rather than just a hair above it.
Alternative: Skip the head count and set up the weighted-average equation directly. Let $J$ be the junior score; the class average is $0.10J + 0.90(83) = 84$. Then $0.10J = 84 - 74.7 = 9.3$, so $J = 93$, matching choice (C).
CCSS standards used (min grade 6)
6.RP.A.3Use ratio and rate reasoning to solve real-world and mathematical problems (Converting $10\%$ and $90\%$ of a $10$-student class into $1$ junior and $9$ seniors.)6.SP.B.5Summarize numerical data sets by reporting number of observations and measures (Using average $\times$ count $=$ total to get $840$ class points and $747$ senior points.)4.NBT.B.4Fluently add and subtract multi-digit whole numbers (Subtracting $840-747=93$ to recover the junior's score.)
⭐ Turn percentages into a real head count, change each average into a total number of points, then subtract the group you know to uncover the one you want.
⭐ Turn percentages into a real head count, change each average into a total number of points, then subtract the group you know to uncover the one you want.
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