AMC 10 · 2021 · #3

Grade 6 arithmetic

Archetype Multiplicative Ratio with Integrality Constraints └ Sub-archetype Percent / Fraction Recovery

percentagelinear-equations-one-varratio-proportionsystems-of-equations convert-to-algebraguess-and-check ↑ Prerequisites: percentagelinear-equations-one-var

📏 Short solution 💡 2 insights

Problem

In an after-school program for juniors and seniors, there is a debate team with an equal number of students from each class on the team. Among the $28$ students in the program, $25\%$ of the juniors as a class and $10\%$ of the seniors as a class are on the debate team. How many juniors are in the program?

Pick an answer.

(A)

(B)

(C)

(D)

~11

(E)

~20

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.

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Toolkit + CCSS Solution

Understand

Restated: An after-school program has $28$ juniors and seniors total. The debate team has the same number of juniors as seniors. That same number is $25\%$ of all juniors and $10\%$ of all seniors. Find how many juniors are in the program.

Givens: Total juniors $+$ seniors $= 28$; $25\% = \frac{1}{4}$ of all juniors are on the team; $10\% = \frac{1}{10}$ of all seniors are on the team; Equal number of juniors and seniors on the team; Answer choices: (A) $5$, (B) $6$, (C) $8$, (D) $11$, (E) $20$

Unknowns: The total number of juniors in the program

Understand

Plan

Primary tool: #3 Eliminate Possibilities

Secondary: #6 Guess and Check, #7 Identify Subproblems

Tool #3 (Eliminate) is fast on multiple choice: juniors must be a multiple of $4$ (so $25\%$ is whole) AND total juniors $\le 28$. Only (C) $8$ and (E) $20$ are multiples of $4$. Tool #6 (Guess and Check) tests each: if juniors $= 8$, then $25\% = 2$ on the team, so seniors on team also $= 2$, and seniors $= 20$ (since $10\%$ of $20 = 2$). Total $= 8 + 20 = 28$ ✓. If juniors $= 20$, then $25\% = 5$, seniors on team $= 5$, seniors $= 50$, total $= 70 \neq 28$ ✗. Tool #7 (Subproblems) provides the structural backbone — "same team size" links the two percentages.

Execute — Answer: C

#3 Eliminate Possibilities 4.OA.B.4 Step 1

Filter the answer choices by divisibility.
$25\%$ of all juniors must be a whole number, so the junior count is a multiple of $4$.
From (A)–(E), only $8$ and $20$ qualify.

$$\text{juniors} \in \{8, 20\}$$

💡 $25\%$ of a whole-people count must be a whole number — Grade 4 factor/multiple thinking.

#6 Guess and Check 6.RP.A.3 Step 2

Test juniors $= 8$.
Then debate team has $\frac{1}{4} \times 8 = 2$ juniors, so also $2$ seniors.
Those $2$ seniors are $10\%$ of all seniors, so seniors $= 2 \div \frac{1}{10} = 20$.

$$\text{juniors on team} = \tfrac{1}{4} \times 8 = 2 \;\Rightarrow\; \text{seniors} = 2 \div \tfrac{1}{10} = 20$$

💡 If part is $10\%$ of whole, whole is part $\times 10$ — Grade 6 percent reasoning.

#6 Guess and Check 2.OA.A.1 Step 3

Check the total.
Juniors $+$ seniors $= 8 + 20 = 28$.
This matches the given total exactly — so juniors $= 8$ works.

$$8 + 20 = 28 \checkmark$$

💡 Add two small whole numbers — Grade 2 word-problem standard.

#3 Eliminate Possibilities 6.RP.A.3 Step 4

Rule out the other candidate.
If juniors $= 20$, team has $\frac{1}{4} \times 20 = 5$ juniors, so $5$ seniors, so seniors $= 5 \times 10 = 50$.
Total would be $20 + 50 = 70$, not $28$ — contradiction.

$$20 + 50 = 70 \neq 28 \;\Rightarrow\; \text{juniors} \neq 20$$

💡 Reject the candidate that violates the total — same percent logic, different scale.

#3 Eliminate Possibilities 6.RP.A.3 Step 5

Only juniors $= 8$ survives.
That is choice (C).

$$\text{juniors} = 8 \;\Rightarrow\; \textbf{(C)}$$

💡 Multiple choice with one survivor — pick it.

[1] #3 4.OA.B.4 Filter the answer choices by divisibility. $25\%$ of all juniors must be a whole

[2] #6 6.RP.A.3 Test juniors $= 8$. Then debate team has $\frac{1}{4} \times 8 = 2$ juniors, so

[3] #6 2.OA.A.1 Check the total. Juniors $+$ seniors $= 8 + 20 = 28$. This matches the given tot

[4] #3 6.RP.A.3 Rule out the other candidate. If juniors $= 20$, team has $\frac{1}{4} \times 20

[5] #3 6.RP.A.3 Only juniors $= 8$ survives. That is choice (C).

Review

Reasonableness: Verify the whole story: $8$ juniors, $20$ seniors, total $28$ ✓. Team size: $25\% \times 8 = 2$ juniors and $10\% \times 20 = 2$ seniors — equal as required ✓. All counts are whole people. Every condition checks.

Alternative: Tool #13 (Convert to Algebra) — let $x =$ team size per class. Then juniors $= 4x$, seniors $= 10x$. Sum: $4x + 10x = 14x = 28$, so $x = 2$ and juniors $= 4x = 8$. Same answer (C) by a single equation.

CCSS standards used (min grade 6)

2.OA.A.1 Solve one- and two-step word problems using addition and subtraction within 100 (Checking that $8 + 20 = 28$ matches the given total.)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Recognizing that the junior count must be a multiple of $4$ for $25\%$ to be whole.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Going back and forth between the team count, $25\%$ of juniors, and $10\%$ of seniors.)

⭐ This AMC 10 problem only needs Grade 6 percent reasoning — once you notice juniors must be a multiple of $4$, just test $8$: a team of $2$ juniors equals $2$ seniors, so seniors $= 20$ and the total is $28$ exactly!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

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