AMC 8 · 2017 · #14

Grade 6 rate-ratioalgebra

Archetype Arithmetic Expression Decomposition └ Sub-archetype Weighted Group Aggregation

percentagemean-median-mode-rangeratio-proportion identify-subproblemsconvert-to-algebra ↑ Prerequisites: fraction-arithmeticpercentage

📏 Medium solution 💡 3 insights

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Problem

Chloe and Zoe got the same homework. Each girl solved half of the problems alone and then the other half together with the partner. Chloe got 80% of her alone problems right and 88% of her total right. Zoe got 90% of her alone problems right. What percent of all her problems did Zoe get right overall?

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The question asks about Zoe, but the data is mostly about Chloe — that gap is the whole problem. Tool #7 (Identify Subproblems) splits it into two clean pieces: (1) use Chloe's two numbers to recover the together-accuracy T%, then (2) average Zoe's alone-accuracy with T% to get Zoe's overall. Tool #13 (Convert to Algebra) handles subproblem 1 with a one-step equation $\frac{1}{2}$ (80) + $\frac{1}{2}$ (T) = 88. Because the two halves are equal in size, the overall percent is just the average of the two half-percents — no fancy weighting needed. Tool #3 (Eliminate Possibilities) is held in reserve to plug the final answer back into Chloe's check.

1STEP 1

Equal-size halves means each girl's overall percent is just the average of her alone-half and together-half percents.

overall% =

\frac{1}{2}

(alone%) +

\frac{1}{2}

(together%)

2STEP 2

Let T be the shared together-half rate (same for both girls), then plug Chloe's numbers into the equal-halves average.

\frac{1}{2}

(80) +

\frac{1}{2}

(T) = 88

3STEP 3

Simplify the left side and isolate T: subtract 40, then double, giving T = 96.

40 +

\frac{T}{2}

= 88 →

\frac{T}{2}

= 48 → T = 96

4STEP 4

Average Zoe's 90% alone rate with the same T = 96 together rate: (90 + 96)/2 = 93.

Zoe overall% =

\frac{1}{2}

(90) +

\frac{1}{2}

(96) =

\frac{90 + 96}{2}

\frac{186}{2}

= 93

5STEP 5

93% is choice (C); back-check by plugging T = 96 into Chloe: $\frac{1}{2}$ (80) + $\frac{1}{2}$ (96) = 88%, matching the given.

93% → (C)

Answer

Zoe got 90% alone and worked with Chloe on the rest. Two heads are better than one, so the together half should beat Chloe's solo 80% — and indeed T = 96% does. Zoe's overall must sit between her alone score 90% and the together score 96%, weighted equally, so it must land at the midpoint 93%. That is strictly above Zoe's alone score (working together helped her) and strictly below Chloe's overall would be too low (88% vs 93%), which makes sense because Zoe is the stronger solo solver. All three of these directional checks pass.

💡Key takeaway

This AMC 8 problem only needs Grade 6 percent and average ideas you already know — when two groups are the same size, the overall percent is just the average of the two!

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