AMC 8 · 2017 · #14

Grade 6 rate-ratioalgebra

Archetype Arithmetic Expression Decomposition

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

📏 Medium solution 💡 3 insights

📘 View easy version →

Problem

Chloe and Zoe are both students in Ms. Demeanor's math class. Last night, they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only $80\%$ of the problems she solved alone, but overall $88\%$ of her answers were correct. Zoe had correct answers to $90\%$ of the problems she solved alone. What was Zoe's overall percentage of correct answers?

$\textbf{(A) }89\qquad\textbf{(B) }92\qquad\textbf{(C) }93\qquad\textbf{(D) }96\qquad\textbf{(E) }98$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: 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?

Givens: Each girl: half the problems alone, half together (equal-size halves).; Chloe alone accuracy = $80\%$.; Chloe overall accuracy = $88\%$.; Zoe alone accuracy = $90\%$.; Answer choices: (A) $89$, (B) $92$, (C) $93$, (D) $96$, (E) $98$ (percent).

Unknowns: Zoe's overall percentage of correct answers across all the homework problems.

Understand

Plan

Primary tool: #7 Identify Subproblems

Secondary: #13 Convert to Algebra, #3 Eliminate Possibilities

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 $\tfrac{1}{2}(80) + \tfrac{1}{2}(T) = 88$. 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.

Execute — Answer: C

#7 Identify Subproblems 6.SP.A.3 Step 1

Use the equal-halves structure.
Because each girl's homework is split into two equal-size halves, her overall percent correct equals the average of her percent correct on each half: $\text{overall} = \tfrac{1}{2}(\text{alone}\%) + \tfrac{1}{2}(\text{together}\%)$.
This is the key fact we will use for both Chloe and Zoe.

$$\text{overall}\% = \tfrac{1}{2}(\text{alone}\%) + \tfrac{1}{2}(\text{together}\%)$$

💡 When two groups are the same size, the average of their percents IS the overall percent — no weighting math is needed.

#13 Convert to Algebra 6.EE.B.7 Step 2

Set up Chloe's equation to find the together-accuracy $T$.
Let $T$ be the percent the girls got right on the 'together' half (this number is the SAME for both girls because they did those problems jointly).
Plug Chloe's numbers into the equal-halves formula.

$$\tfrac{1}{2}(80) + \tfrac{1}{2}(T) = 88$$

💡 Translating 'Chloe's overall is 88%' into a single equation in $T$ is a one-variable Grade 6 equation.

#13 Convert to Algebra 6.EE.B.7 Step 3

Solve Chloe's equation for $T$.
Simplify the left side, then isolate $T$.

$$40 + \tfrac{T}{2} = 88 \;\Rightarrow\; \tfrac{T}{2} = 48 \;\Rightarrow\; T = 96$$

💡 A one-step subtract-and-double gives $T$ — exactly the Grade 6 'solve $px + q = r$' move.

#7 Identify Subproblems 6.SP.A.3 Step 4

Apply the same equal-halves formula to Zoe.
Zoe's alone accuracy is $90\%$, and her together accuracy is the same $T = 96\%$ we just found (same problems, same outcome).
Average the two.

$$\text{Zoe overall}\% = \tfrac{1}{2}(90) + \tfrac{1}{2}(96) = \tfrac{90 + 96}{2} = \tfrac{186}{2} = 93$$

💡 Same equal-halves average as before, now with Zoe's numbers.

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

Match to the answer choices.
$93\%$ is choice (C).
Quick sanity check via Tool #3 (Eliminate): plug $T = 96$ back into Chloe — $\tfrac{1}{2}(80) + \tfrac{1}{2}(96) = 40 + 48 = 88\%$, which matches the given Chloe overall.

$$93\% \;\Rightarrow\; \textbf{(C)}$$

💡 Identifying the matching choice and back-checking the recovered together-rate against Chloe's data is the standard multiple-choice verification step.

[1] #7 6.SP.A.3 Use the equal-halves structure. Because each girl's homework is split into two e

[2] #13 6.EE.B.7 Set up Chloe's equation to find the together-accuracy $T$. Let $T$ be the percen

[3] #13 6.EE.B.7 Solve Chloe's equation for $T$. Simplify the left side, then isolate $T$.

[4] #7 6.SP.A.3 Apply the same equal-halves formula to Zoe. Zoe's alone accuracy is $90\%$, and

[5] #3 6.RP.A.3 Match to the answer choices. $93\%$ is choice (C). Quick sanity check via Tool #

Review

Reasonableness: 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.

Alternative: Tool #9 (Solve an Easier Related Problem) — pick a friendly $N = 100$ problems. Chloe alone: $50$ problems, $80\%$ correct $= 40$ right. Chloe overall: $88\%$ of $100 = 88$ right. So Chloe got $88 - 40 = 48$ right on the together half of $50$, meaning together accuracy is $\tfrac{48}{50} = 96\%$. Zoe alone: $90\%$ of $50 = 45$ right. Zoe together: $96\%$ of $50 = 48$ right. Zoe total: $45 + 48 = 93$ out of $100 = 93\%$. Same answer (C), with no algebra at all — just plain counting.

CCSS standards used (min grade 6)

6.SP.A.3 Recognize that a measure of center summarizes all its values with a single number (Using the fact that when two groups are equal in size, the overall percent equals the simple average of the two half-percents — applied once to set up Chloe's equation and once to compute Zoe's overall.)
6.EE.B.7 Solve real-world problems by writing and solving equations of the form px = q (Writing Chloe's condition as $\tfrac{1}{2}(80) + \tfrac{1}{2}(T) = 88$ and solving the one-variable linear equation for $T = 96$.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Interpreting percentages as part-to-whole ratios so $80\%$ of half the problems, $T\%$ of half the problems, and $88\%$ of all the problems are comparable quantities — and back-checking the final $93\%$ against the choices.)

⭐ 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 6)

More like this