Competition · AMC preparation · step 4 of 4
AMC 8 · 2017 · #14
Grade 6 rate-ratioalgebraPick an answer.
AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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 1/2(80) + 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.
Use the half-and-half rule
Equal-size halves means each girl's overall percent is just the average of her alone-half and together-half percents.
When two groups are the same size, the average of their percents IS the overall percent — no weighting math is needed.
Because each girl's homework is a half solved alone and an equal-size half solved together, her overall percent correct is the plain average of her alone-percent and her together-percent.
▸ Why?
Her overall percent measures her total correct answers against all her problems, and that total is the correct answers on the alone half plus the correct answers on the together half.
▸ Why?
The alone half and the together half together cover every problem with none skipped and none counted twice, so their two correct counts add back to the whole count of correct answers.
▸ Why?
Each girl's score on a half is stated as a percent, and a percent of that half's problem count is just that many hundredths of the count — a definite number of correct answers.
▸ Why?
Both halves hold the same number of problems, so that shared size pulls out as a common factor of the two contributions and then cancels against the total, leaving each rate counted equally.
▸ Why?
The same half-size multiplies each half's rate, so it can be taken outside the sum as one common factor over the two rates.
▸ Why?
That pulled-out half-size, divided by the identical size sitting in the total, reduces to one, because dividing a quantity by itself undoes the multiplication.
Set up Chloe's equation
Let T be the shared together-half rate (same for both girls), then plug Chloe's numbers into the equal-halves average.
Translating 'Chloe's overall is 88%' into a single equation in T is a one-variable Grade 6 equation.
6.EE.B.7Convert To AlgebraSolve for the shared accuracy
Simplify the left side and isolate T: subtract 40, then double, giving T = 96.
A one-step subtract-and-double gives T — exactly the Grade 6 'solve px + q = r' move.
6.EE.B.7Convert To AlgebraApply the rule to Zoe
Average Zoe's 90% alone rate with the same T = 96 together rate: (90 + 96)/2 = 93.
Same equal-halves average as before, now with Zoe's numbers.
6.SP.A.3Identify SubproblemsMatch to the choices
93% is choice (C); back-check by plugging T = 96 into Chloe: (80) + (96) = 88%, matching the given.
Identifying the matching choice and back-checking the recovered together-rate against Chloe's data is the standard multiple-choice verification step.
6.RP.A.3Eliminate PossibilitiesThis 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!
- Use the half-and-half rule
- Set up Chloe's equation
- Solve for the shared accuracy
- Apply the rule to Zoe
- Match to the choices
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