AMC 8 · 2001 · #13
Grade 6 arithmeticrate-ratioPick an answer.
AMC 8 2001 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
The question chains two clean steps: first decide how many students prefer cherry pie, then convert that count into a slice of the 360° circle. Tool #7 (Break Into Subproblems) keeps the two ideas separate. Subproblem (a) is pure arithmetic — subtract the three known counts from 36, then halve. Subproblem (b) is one ratio move — take the cherry fraction of 36 and multiply by 360°. Keeping them apart avoids the common slip of multiplying by 360° before the halving step.
Subtract the three known groups from 36 to see how many students are left for cherry and lemon: 10 remain.
Take the whole, peel off each known group, and what is left must be the cherry-plus-lemon group.
4.OA.A.3Identify SubproblemsSplit the 10 remaining students in half, so cherry gets 5 students.
"Half prefer cherry and half prefer lemon" is just dividing the leftover by 2.
4.OA.A.3Identify SubproblemsCherry is 5 of the 36 students, so its slice is that fraction of 360°: 50°.
Because 36 × 10 = 360, each student is worth exactly 10° on the pie graph, so 5 cherry students take 50°.
6.RP.A.3Identify SubproblemsWhen the class size divides 360° cleanly, each student is worth the same number of degrees on the pie graph. Here 360 ÷ 36 = 10° per student, so the 5 cherry fans take 50° — answer (D).