AMC 8 · 2023 · #10

Grade 5 arithmetic
Archetype Arithmetic Expression Decomposition └ Sub-archetype Linear Rate Accumulation
fraction-arithmeticcomplementary-counting complementary-countingidentify-subproblems ↑ Prerequisites: fraction-arithmeticfraction-multiplication
📏 Short solution 💡 2 insights
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Problem
Harold baked one whole plum pie and ate 14\frac{1}{4} of it. A moose then ate 13\frac{1}{3} of what Harold left. A porcupine then ate 13\frac{1}{3} of what the moose left. What fraction of the ORIGINAL pie is still there after the porcupine leaves?

Pick an answer.

(A)
$\frac{1}{12}$
(B)
$\frac{1}{6}$
(C)
$\frac{1}{4}$
(D)
$\frac{1}{3}$
(E)
$\frac{5}{12}$

AMC 8 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Change Focus / Count the Complement

The problem keeps asking "how much was eaten?" but what we actually want is "how much is LEFT." Tool #16 says: flip the question — if an eater takes 13\frac{1}{3}, they LEAVE 1-13\frac{1}{3}=23\frac{2}{3}, so we can just multiply leftover fractions instead of tracking eaten amounts and subtracting. Tool #7 then splits the chain of events into three small subproblems (Harold's leftover, moose's leftover, porcupine's leftover), and we multiply the three "leftover" fractions to get the final answer.

1STEP 1

Flip Harold's bite into a leftover: eating 14\frac{1}{4} means he LEAVES 34\frac{3}{4} of the pie — Tool #16's complement move.

1-14\frac{1}{4}=34\frac{3}{4}
2STEP 2

Same flip for the moose: he leaves 23\frac{2}{3}, so 23\frac{2}{3}×34\frac{3}{4} = 12\frac{1}{2} of the original pie remains (Tool #7 subproblem 2).

23\frac{2}{3}×34\frac{3}{4}=612\frac{6}{12}=12\frac{1}{2}
3STEP 3

Once more for the porcupine: he leaves 23\frac{2}{3}, so 23\frac{2}{3}×12\frac{1}{2} = 13\frac{1}{3} of the original pie is left on the plate.

23\frac{2}{3}×12\frac{1}{2}=26\frac{2}{6}=13\frac{1}{3}
4STEP 4

Bundle all three: multiply each eater's leftover fraction in one shot — Tool #7's combine step.

(1-14\frac{1}{4})(1-13\frac{1}{3})(1-13\frac{1}{3})=34\frac{3}{4}·23\frac{2}{3}·23\frac{2}{3}=1236\frac{12}{36}=13\frac{1}{3} → (D)
Answer
13\frac{1}{3}
Does 13\frac{1}{3} make sense? Harold alone leaves 34\frac{3}{4}. Each animal then keeps 23\frac{2}{3} of what arrives, so the pie shrinks by a factor of 23\frac{2}{3} twice: 34\frac{3}{4}→12\frac{1}{2}→13\frac{1}{3}. The values are decreasing as expected and end above 14\frac{1}{4} (since Harold left 34\frac{3}{4} and the animals together couldn't eat all of that). Answer (D) 13\frac{1}{3} is consistent.
💡Key takeaway

This AMC 8 problem only needs Grade 5 fraction-times-fraction multiplication you already know!

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