AMC 10 · 2022 · #1

Grade 5 arithmetic
fraction-arithmeticfraction-multiplication identify-subproblemswork-backwards ↑ Prerequisites: fraction-arithmetic
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Problem
Evaluate a three-level continued fraction and match the value to one of the five answer choices. Every level has three as its whole part.

Pick an answer.

(A)
$\frac{31}{10}$
(B)
$\frac{49}{15}$
(C)
$\frac{33}{10}$
(D)
$\frac{109}{33}$
(E)
$\frac{15}{4}$
How to solve
Strategy Identify Subproblems

The expression has three nested layers, so we cannot evaluate it left-to-right. Tool #7 (Identify Subproblems) splits the giant fraction into three identical small jobs: "compute 3 + 1/(something already known)". Tool #11 (Work Backwards) tells us where to start — at the deepest layer, then climb outward, replacing each □ with the value we just found. Algebra (#13) is overkill since there are no unknowns; the layered structure tells us the path.

1STEP 1

Evaluate the innermost level

Start from the innermost level.

3 + 1/3 = 9/3 + 1/3 = 10/3
2STEP 2

Take the reciprocal

A reciprocal is just a flip.

1/ 10/3 = 1 ÷ 10/3 = 1 × 3/10 = 3/10
3STEP 3

Move out one level

Add three to move out a level.

3 + 3/10 = 30/10 + 3/10 = 33/10
4STEP 4

Finish the last level

Finishing gives one hundred nine over thirty-three.

3 + 10/33 = 99/33 + 10/33 = 109/33 → (D)
Answer
109/33
Sanity check the size. Each layer is roughly 3 + 1/(a bit more than 3), so the whole thing should be a little more than 3 but well under 4. 109/33 ≈ 3.303 fits that exactly. Choice (C) 33/10 = 3.3 is the middle layer's value (a common trap if you stop one level too early); (D) is the genuine outer value, matching the answer.
💡Key takeaway

This AMC 12 problem only needs Grade 5 "add fractions with unlike denominators and flip a fraction" — work from the deepest 1/3 outward, one layer at a time.