AMC 10 · 2022 · #1

Grade 5 arithmetic

Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation

fraction-arithmeticfraction-multiplication identify-subproblemswork-backwards ↑ Prerequisites: fraction-arithmetic

📏 Short solution 💡 1 insight

📘 View easy version →

Problem

What is the value of $3+\frac{1}{3+\frac{1}{3+\frac13}}?$

Pick an answer.

(A)

$\frac{31}{10}$

(B)

$\frac{49}{15}$

(C)

$\frac{33}{10}$

(D)

$\frac{109}{33}$

(E)

$\frac{15}{4}$

AMC 10 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

View mode:

Toolkit + CCSS Solution

Understand

Restated: Evaluate the three-level continued fraction $3+\dfrac{1}{3+\dfrac{1}{3+\frac{1}{3}}}$ and match the value to one of the five answer choices.

Givens: A continued fraction with three nested layers, each of the form $3 + \dfrac{1}{\square}$; The innermost layer is $3 + \dfrac{1}{3}$; Answer choices: (A) $\dfrac{31}{10}$, (B) $\dfrac{49}{15}$, (C) $\dfrac{33}{10}$, (D) $\dfrac{109}{33}$, (E) $\dfrac{15}{4}$

Unknowns: The exact value of the full continued fraction as a single fraction

Understand

Restated: Evaluate the three-level continued fraction $3+\dfrac{1}{3+\dfrac{1}{3+\frac{1}{3}}}$ and match the value to one of the five answer choices.

Plan

Primary tool: #7 Identify Subproblems

Secondary: #11 Work Backwards

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 $3 + \dfrac{1}{\text{something already known}}$". Tool #11 (Work Backwards) tells us where to start — at the deepest layer, then climb outward, replacing each $\square$ with the value we just found. Algebra (#13) is overkill since there are no unknowns; the layered structure tells us the path.

Execute — Answer: D

#11 Work Backwards 5.NF.A.1 Step 1

Compute the innermost layer $3 + \dfrac{1}{3}$.
Write $3$ as $\dfrac{9}{3}$ to get a common denominator and add.

$$3 + \dfrac{1}{3} = \dfrac{9}{3} + \dfrac{1}{3} = \dfrac{10}{3}$$

💡 Adding a whole number and a unit fraction is the standard Grade 5 "unlike denominators" job — rename $3$ as $\dfrac{9}{3}$ and add.

#7 Identify Subproblems 5.NF.B.3 Step 2

Replace the innermost $\square$ with $\dfrac{10}{3}$.
The middle layer becomes $3 + \dfrac{1}{10/3}$.
Dividing $1$ by $\dfrac{10}{3}$ is the same as multiplying by its reciprocal, so $\dfrac{1}{10/3} = \dfrac{3}{10}$.

$$\dfrac{1}{\,10/3\,} = 1 \div \dfrac{10}{3} = 1 \times \dfrac{3}{10} = \dfrac{3}{10}$$

💡 "$1$ divided by a fraction" is the same as flipping that fraction — Grade 5 interpretation of a fraction as a division.

#11 Work Backwards 5.NF.A.1 Step 3

Now finish the middle layer: $3 + \dfrac{3}{10}$.
Rewrite $3$ as $\dfrac{30}{10}$ and add.

$$3 + \dfrac{3}{10} = \dfrac{30}{10} + \dfrac{3}{10} = \dfrac{33}{10}$$

💡 Same Grade 5 "common denominator" trick — the middle layer collapses to a tidy fraction.

#7 Identify Subproblems 5.NF.A.1 Step 4

Climb to the outer layer.
Replace its $\square$ with $\dfrac{33}{10}$.
Flip to get $\dfrac{1}{33/10} = \dfrac{10}{33}$, then add to $3 = \dfrac{99}{33}$.

$$3 + \dfrac{10}{33} = \dfrac{99}{33} + \dfrac{10}{33} = \dfrac{109}{33} \;\Rightarrow\; \textbf{(D)}$$

💡 One last "flip and add over a common denominator" finishes the climb — pure Grade 5 fraction work.

[1] #11 5.NF.A.1 Compute the innermost layer $3 + \dfrac{1}{3}$. Write $3$ as $\dfrac{9}{3}$ to g

[2] #7 5.NF.B.3 Replace the innermost $\square$ with $\dfrac{10}{3}$. The middle layer becomes $

[3] #11 5.NF.A.1 Now finish the middle layer: $3 + \dfrac{3}{10}$. Rewrite $3$ as $\dfrac{30}{10}

[4] #7 5.NF.A.1 Climb to the outer layer. Replace its $\square$ with $\dfrac{33}{10}$. Flip to g

Review

Reasonableness: Sanity check the size. Each layer is roughly $3 + \dfrac{1}{\text{a bit more than }3}$, so the whole thing should be a little more than $3$ but well under $4$. $\dfrac{109}{33} \approx 3.303$ fits that exactly. Choice (C) $\dfrac{33}{10} = 3.3$ 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.

Alternative: Tool #3 (Eliminate Possibilities) plus a decimal estimate. Compute each choice's decimal: (A) $3.1$, (B) $\approx 3.27$, (C) $3.3$, (D) $\approx 3.303$, (E) $3.75$. A quick mental pass — innermost $\approx 3.33$, middle $\approx 3 + 1/3.33 \approx 3.3$, outer $\approx 3 + 1/3.3 \approx 3.303$ — lands squarely on (D).

CCSS standards used (min grade 5)

5.NF.A.1 Add and subtract fractions with unlike denominators (Adding $3 + \dfrac{1}{3}$, then $3 + \dfrac{3}{10}$, then $3 + \dfrac{10}{33}$ by rewriting the whole number with the right common denominator.)
5.NF.B.3 Interpret a fraction as division of the numerator by the denominator (Reading $\dfrac{1}{10/3}$ as $1 \div \dfrac{10}{3} = \dfrac{3}{10}$ at every layer of the climb.)

⭐ This AMC 10 problem only needs Grade 5 "add fractions with unlike denominators and flip a fraction" — work from the deepest $\dfrac{1}{3}$ outward, one layer at a time.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 5)

More like this