AMC 10 · 2020 · #1

Grade 5 arithmetic

Archetype Arithmetic Expression Decomposition

fraction-arithmeticlinear-equations-one-varwork-backwards work-backwards ↑ Prerequisites: fraction-arithmetic

📏 Short solution 💡 1 insight

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Problem

What value of $x$ satisfies
$x- \frac{3}{4} = \frac{5}{12} - \frac{1}{3}?$

Pick an answer.

(A)

${-}\frac{2}{3}$

(B)

$\frac{7}{36}$

(C)

$\frac{7}{12}$

(D)

$\frac{2}{3}$

(E)

$\frac{5}{6}$

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Toolkit + CCSS Solution

Understand

Restated: Find the number $x$ that makes $x - \frac{3}{4} = \frac{5}{12} - \frac{1}{3}$ true.

Givens: The left side is $x - \frac{3}{4}$; The right side is $\frac{5}{12} - \frac{1}{3}$; Answer choices: (A) $-\frac{2}{3}$, (B) $\frac{7}{36}$, (C) $\frac{7}{12}$, (D) $\frac{2}{3}$, (E) $\frac{5}{6}$

Unknowns: The single number $x$

Understand

Restated: Find the number $x$ that makes $x - \frac{3}{4} = \frac{5}{12} - \frac{1}{3}$ true.

Plan

Primary tool: #11 Work Backwards

Secondary: #3 Eliminate Possibilities

Tool #11 (Work Backwards) fits perfectly: $x$ went through one operation (subtract $\frac{3}{4}$) to reach the right-hand value, so we undo it by adding $\frac{3}{4}$. The arithmetic on the right is a small fraction sum on the common denominator $12$. Tool #3 (Eliminate) is a quick sanity check — the right side is clearly small and positive after adding back $\frac{3}{4}$, so negative answer (A) and tiny $\frac{7}{36}$ (B) can be eliminated by feel before the final compute.

Execute — Answer: E

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

First compute the right side.
The denominators $12$ and $3$ have a common denominator of $12$, so rewrite $\frac{1}{3} = \frac{4}{12}$ and subtract.

$$\dfrac{5}{12} - \dfrac{1}{3} = \dfrac{5}{12} - \dfrac{4}{12} = \dfrac{1}{12}$$

💡 Same-denominator subtraction is straightforward — rewrite $\frac{1}{3}$ over $12$ first.

#11 Work Backwards 1.OA.D.8 Step 2

Undo the $-\frac{3}{4}$ on $x$ by adding $\frac{3}{4}$ to both sides.
The equation becomes $x = \frac{1}{12} + \frac{3}{4}$.

$$x - \dfrac{3}{4} = \dfrac{1}{12} \;\Rightarrow\; x = \dfrac{1}{12} + \dfrac{3}{4}$$

💡 If a number minus $\frac{3}{4}$ gives $\frac{1}{12}$, then the number is $\frac{1}{12}$ plus $\frac{3}{4}$ — Grade 1 unknown-addend logic with fractions.

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

Add $\frac{1}{12} + \frac{3}{4}$ on the common denominator $12$: $\frac{3}{4} = \frac{9}{12}$, so $x = \frac{1}{12} + \frac{9}{12} = \frac{10}{12} = \frac{5}{6}$.

$$x = \dfrac{1}{12} + \dfrac{9}{12} = \dfrac{10}{12} = \dfrac{5}{6} \;\Rightarrow\; \textbf{(E)}$$

💡 Rewrite over the common denominator $12$ and simplify by dividing top and bottom by $2$.

#3 Eliminate Possibilities 4.NF.A.2 Step 4

Quick elimination check.
The right side $\frac{1}{12} + \frac{3}{4}$ is clearly close to $\frac{3}{4} \approx 0.83$, so $x$ must be near $\frac{5}{6} \approx 0.83$.
Choices (A) negative, (B) $\approx 0.19$, (C) $\approx 0.58$, (D) $\approx 0.67$ all miss; only (E) $\frac{5}{6}$ matches.

$$\dfrac{1}{12} + \dfrac{3}{4} \approx 0.83 \;\Rightarrow\; \textbf{(E)}\ \dfrac{5}{6}$$

💡 Comparing fractions to $\frac{3}{4}$ rules out four of the five choices instantly.

[1] #11 5.NF.A.1 First compute the right side. The denominators $12$ and $3$ have a common denomi

[2] #11 1.OA.D.8 Undo the $-\frac{3}{4}$ on $x$ by adding $\frac{3}{4}$ to both sides. The equati

[3] #11 5.NF.A.1 Add $\frac{1}{12} + \frac{3}{4}$ on the common denominator $12$: $\frac{3}{4} =

[4] #3 4.NF.A.2 Quick elimination check. The right side $\frac{1}{12} + \frac{3}{4}$ is clearly

Review

Reasonableness: Plug $x = \frac{5}{6}$ back in: $\frac{5}{6} - \frac{3}{4} = \frac{10}{12} - \frac{9}{12} = \frac{1}{12}$, and the right side was $\frac{5}{12} - \frac{4}{12} = \frac{1}{12}$. Both sides match, so (E) is correct.

Alternative: Tool #3 (Eliminate) alone — plug each choice into the left side and stop when it equals $\frac{1}{12}$. For (E): $\frac{5}{6} - \frac{3}{4} = \frac{10 - 9}{12} = \frac{1}{12}\ \checkmark$. This is the AMC multiple-choice instinct: when the work is short, just verify the answer choices directly.

CCSS standards used (min grade 5)

1.OA.D.8 Determine the unknown whole number in an addition or subtraction equation (Recognizing that $x - \frac{3}{4} = \frac{1}{12}$ means $x = \frac{1}{12} + \frac{3}{4}$ — the unknown-addend idea applied to fractions.)
4.NF.A.2 Compare two fractions with different numerators and different denominators (Eliminating answer choices by comparing their sizes to $\frac{3}{4}$.)
5.NF.A.1 Add and subtract fractions with unlike denominators (Computing $\frac{5}{12} - \frac{1}{3}$ and $\frac{1}{12} + \frac{3}{4}$ by rewriting over the common denominator $12$.)

⭐ This AMC 10 problem only needs Grade 5 "add and subtract fractions with unlike denominators" you already know — write everything over $12$, undo the $-\frac{3}{4}$ by adding it back, and the answer falls out as $\frac{5}{6}$.

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: E

Review

CCSS standards used (min grade 5)

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