AMC 10 · 2023 · #2

Grade 7 arithmetic

linear-equations-one-varfraction-arithmeticfraction-multiplication convert-to-algebraidentify-subproblems ↑ Prerequisites: fraction-arithmeticlinear-equations-one-var

📏 Medium solution 💡 2 insights

Problem

The weight of $\frac{1}{3}$ of a large pizza together with $3 \frac{1}{2}$ cups of orange slices is the same as the weight of $\frac{3}{4}$ of a large pizza together with $\frac{1}{2}$ cup of orange slices. A cup of orange slices weighs $\frac{1}{4}$ of a pound. What is the weight, in pounds, of a large pizza?

Pick an answer.

(A)

$1\frac{4}{5}$

(B)

(C)

$2\frac{2}{5}$

(D)

(E)

$3\frac{3}{5}$

AMC 10 2023 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.

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Toolkit + CCSS Solution

Understand

Restated: A balance equation: $\tfrac{1}{3}$ of a pizza plus $3\tfrac{1}{2}$ cups of orange slices weighs the same as $\tfrac{3}{4}$ of a pizza plus $\tfrac{1}{2}$ cup of orange slices. One cup of orange slices weighs $\tfrac{1}{4}$ pound. Find the weight of one whole pizza in pounds.

Givens: $\tfrac{1}{3}$ pizza $+\,3\tfrac{1}{2}$ cups orange $=$ $\tfrac{3}{4}$ pizza $+\,\tfrac{1}{2}$ cup orange (equal weights); $1$ cup of orange slices weighs $\tfrac{1}{4}$ lb; Answer choices: (A) $1\tfrac{4}{5}$, (B) $2$, (C) $2\tfrac{2}{5}$, (D) $3$, (E) $3\tfrac{3}{5}$

Unknowns: The weight of one whole large pizza, in pounds

Understand

Plan

Primary tool: #13 Convert to Algebra

Secondary: #7 Identify Subproblems

Two combined weights are stated to be equal — this is exactly the trigger for Tool #13 (Convert to Algebra): let $p$ be the pizza weight and turn the balance sentence into an equation. Tool #7 (Identify Subproblems) splits the work cleanly: first reduce the orange-slice quantities to numerical pounds (just arithmetic with $\tfrac{1}{4}$), then solve the resulting linear equation in $p$ alone. This keeps each step a one-thought move and avoids the common error of carrying mixed numbers through algebra.

Execute — Answer: A

#13 Convert to Algebra 6.EE.B.7 Step 1

Let $p$ be the weight of one whole pizza, in pounds.
Translate the balance sentence directly: left pan equals right pan.
The orange-slice quantity on each side becomes (cups) $\times \tfrac{1}{4}$ pound per cup.

$$\tfrac{1}{3}p + \tfrac{7}{2} \cdot \tfrac{1}{4} = \tfrac{3}{4}p + \tfrac{1}{2} \cdot \tfrac{1}{4}$$

💡 Letting the unknown weight be $p$ and writing each pan's total as an expression turns a word balance into the Grade 6 equation form $\text{stuff in } p = \text{stuff in } p$.

#7 Identify Subproblems 5.NF.B.4 Step 2

Subproblem A: compute the two orange-slice terms.
Multiply the cup counts by $\tfrac{1}{4}$ pound per cup to get the actual orange-slice weights in pounds.

$$\tfrac{7}{2} \cdot \tfrac{1}{4} = \tfrac{7}{8}, \quad \tfrac{1}{2} \cdot \tfrac{1}{4} = \tfrac{1}{8}$$

💡 Multiplying two fractions $\tfrac{a}{b} \cdot \tfrac{c}{d} = \tfrac{ac}{bd}$ is the Grade 5 fraction-times-fraction skill — no variables involved here, just numbers.

#13 Convert to Algebra 7.EE.B.4 Step 3

Substitute the simplified weights and collect the $p$ terms on one side, constants on the other.
Subtract $\tfrac{1}{3}p$ from both sides and $\tfrac{1}{8}$ from both sides.

$$\tfrac{1}{3}p + \tfrac{7}{8} = \tfrac{3}{4}p + \tfrac{1}{8} \;\Rightarrow\; \tfrac{7}{8} - \tfrac{1}{8} = \tfrac{3}{4}p - \tfrac{1}{3}p$$

💡 Moving like terms to opposite sides of an equation is the standard Grade 7 "solve a linear equation" move.

#13 Convert to Algebra 5.NF.A.1 Step 4

Combine on each side.
Left: $\tfrac{7}{8} - \tfrac{1}{8} = \tfrac{6}{8} = \tfrac{3}{4}$.
Right: use common denominator $12$ to get $\tfrac{3}{4}p - \tfrac{1}{3}p = \tfrac{9}{12}p - \tfrac{4}{12}p = \tfrac{5}{12}p$.

$$\tfrac{3}{4} = \tfrac{5}{12}p$$

💡 Subtracting fractions with unlike denominators by rewriting with a common denominator is the Grade 5 fraction-add/subtract standard.

#13 Convert to Algebra 6.EE.B.7 Step 5

Isolate $p$ by multiplying both sides by the reciprocal $\tfrac{12}{5}$.
Then convert to a mixed number to match the answer choices.

$$p = \tfrac{3}{4} \cdot \tfrac{12}{5} = \tfrac{36}{20} = \tfrac{9}{5} = 1\tfrac{4}{5} \;\Rightarrow\; \textbf{(A)}$$

💡 Dividing both sides by $\tfrac{5}{12}$ (i.e. multiplying by $\tfrac{12}{5}$) finishes the Grade 6 "$px = q$" equation in one move.

[1] #13 6.EE.B.7 Let $p$ be the weight of one whole pizza, in pounds. Translate the balance sente

[2] #7 5.NF.B.4 Subproblem A: compute the two orange-slice terms. Multiply the cup counts by $\t

[3] #13 7.EE.B.4 Substitute the simplified weights and collect the $p$ terms on one side, constan

[4] #13 5.NF.A.1 Combine on each side. Left: $\tfrac{7}{8} - \tfrac{1}{8} = \tfrac{6}{8} = \tfrac

[5] #13 6.EE.B.7 Isolate $p$ by multiplying both sides by the reciprocal $\tfrac{12}{5}$. Then co

Review

Reasonableness: Plug $p = \tfrac{9}{5}$ back into both pans. Left: $\tfrac{1}{3} \cdot \tfrac{9}{5} + \tfrac{7}{8} = \tfrac{3}{5} + \tfrac{7}{8} = \tfrac{24}{40} + \tfrac{35}{40} = \tfrac{59}{40}$. Right: $\tfrac{3}{4} \cdot \tfrac{9}{5} + \tfrac{1}{8} = \tfrac{27}{20} + \tfrac{1}{8} = \tfrac{54}{40} + \tfrac{5}{40} = \tfrac{59}{40}$. Both pans balance — the answer $1\tfrac{4}{5}$ lb is correct. Magnitude makes sense: a pizza weighing under $2$ pounds is small but possible, and the answer matches choice (A) exactly.

Alternative: Tool #6 (Guess and Check) on the five answer choices. Test (A) $1\tfrac{4}{5}$: left pan $= \tfrac{1}{3}(1.8) + 3.5(0.25) = 0.6 + 0.875 = 1.475$; right pan $= \tfrac{3}{4}(1.8) + 0.5(0.25) = 1.35 + 0.125 = 1.475$. Equal — (A) works on the first try. The other choices can be ruled out because they would make $\tfrac{3}{4}p$ much bigger than $\tfrac{1}{3}p$ without enough orange-slice difference to compensate.

CCSS standards used (min grade 7)

5.NF.A.1 Add and subtract fractions with unlike denominators (Subtracting $\tfrac{1}{3}p$ from $\tfrac{3}{4}p$ via the common denominator $12$ to get $\tfrac{5}{12}p$, and subtracting $\tfrac{1}{8}$ from $\tfrac{7}{8}$ to get $\tfrac{3}{4}$.)
5.NF.B.4 Multiply a fraction by a fraction (Multiplying $\tfrac{7}{2} \cdot \tfrac{1}{4} = \tfrac{7}{8}$ and $\tfrac{1}{2} \cdot \tfrac{1}{4} = \tfrac{1}{8}$ to convert cups of orange slices into pounds.)
6.EE.B.7 Solve real-world problems by writing and solving equations of the form px = q (Writing the balance "left pan $=$ right pan" as an equation in $p$ and finishing with $\tfrac{5}{12}p = \tfrac{3}{4} \Rightarrow p = \tfrac{9}{5}$.)
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities (Collecting the $p$ terms on one side and the constants on the other to get a clean linear equation in $p$ alone.)

⭐ This AMC 10 problem only needs Grade 7 "build and solve a linear equation" — call the pizza $p$, write each pan as pounds, and balance.