Competition · AMC preparation · step 4 of 4

AMC 10 · 2023A · #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

A balance equation:

\frac{1}{3}

of a pizza plus 3

\frac{1}{2}

cups of orange slices weighs the same as

\frac{3}{4}

of a pizza plus

\frac{1}{2}

cup of orange slices. One cup of orange slices weighs

\frac{1}{4}

pound. Find the weight of one whole pizza in pounds.

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.

How to solve

Strategy Convert to Algebra

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 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.

1STEP 1

Name the pizza weight

Let p be the pizza's weight in pounds, then rewrite left pan = right pan with each orange term as (cups) × $\frac{1}{4}$ pound.

1/3p + 7/2 · 1/4 = 3/4p + 1/2 · 1/4

2STEP 2

Compute the orange weights

Subproblem A: turn the cup counts into pounds — $\frac{7}{8}$ pound on the left, $\frac{1}{8}$ pound on the right.

7/2 · 1/4 = 7/8, 1/2 · 1/4 = 1/8

3STEP 3

Collect the p terms

Substitute those weights, then subtract $\frac{1}{3}$ p and $\frac{1}{8}$ from both sides to gather p-terms apart from the constants.

1/3p + 7/8 = 3/4p + 1/8 → 7/8 - 1/8 = 3/4p - 1/3p

4STEP 4

Combine each side

Combine each side: the left is $\frac{3}{4}$ and the right becomes $\frac{5}{12}$ p using common denominator 12.

3/4 = 5/12p

5STEP 5

Multiply by the reciprocal

Multiply both sides by the reciprocal $\frac{12}{5}$ to get p = $\frac{9}{5}$ , then write it as a mixed number to match choice (A).

p = 3/4 · 12/5 = 36/20 = 9/5 = 1 4/5 → (A)

Answer

1 4/5

Plug p =

\frac{9}{5}

back into both pans. Left:

\frac{1}{3}

\frac{9}{5}

\frac{7}{8}

\frac{3}{5}

\frac{7}{8}

\frac{24}{40}

\frac{35}{40}

\frac{59}{40}

. Right:

\frac{3}{4}

\frac{9}{5}

\frac{1}{8}

\frac{27}{20}

\frac{1}{8}

\frac{54}{40}

\frac{5}{40}

\frac{59}{40}

. Both pans balance — the answer 1

\frac{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.

💡Key takeaway

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.

Name the pizza weight
Compute the orange weights
Collect the p terms
Combine each side
Multiply by the reciprocal

A parent dashboard for the family lives at sensimlab.com.