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
A balance equation: 13\frac{1}{3} of a pizza plus 3 12\frac{1}{2} cups of orange slices weighs the same as 34\frac{3}{4} of a pizza plus 12\frac{1}{2} cup of orange slices. One cup of orange slices weighs 14\frac{1}{4} pound. Find the weight of one whole pizza in pounds.

Pick an answer.

(A)
$1\frac{4}{5}$
(B)
2
(C)
$2\frac{2}{5}$
(D)
3
(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 14\frac{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

Let p be the pizza's weight in pounds, then rewrite left pan = right pan with each orange term as (cups) × 14\frac{1}{4} pound.

13\frac{1}{3}p + 72\frac{7}{2} · 14\frac{1}{4} = 34\frac{3}{4}p + 12\frac{1}{2} · 14\frac{1}{4}
2STEP 2

Subproblem A: turn the cup counts into pounds — 78\frac{7}{8} pound on the left, 18\frac{1}{8} pound on the right.

72\frac{7}{2} · 14\frac{1}{4} = 78\frac{7}{8}, 12\frac{1}{2} · 14\frac{1}{4} = 18\frac{1}{8}
3STEP 3

Substitute those weights, then subtract 13\frac{1}{3}p and 18\frac{1}{8} from both sides to gather p-terms apart from the constants.

13\frac{1}{3}p + 78\frac{7}{8} = 34\frac{3}{4}p + 18\frac{1}{8} → 78\frac{7}{8} - 18\frac{1}{8} = 34\frac{3}{4}p - 13\frac{1}{3}p
4STEP 4

Combine each side: the left is 34\frac{3}{4} and the right becomes 512\frac{5}{12}p using common denominator 12.

34\frac{3}{4} = 512\frac{5}{12}p
5STEP 5

Multiply both sides by the reciprocal 125\frac{12}{5} to get p = 95\frac{9}{5}, then write it as a mixed number to match choice (A).

p = 34\frac{3}{4} · 125\frac{12}{5} = 3620\frac{36}{20} = 95\frac{9}{5} = 1 45\frac{4}{5} → (A)
Answer
1 45\frac{4}{5}
Plug p = 95\frac{9}{5} back into both pans. Left: 13\frac{1}{3} · 95\frac{9}{5} + 78\frac{7}{8} = 35\frac{3}{5} + 78\frac{7}{8} = 2440\frac{24}{40} + 3540\frac{35}{40} = 5940\frac{59}{40}. Right: 34\frac{3}{4} · 95\frac{9}{5} + 18\frac{1}{8} = 2720\frac{27}{20} + 18\frac{1}{8} = 5440\frac{54}{40} + 540\frac{5}{40} = 5940\frac{59}{40}. Both pans balance — the answer 1 45\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.