AMC 8 · 2022 · #21

Grade 7 algebrarate-ratio

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

percentageratio-proportionsystems-of-equations bound-inequality-then-enumerateconvert-to-algebra ↑ Prerequisites: percentagelinear-equations-two-var

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Steph and Candace each play one game in two halves. Steph hits 15 of 20 shots in the first half and 10 of 10 in the second. Candace takes 12 shots in the first half and 18 in the second, with x and y baskets respectively. In each half Steph's shooting percentage is strictly higher than Candace's, yet their overall percentages turn out equal. We want y - x, the number of extra baskets Candace made in the second half compared to the first.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2022 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Convert to Algebra

The problem hands us two unknowns (x, y) tied together by one equation (equal overall percentages) and two inequalities (per-half comparisons), which is the textbook trigger for Tool #13 (Convert to Algebra). Tool #15 (Organize in More Ways) helps us first lay the data into a 2 × 3 table so the key observation — that both players take 30 total shots — jumps out and converts the equal-percentage statement into the clean equation x + y = 25. Tool #3 (Eliminate) then serves as the final squeeze: combining x ≤ 8 from one half with y ≤ 17 (so x ≥ 8) from the other leaves a single integer pair, eliminating every other possibility.

1STEP 1

Lay the shot data in a 2×3 table; the key jumps out — both players take the same 30 total shots (20+10 = 12+18).

20 + 10 = 30, 12 + 18 = 30

2STEP 2

Equal overall percentages over the same denominator force equal numerators: x + y = 25.

\frac{25}{30}

\frac{x+y}{30}

⟹ x + y = 25

3STEP 3

Turn each half's strict comparison into an inequality: $\frac{x}{12}$ ＜ $\frac{3}{4}$ gives x ≤ 8, and $\frac{y}{18}$ ＜ 1 gives y ≤ 17.

\frac{x}{12}

＜

\frac{3}{4}

→ x ＜ 9 → x ≤ 8

\frac{y}{18}

＜ 1 → y ＜ 18 → y ≤ 17

4STEP 4

Combine x + y = 25 with y ≤ 17 to force x ≥ 8; together with x ≤ 8 this pins x = 8 and y = 17.

x ≥ 8 and x ≤ 8 → x = 8, y = 17

5STEP 5

The question wants y - x, so subtract: y - x = 9, which is choice (C).

y - x = 17 - 8 = 9 → (C)

Answer

Check the scenario with (x, y) = (8, 17). First half: Candace shot

\frac{8}{12}

≈ 66.7% vs Steph's 75% — Steph higher, as required. Second half: Candace shot

\frac{17}{18}

≈ 94.4% vs Steph's 100% — Steph higher again, as required. Overall: Candace

\frac{25}{30}

≈ 83.3% vs Steph

\frac{25}{30}

— equal, as required. Every condition holds, the answer 9 matches choice (C), and 9 sits in the middle of the listed range 7 to 11, so no extreme-value red flags.

💡Key takeaway

This AMC 8 problem only needs Grade 7 equations and inequalities you already know — set up two simple inequalities from the comparisons and squeeze the answer!

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