AMC 8 · 2022 · #21

Grade 7 algebrarate-ratio

Archetype Small-Domain Constrained Search

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

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Steph scored $15$ baskets out of $20$ attempts in the first half of a game, and $10$ baskets out of $10$ attempts in the second half. Candace took $12$ attempts in the first half and $18$ attempts in the second. In each half, Steph scored a higher percentage of baskets than Candace. Surprisingly they ended with the same overall percentage of baskets scored. How many more baskets did Candace score in the second half than in the first?

$\textbf{(A) } 7\qquad\textbf{(B) } 8\qquad\textbf{(C) } 9\qquad\textbf{(D) } 10\qquad\textbf{(E) } 11$

Pick an answer.

(A)

(B)

(C)

(D)

(E)

View mode:

Toolkit + CCSS Solution

Understand

Restated: 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.

Givens: Steph: $15/20$ in the first half, $10/10$ in the second half (so $25/30$ overall); Candace: $x/12$ in the first half, $y/18$ in the second half (so $(x+y)/30$ overall); Steph's first-half percentage $>$ Candace's first-half percentage; Steph's second-half percentage $>$ Candace's second-half percentage; Steph's overall percentage $=$ Candace's overall percentage; $x$ and $y$ are non-negative integers; Answer choices: (A) $7$, (B) $8$, (C) $9$, (D) $10$, (E) $11$

Unknowns: The value of $y - x$ (extra baskets Candace made in the second half)

Understand

Plan

Primary tool: #13 Convert to Algebra

Secondary: #15 Organize Information in More Ways, #3 Eliminate Possibilities

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 $2 \times 3$ 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 $x \le 8$ from one half with $y \le 17$ (so $x \ge 8$) from the other leaves a single integer pair, eliminating every other possibility.

Execute — Answer: C

#15 Organize Information in More Ways 6.RP.A.1 Step 1

Reorganize the shot data into a $2 \times 3$ table (player by half plus total) so the parallel structure is visible at a glance.
| Player | 1st half | 2nd half | Total | |---------|-----------|-----------|----------| | Steph | $15/20$ | $10/10$ | $25/30$ | | Candace | $x/12$ | $y/18$ | $(x+y)/30$ | The crucial pattern: both players' totals have the same denominator $30$ (since $20+10=12+18=30$).

$$20 + 10 = 30, \quad 12 + 18 = 30$$

💡 Putting the data in a table is the Tool #15 move — it makes the matching $30$-shot totals stand out, which is the whole key to the problem.

#13 Convert to Algebra 6.RP.A.3 Step 2

Translate "equal overall percentages" into an equation.
Because both denominators are $30$, equal fractions force equal numerators, giving the first equation in our system.

$$\dfrac{25}{30} = \dfrac{x+y}{30} \;\Longrightarrow\; x + y = 25$$

💡 Two ratios with the same denominator are equal only when the numerators match — that is Grade 6 ratio reasoning.

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

Translate each per-half comparison into an inequality and solve.
First half: $\dfrac{x}{12} < \dfrac{15}{20} = \dfrac{3}{4}$, so multiplying both sides by $12$ gives $x < 9$.
Since $x$ must be a whole number, $x \le 8$.
Second half: $\dfrac{y}{18} < \dfrac{10}{10} = 1$, so $y < 18$, meaning $y \le 17$.

$\dfrac{x}{12} < \dfrac{3}{4} \;\Rightarrow\; x < 9 \;\Rightarrow\; x \le 8$ $\dfrac{y}{18} < 1 \;\Rightarrow\; y < 18 \;\Rightarrow\; y \le 17$

💡 Building and solving simple inequalities for the unknowns is exactly the Grade 7 "construct and solve inequalities" standard.

#3 Eliminate Possibilities 7.EE.B.4 Step 4

Combine the equation with the two inequalities to pin down $x$ and $y$.
From $x + y = 25$ and $y \le 17$, we get $x = 25 - y \ge 25 - 17 = 8$.
Together with $x \le 8$, this squeezes $x = 8$ exactly, hence $y = 17$.

$$x \ge 8 \text{ and } x \le 8 \;\Rightarrow\; x = 8, \; y = 17$$

💡 The two opposite-direction bounds eliminate every value except one — Tool #3's "squeeze to a single survivor" idea applied to integers.

#13 Convert to Algebra 4.NBT.B.4 Step 5

Answer the question: how many more baskets did Candace make in the second half than in the first?

$$y - x = 17 - 8 = 9 \;\Rightarrow\; \textbf{(C)}$$

💡 Once both numbers are known, the answer is a single Grade 4 subtraction.

[1] #15 6.RP.A.1 Reorganize the shot data into a $2 \times 3$ table (player by half plus total) s

[2] #13 6.RP.A.3 Translate "equal overall percentages" into an equation. Because both denominator

[3] #13 7.EE.B.4 Translate each per-half comparison into an inequality and solve. First half: $\d

[4] #3 7.EE.B.4 Combine the equation with the two inequalities to pin down $x$ and $y$. From $x

[5] #13 4.NBT.B.4 Answer the question: how many more baskets did Candace make in the second half t

Review

Reasonableness: Check the scenario with $(x, y) = (8, 17)$. First half: Candace shot $8/12 \approx 66.7\%$ vs Steph's $75\%$ — Steph higher, as required. Second half: Candace shot $17/18 \approx 94.4\%$ vs Steph's $100\%$ — Steph higher again, as required. Overall: Candace $25/30 \approx 83.3\%$ vs Steph $25/30$ — 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.

Alternative: Tool #6 (Guess and Check) on the choices: the candidates for $y - x$ are $7, 8, 9, 10, 11$. Combined with $x + y = 25$, each choice fixes $(x, y)$ uniquely — for instance $y - x = 9$ gives $(x, y) = (8, 17)$. Testing each pair against $x \le 8$ and $y \le 17$: only $(8, 17)$ satisfies both bounds with equality; $7 \to (9, 16)$ fails $x \le 8$, while $10, 11 \to (7.5, 17.5), (7, 18)$ either are non-integers or fail $y \le 17$. Choice (C) is the only survivor.

CCSS standards used (min grade 7)

4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Computing the final answer $y - x = 17 - 8 = 9$ once both basket counts are known.)
6.RP.A.1 Understand the concept of a ratio and use ratio language (Reorganizing Steph's and Candace's shot data into a table of baskets-to-attempts ratios so the matching $30$-attempt totals become visible.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Recognizing that equal ratios $25/30 = (x+y)/30$ with a common denominator force equal numerators, giving $x + y = 25$.)
7.EE.B.4 Use variables to represent quantities and construct simple equations and inequalities (Setting up and solving the per-half inequalities $x/12 < 3/4$ and $y/18 < 1$, then combining them with $x + y = 25$ to pin down $x = 8$ and $y = 17$.)

⭐ 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!

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 7)

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