Track / Stage 3 / 49 of 260 #49 of 2000

Problem 49

AMC 10/12, early questions

Algebra Difficulty 3.1 Find the answer

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?

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

Your answer

Official solution

Let $x$ be the number of shots that Candace made in the first half, and let $y$ be the number of shots Candace made in the second half. Since Candace and Steph took the same number of attempts, with an equal percentage of baskets scored, we have $x+y=10+15=25.$ In addition, we have the following inequalities: $\frac{x}{12}<\frac{15}{20} \implies x<9,$ and $\frac{y}{18}<\frac{10}{10} \implies y<18.$ Pairing this up with $x+y=25$ we see the only possible solution is $(x,y)=(8,17),$ for an answer of $17-8 = \boxed{\textbf{(C) } 9}.$
~wamofan

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.