2013 AMC 10B Problems/Problem 10

Problem

A basketball team's players were successful on 50% of their two-point shots and 40% of their three-point shots, which resulted in 54 points. They attempted 50% more two-point shots than three-point shots. How many three-point shots did they attempt?

$\textbf{(A) }10\qquad\textbf{(B) }15\qquad\textbf{(C) }20\qquad\textbf{(D) }25\qquad\textbf{(E) }30$

Solution

Let $x$ be the number of two point shots attempted and $y$ the number of three point shots attempted. Because each two point shot is worth two points and the team made 50% and each three point shot is worth 3 points and the team made 40%, $0.5(2x)+0.4(3y)=54$ or $x+1.2y=54$ . Because the team attempted 50% more two point shots then threes, $x=1.5y$ . Substituting $1.5y$ for $x$ in the first equation gives $1.5y+1.2y=54$ , which equals $2.7y=54$ so $y=$ $\boxed{\textbf{(C) }20}$

Video Solutions

i) https://youtu.be/lKGz-hay06I?si=chrn0Bk-kUH0WQWX [By Daniel Jin] (A full walkthrough of $ALL$ AMC 10B Problems!)

ii) https://youtu.be/FVSDxtVz6MQ?si=-M1E4e1CzBlimglD [By CanadaMath] (A walkthrough of AMC 10B Problems to 10!)

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2013 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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All AMC 10 Problems and Solutions

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2013 AMC 10B Problems/Problem 10

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