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Difference between revisions of "2013 AMC 10B Problems/Problem 10"

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==Solution==
 
==Solution==
 
Let <math>x</math> be the number of two point shots attempted and <math>y</math> 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%, <math>0.5(2x)+0.4(3y)=54</math> or <math>x+1.2y=54</math>. Because the team attempted 50% more two point shots then threes, <math>x=1.5y</math>. Substituting <math>1.5y</math> for <math>x</math> in the first equation gives <math>1.5y+1.2y=54</math>, which equals <math>2.7y=54</math> so <math>y=</math> <math>\boxed{\textbf{(C) }20}</math>
 
Let <math>x</math> be the number of two point shots attempted and <math>y</math> 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%, <math>0.5(2x)+0.4(3y)=54</math> or <math>x+1.2y=54</math>. Because the team attempted 50% more two point shots then threes, <math>x=1.5y</math>. Substituting <math>1.5y</math> for <math>x</math> in the first equation gives <math>1.5y+1.2y=54</math>, which equals <math>2.7y=54</math> so <math>y=</math> <math>\boxed{\textbf{(C) }20}</math>
 +
 +
==Video Solutions==
 +
i) https://youtu.be/lKGz-hay06I?si=chrn0Bk-kUH0WQWX [By Daniel Jin] (A full walkthrough of <math>ALL</math> AMC 10B Problems!)
 +
ii) https://youtu.be/FVSDxtVz6MQ?si=-M1E4e1CzBlimglD [By CanadaMath] (A walkthrough of AMC 10B Problems to 10!)
  
 
== See also ==
 
== See also ==
 
{{AMC10 box|year=2013|ab=B|num-b=9|num-a=11}}
 
{{AMC10 box|year=2013|ab=B|num-b=9|num-a=11}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 08:44, 19 December 2024

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!)

See also

2013 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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