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Difference between revisions of "2013 AMC 10A Problems/Problem 9"

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==Problem==
 
==Problem==
  
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In a recent basketball game, Shenille attempted only three-point shots and two-point shots.  She was successful on 20% of her three-point shots and 30% of her two-point shots.  Shenille attempted 30 shots.  How many points did she score?
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In a recent basketball game, Shenille attempted only three-point shots and two-point shots.  She was successful on <math>20\%</math> of her three-point shots and <math>30\%</math> of her two-point shots.  Shenille attempted <math>30</math> shots.  How many points did she score?
  
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<math> \textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18  \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36 </math>
  
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<math> \textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36 </math>
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==Solution 1==
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Let the number of attempted three-point shots be <math>x</math> and the number of attempted two-point shots be <math>y</math>. We know that <math>x+y=30</math>, and we need to evaluate <math>3(0.2x) + 2(0.3y)</math>, as we know that the three-point shots are worth <math>3</math> points and that she made <math>20</math>% of them and that the two-point shots are worth <math>2</math> and that she made <math>30</math>% of them.
  
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==Solution==
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Simplifying, we see that this is equal to <math>0.6x + 0.6y = 0.6(x+y)</math>.  Plugging in <math>x+y=30</math>, we get <math>0.6(30) = \boxed{\textbf{(B) }18}</math>
  
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Let the number of attempted three-point shots made be <math>x</math> and the number of attempted two-point shots be <math>y</math>.  We know that <math>x+y=30</math>, and we need to evaluate <math>(0.2\cdot3)x +(0.3\cdot2)y</math>, as we know that the
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==Solution 2 (cheap)==
  
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three-point shots are worth 3 points and that she made 20% of them and that the two-point shots are worth 2 and that she made 30% of them.
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The problem statement implies that it doesn't matter how many two-point shots or three-point shots are attempted. If we assume that Shenille only attempts three-pointers, then she makes <math>0.2 \cdot 30 = 6</math> shots, which are worth <math>6 \cdot 3 = \boxed{\textbf{(B) }18}</math> points. If we assume Shenille only attempts two-pointers, then she makes <math>0.3 \cdot 30 = 9</math> shots, which are worth <math>9 \cdot 2 = 18</math> points.
  
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==See Also==
  
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Simplifying, we see that this is equal to <math>0.6x + 0.6y = 0.6(x+y)</math>.  Plugging in <math>x+y=30</math>, we get <math>0.6(30) = 18</math>, <math>\textbf{(B)}</math>.
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{{AMC10 box|year=2013|ab=A|num-b=8|num-a=10}}
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{{AMC12 box|year=2013|ab=A|num-b=5|num-a=7}}
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{{MAA Notice}}

Revision as of 12:59, 26 July 2021

Problem

In a recent basketball game, Shenille attempted only three-point shots and two-point shots. She was successful on $20\%$ of her three-point shots and $30\%$ of her two-point shots. Shenille attempted $30$ shots. How many points did she score?

$\textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18 \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36$

Solution 1

Let the number of attempted three-point shots be $x$ and the number of attempted two-point shots be $y$. We know that $x+y=30$, and we need to evaluate $3(0.2x) + 2(0.3y)$, as we know that the three-point shots are worth $3$ points and that she made $20$% of them and that the two-point shots are worth $2$ and that she made $30$% of them.

Simplifying, we see that this is equal to $0.6x + 0.6y = 0.6(x+y)$. Plugging in $x+y=30$, we get $0.6(30) = \boxed{\textbf{(B) }18}$

Solution 2 (cheap)

The problem statement implies that it doesn't matter how many two-point shots or three-point shots are attempted. If we assume that Shenille only attempts three-pointers, then she makes $0.2 \cdot 30 = 6$ shots, which are worth $6 \cdot 3 = \boxed{\textbf{(B) }18}$ points. If we assume Shenille only attempts two-pointers, then she makes $0.3 \cdot 30 = 9$ shots, which are worth $9 \cdot 2 = 18$ points.

See Also

2013 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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
2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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 12 Problems and Solutions

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