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

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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?
 
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?
 
  
 
<math> \textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18  \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36 </math>
 
<math> \textbf{(A)}\ 12 \qquad\textbf{(B)}\ 18  \qquad\textbf{(C)}\ 24 \qquad\textbf{(D)}\ 30 \qquad\textbf{(E)}\ 36 </math>
  
==Solution==
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==Solution 1==
  
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 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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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.
  
 
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>
 
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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==Solution 2 (cheap)==  
 
==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 <math>0.2 \cdot 30 = 6</math> shots, which are worth <math>6 \cdot 3 = \boxed{\textbf{(B) }18}</math> points.
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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.
 
 
-- I-_-I
 
  
 
==See Also==
 
==See Also==

Latest revision as of 11:59, 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 (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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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