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Difference between revisions of "2013 AMC 12A Problems/Problem 6"

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<math>18</math>, which is <math>B</math>
 
<math>18</math>, which is <math>B</math>
  
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==Alternative Solution==
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==Solution 2==
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Since the problem doesn't specify the number of 3-point shots she attempted, it can be assumed that number doesn't matter, so let it be <math>0</math>. Then, she must have made <math>30</math> 2-point shots. So, her score must be:
+
Since the problem doesn't specify the number of 3-point shots she attempted, it can be assumed that number doesn't matter, so let it be <math>0</math>. Then, she must have attempted <math>30</math> 2-point shots. So, her score must be:
  
 
<math>\frac{3}{10}*30*2</math>,which is <math>B</math>.
 
<math>\frac{3}{10}*30*2</math>,which is <math>B</math>.
 +
 +
==Solution 3==
 +
(similar to Solution 1, however a slightly more obvious way)
 +
 +
Say that
 +
 +
x = # of 2-pt shots
 +
 +
y = # of 3-pt shots
 +
 +
Because the total number of shots is <math>30</math>, <math>x + y = 30</math>
 +
 +
However, Shenille was only successful on <math>20\%</math> of the 3-pt shots, and <math>30\%</math> of the 2-pt shots, so
 +
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<math>0.2x + 0.3y</math> = #number of successful shots
 +
 +
For each successful shot, there is an associated number of points with it.
 +
 +
Therefore, <math>0.2(x)(3) + (0.3)(y)(2)</math> = her score
 +
 +
this evaluates to <math>0.6 (x + y)</math> = her score
 +
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<math>x + y</math> is already determined to be 30, so her score is <math>0.6 (30) = \boxed{\textbf{(B)}\ 18}</math>
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~amuppalla
  
 
==Additional note==
 
==Additional note==
  
 
It is also reasonably easy to find all possibilities for the number of two-point and three-point shots she made. Just note that both numbers of successful throws have to be integers. For "<math>30\%</math> of her two-point shots" to be an integer we need the number of two-point shots to be divisible by 10. This only leaves four possibilities for the number of two-point shots: 0, 10, 20, or 30. Each of them also works for the three-point shots, and as shown above, for each of them the total number of points scored is the same.
 
It is also reasonably easy to find all possibilities for the number of two-point and three-point shots she made. Just note that both numbers of successful throws have to be integers. For "<math>30\%</math> of her two-point shots" to be an integer we need the number of two-point shots to be divisible by 10. This only leaves four possibilities for the number of two-point shots: 0, 10, 20, or 30. Each of them also works for the three-point shots, and as shown above, for each of them the total number of points scored is the same.
 +
 +
==Video Solution==
 +
 +
https://youtu.be/CCjcMVtkVaQ
 +
~sugar_rush
  
 
== See also ==
 
== See also ==

Latest revision as of 10:39, 10 March 2024

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

Let the number of 3-point shots attempted be $x$. Since she attempted 30 shots, the number of 2-point shots attempted must be $30 - x$.

Since she was successful on $20\%$, or $\frac{1}{5}$, of her 3-pointers, and $30\%$, or $\frac{3}{10}$, of her 2-pointers, then her score must be

$\frac{1}{5}*3x + \frac{3}{10}*2(30-x)$


$\frac{3}{5}*x + \frac{3}{5}(30-x)$


$\frac{3}{5}(x+30-x)$


$\frac{3}{5}*30$


$18$, which is $B$

Solution 2

Since the problem doesn't specify the number of 3-point shots she attempted, it can be assumed that number doesn't matter, so let it be $0$. Then, she must have attempted $30$ 2-point shots. So, her score must be:

$\frac{3}{10}*30*2$,which is $B$.

Solution 3

(similar to Solution 1, however a slightly more obvious way)

Say that

x = # of 2-pt shots

y = # of 3-pt shots

Because the total number of shots is $30$, $x + y = 30$

However, Shenille was only successful on $20\%$ of the 3-pt shots, and $30\%$ of the 2-pt shots, so

$0.2x + 0.3y$ = #number of successful shots

For each successful shot, there is an associated number of points with it.

Therefore, $0.2(x)(3) + (0.3)(y)(2)$ = her score

this evaluates to $0.6 (x + y)$ = her score

$x + y$ is already determined to be 30, so her score is $0.6 (30) = \boxed{\textbf{(B)}\ 18}$

~amuppalla

Additional note

It is also reasonably easy to find all possibilities for the number of two-point and three-point shots she made. Just note that both numbers of successful throws have to be integers. For "$30\%$ of her two-point shots" to be an integer we need the number of two-point shots to be divisible by 10. This only leaves four possibilities for the number of two-point shots: 0, 10, 20, or 30. Each of them also works for the three-point shots, and as shown above, for each of them the total number of points scored is the same.

Video Solution

https://youtu.be/CCjcMVtkVaQ ~sugar_rush

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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