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

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
+== Problem==
+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>
+==Solution==
 Let the number of 3-point shots attempted be <math>x</math>. Since she attempted 30 shots, the number of 2-point shots attempted must be <math>30 - x</math>.
@@ Line 15: / Line 22: @@
-<math>18</math>
+<math>18</math>, which is <math>B</math>
+==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 <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>.
+==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
+<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
+<math>x + y</math> is already determined to be 30, so her score is <math>0.6 (30) = \boxed{\textbf{(B)}\ 18}</math>
+~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 "<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 ==
+{{AMC10 box|year=2013|ab=A|num-b=8|num-a=10}}
+{{AMC12 box|year=2013|ab=A|num-b=5|num-a=7}}
+{{MAA Notice}}

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