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

m (Alternative Solution)
Line 32: Line 32:
  
 
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 ==

Revision as of 17:00, 24 November 2020

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 $B$

Alternative Solution

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 $B$.

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 (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

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