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

m (Solution: fixed strange paragraphing)
Line 8: Line 8:
 
==Solution==
 
==Solution==
  
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  
+
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.
 
 
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 <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>

Revision as of 13:24, 27 January 2015

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 attempted three-point shots made be $x$ and the number of attempted two-point shots be $y$. We know that $x+y=30$, and we need to evaluate $(0.2\cdot3)x +(0.3\cdot2)y$, 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 $0.6(30) = \boxed{\textbf{(B) }18}$

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

Invalid username
Login to AoPS