Difference between revisions of "1994 AHSME Problems/Problem 16"
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<math> \textbf{(A)}\ 8 \qquad\textbf{(B)}\ 22 \qquad\textbf{(C)}\ 36 \qquad\textbf{(D)}\ 57 \qquad\textbf{(E)}\ 71 </math> | <math> \textbf{(A)}\ 8 \qquad\textbf{(B)}\ 22 \qquad\textbf{(C)}\ 36 \qquad\textbf{(D)}\ 57 \qquad\textbf{(E)}\ 71 </math> | ||
==Solution== | ==Solution== | ||
| + | Let <math>r</math> and <math>b</math> be the number of red and blue marbles originally in the bag respectively. After <math>1</math> red marble is removed, there are <math>r+b-1</math> marbles left in the bag and <math>r-1</math> red marbles left. So <cmath>\frac{r-1}{r+b-1}=\frac{1}{7}.</cmath> When <math>2</math> blue marbles are removed, there are <math>r</math> red marbles and <math>r+b-2</math> total marbles left in the bag. So <cmath>\frac{r}{r+b-2}=\frac{1}{5}.</cmath> Cross multiplying for each yields <cmath>\begin{align*}7r-7=r+b-1&\implies 7r-6=r+b\\ 5r=r+b-2&\implies 5r+2=r+b.\end{align*}</cmath> We can equate each of these expressions to yields <cmath>7r-6=5r+2\implies 2r=8\implies r=4\implies b=18.</cmath> Therefore, the total number of marbles is <cmath>r+b=4+18=\boxed{\textbf{(B) }22.}</cmath> | ||
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| + | --Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician] | ||
Revision as of 19:56, 20 July 2014
Problem
Some marbles in a bag are red and the rest are blue. If one red marble is removed, then one-seventh of the remaining marbles are red. If two blue marbles are removed instead of one red, then one-fifth of the remaining marbles are red. How many marbles were in the bag originally?
Solution
Let 
and 
be the number of red and blue marbles originally in the bag respectively. After 
red marble is removed, there are 
marbles left in the bag and 
red marbles left. So ![\[\frac{r-1}{r+b-1}=\frac{1}{7}.\]](http://latex.artofproblemsolving.com/2/2/b/22bacd28f2461299b9bc6fd9fc16797b6de19cf2.png)
When 
blue marbles are removed, there are red marbles and 
total marbles left in the bag. So ![\[\frac{r}{r+b-2}=\frac{1}{5}.\]](http://latex.artofproblemsolving.com/0/9/2/092b74f810773f5f0a97b58d484d17c83b7e9208.png)
Cross multiplying for each yields 
We can equate each of these expressions to yields ![\[7r-6=5r+2\implies 2r=8\implies r=4\implies b=18.\]](http://latex.artofproblemsolving.com/f/4/6/f46b8d25e0edb5db27488de838bf9fea35bf6606.png)
Therefore, the total number of marbles is ![\[r+b=4+18=\boxed{\textbf{(B) }22.}\]](http://latex.artofproblemsolving.com/4/7/3/473054172021df69b13e9a8256b85b33bf69a3f4.png)
--Solution by TheMaskedMagician
