Difference between revisions of "1994 AHSME Problems/Problem 16"

(Created page with "==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 ...")
 
(Solution)
 
Line 4: Line 4:
 
<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>
 +
 +
--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician]

Latest revision as of 20:56, 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?

$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 22 \qquad\textbf{(C)}\ 36 \qquad\textbf{(D)}\ 57 \qquad\textbf{(E)}\ 71$

Solution

Let $r$ and $b$ be the number of red and blue marbles originally in the bag respectively. After $1$ red marble is removed, there are $r+b-1$ marbles left in the bag and $r-1$ red marbles left. So \[\frac{r-1}{r+b-1}=\frac{1}{7}.\] When $2$ blue marbles are removed, there are $r$ red marbles and $r+b-2$ total marbles left in the bag. So \[\frac{r}{r+b-2}=\frac{1}{5}.\] Cross multiplying for each yields \begin{align*}7r-7=r+b-1&\implies 7r-6=r+b\\ 5r=r+b-2&\implies 5r+2=r+b.\end{align*} We can equate each of these expressions to yields \[7r-6=5r+2\implies 2r=8\implies r=4\implies b=18.\] Therefore, the total number of marbles is \[r+b=4+18=\boxed{\textbf{(B) }22.}\]

--Solution by TheMaskedMagician

Invalid username
Login to AoPS