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

Latest revision as of 01:54, 28 May 2021

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

1994 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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

@@ Line 7: / Line 7: @@
 --Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician]
+==See Also==
+{{AHSME box|year=1994|num-b=15|num-a=17}}
+{{MAA Notice}}

Page

Toolbox

Search

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

Latest revision as of 01:54, 28 May 2021

Problem

Solution

See Also