# 2000 AMC 12 Problems/Problem 3

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The following problem is from both the 2000 AMC 12 #3 and 2000 AMC 10 #3, so both problems redirect to this page.

## Problem

Each day, Jenny ate $20\%$ of the jellybeans that were in her jar at the beginning of that day. At the end of the second day, $32$ remained. How many jellybeans were in the jar originally? $\textbf{(A)} \ 40 \qquad \textbf{(B)} \ 50 \qquad \textbf{(C)} \ 55 \qquad \textbf{(D)} \ 60 \qquad \textbf{(E)} \ 75$

## Solution 1 (Algebra)

We can begin by labeling the number of initial jellybeans $x$. If she ate $20\%$ of the jellybeans, then $80\%$ is remaining. Hence, after day 1, there are: $0.8 * x$

After day 2, there are: $0.8 * 0.8 * x$ or $0.64x$ jellybeans. $0.64x = 32$, so $x = \boxed{(B) 50}$

Solution By: armang32324

Testing the answers choices out, we see that the answer is $\boxed{B}$.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 