2000 AMC 12 Problems/Problem 3

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

Solution 2 (answer choices)

Testing the answers choices out, we see that the answer is $\boxed{B}$.

Video Solution by Daily Dose of Math

https://youtu.be/qJYI_qOMyTo?si=jFa9C5BObcWaI_r6

~Thesmartgreekmathdude

See also

2000 AMC 12 (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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
2000 AMC 10 (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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

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