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 $\boxed{B}$.

Solution 3

We can work backwards to figure out what the starting amount is. If she ate $20\%$ of the jellybeans, $80\% = \frac{4}{5}$ of the jellybeans were remaining at the end of each day. This means that we can multiply by $\frac{1}{80\%} = \frac{5}{4}$ to get the amount of jellybeans that she had at the end of the previous day. So we have:

$32 \cdot \frac{5}{4} \cdot \frac{5}{4} = 40 \cdot \frac{5}{4} = \boxed{(B)   50}$

~anabel.disher

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

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