2000 AMC 12 Problems/Problem 3

Revision as of 14:40, 25 November 2022 by Z-fg (talk | contribs) (Solution 1)
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

Since Jenny eats $20\%$ of her jelly beans per day, $80\%=\frac{4}{5}$ of her jelly beans remain after one day.

Let $x$ be the number of jelly beans in the jar originally.

\[\frac{4}{5}\cdot x=32\], so

\[\frac{4}{5}\cdot x=32\], so

\[x=\frac{5}{4}\cdot32= 50 \Rightarrow \boxed{B} .\]

Solution 2 (answer choices)

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

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