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

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.

$\frac{4}{5}\cdot\frac{4}{5}\cdot x=32$

$\frac{16}{25}\cdot x=32$

$x=\frac{25}{16}\cdot32= 50 \Rightarrow \boxed{B}$

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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All AMC 12 Problems and Solutions

2000 AMC 10 (Problems • Answer Key • Resources)
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.

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2000 AMC 12 Problems/Problem 3

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