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2000 AMC 12 Problems/Problem 3

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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?

\mathrm{(A) \ 40 } \qquad \mathrm{(B) \ 50 } \qquad \mathrm{(C) \ 55 } \qquad \mathrm{(D) \ 60 } \qquad \mathrm{(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 B

See also

2000 AMC 12 (ProblemsResources)
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
Want to learn how to tackle those tough AMC/AIME/Olympiad counting and probability problems? Check out Art of Problem Solving's Intermediate Counting & Probability by David Patrick.
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