Difference between revisions of "2000 AMC 12 Problems/Problem 3"

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== See also ==
 
== See also ==
* [[2000 AMC 12 Problems]]
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{{AMC12 box|year=2000|num-b=2|num-a=4}}
*[[2000 AMC 12 Problems/Problem 2|Previous Problem]]
 
*[[2000 AMC 12 Problems/Problem 4|Next problem]]
 
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 13:24, 17 October 2007

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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All AMC 12 Problems and Solutions