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

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(Solution)
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== Solution ==
 
== Solution ==
Since Jenny eats <math>20%</math> of her jelly beans per day, <math>80%=\frac{4}{5}</math> of her jelly beans remain after one day.  
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Since Jenny eats <math>20\%</math> of her jelly beans per day, <math>80\%=\frac{4}{5}</math> of her jelly beans remain after one day.  
  
 
Let <math>x</math> be the number of jelly beans in the jar originally.  
 
Let <math>x</math> be the number of jelly beans in the jar originally.  
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<math>\frac{16}{25}\cdot x=32</math>
 
<math>\frac{16}{25}\cdot x=32</math>
  
<math>\displaystyle x=\frac{25}{16}\cdot32= 50 \Rightarrow B </math>
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<math>x=\frac{25}{16}\cdot32= 50 \Rightarrow B </math>
  
 
== See also ==
 
== See also ==

Revision as of 14:20, 17 October 2007

Problem

Each day, Jenny ate $20%$ (Error compiling LaTeX. Unknown error_msg) 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