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

(Solution)
(Problem)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
Each day, Jenny ate <math>20%</math> of the jellybeans that were in her jar at the beginning of that day. At the end of the second day, <math>32</math> remained. How many jellybeans were in the jar originally?
+
Each day, Jenny ate <math>20\%</math> of the jellybeans that were in her jar at the beginning of that day. At the end of the second day, <math>32</math> remained. How many jellybeans were in the jar originally?
  
 
<math> \mathrm{(A) \ 40 } \qquad \mathrm{(B) \ 50 } \qquad \mathrm{(C) \ 55 } \qquad \mathrm{(D) \ 60 } \qquad \mathrm{(E) \ 75 }  </math>
 
<math> \mathrm{(A) \ 40 } \qquad \mathrm{(B) \ 50 } \qquad \mathrm{(C) \ 55 } \qquad \mathrm{(D) \ 60 } \qquad \mathrm{(E) \ 75 }  </math>

Revision as of 13:21, 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