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

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{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #3]] and [[2000 AMC 10 Problems|2000 AMC 10 #3]]}}
 
== 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?
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== See also ==
 
== See also ==
 
{{AMC12 box|year=2000|num-b=2|num-a=4}}
 
{{AMC12 box|year=2000|num-b=2|num-a=4}}
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{{AMC10 box|year=2000|num-b=2|num-a=4}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Revision as of 22:32, 26 November 2011

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?

$\mathrm{(A) \ 40 } \qquad \mathrm{(B) \ 50 } \qquad \mathrm{(C) \ 55 } \qquad \mathrm{(D) \ 60 } \qquad \mathrm{(E) \ 75 }$

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
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 12 Problems and Solutions
2000 AMC 10 (ProblemsAnswer KeyResources)
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