Difference between revisions of "2000 AMC 12 Problems/Problem 3"
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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. | ||
| − | <cmath> | + | <cmath>\frac{4}{5}\cdot x=32</cmath>, so |
<cmath>\frac{16}{25}\cdot x=32</cmath>, so | <cmath>\frac{16}{25}\cdot x=32</cmath>, so | ||
Revision as of 15:38, 25 November 2022
- 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 
of the jellybeans that were in her jar at the beginning of that day. At the end of the second day, 
remained. How many jellybeans were in the jar originally?
Solution 1
Since Jenny eats of her jelly beans per day, 
of her jelly beans remain after one day.
Let 
be the number of jelly beans in the jar originally.
![\[\frac{4}{5}\cdot x=32\]](http://latex.artofproblemsolving.com/6/1/8/618e5c16fde4722b4dd45dad953566ee9804ab27.png)
, so
![\[\frac{16}{25}\cdot x=32\]](http://latex.artofproblemsolving.com/e/1/8/e180fb8b3c4e40b34b85ca8f667ad8d8f2bd6448.png)
, so
![\[x=\frac{25}{16}\cdot32= 50 \Rightarrow \boxed{B} .\]](http://latex.artofproblemsolving.com/8/b/2/8b24028859e3f93a39428a5f28c14aa20c473c36.png)
Solution 2 (answer choices)
Testing the answers choices out, we see that the answer is 
.
See also
| 2000 AMC 12 (Problems • Answer Key • Resources) | |
| 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 (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
