Difference between revisions of "2000 AMC 12 Problems/Problem 3"
(→Problem) |
|||
| (14 intermediate revisions by 10 users not shown) | |||
| Line 1: | Line 1: | ||
| + | {{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? | ||
| − | <math> \ | + | <math> \textbf{(A)} \ 40 \qquad \textbf{(B)} \ 50 \qquad \textbf{(C)} \ 55 \qquad \textbf{(D)} \ 60 \qquad \textbf{(E)} \ 75 </math> |
| − | == Solution == | + | == Solution 1 (Algebra)== |
| − | + | We can begin by labeling the number of initial jellybeans <math>x</math>. If she ate <math>20\%</math> of the jellybeans, then <math>80\%</math> is remaining. Hence, after day 1, there are: | |
| + | <math>0.8 * x</math> | ||
| − | + | After day 2, there are: | |
| + | <math>0.8 * 0.8 * x</math> or <math>0.64x</math> jellybeans. <math>0.64x = 32</math>, so <math>x = \boxed{(B) 50}</math> | ||
| − | + | Solution By: armang32324 | |
| − | <math>\ | + | == Solution 2 (answer choices) == |
| + | Testing the answers choices out, we see that the answer is <math>\boxed{B}</math>. | ||
| − | + | ==Video Solution by Daily Dose of Math== | |
| + | |||
| + | https://youtu.be/qJYI_qOMyTo?si=jFa9C5BObcWaI_r6 | ||
| + | |||
| + | ~Thesmartgreekmathdude | ||
== See also == | == See also == | ||
| − | + | {{AMC12 box|year=2000|num-b=2|num-a=4}} | |
| − | + | {{AMC10 box|year=2000|num-b=2|num-a=4}} | |
| − | |||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 23:36, 14 July 2024
- The following problem is from both the 2000 AMC 12 #3 and 2000 AMC 10 #3, so both problems redirect to this page.
Contents
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 (Algebra)
We can begin by labeling the number of initial jellybeans 
. If she ate of the jellybeans, then 
is remaining. Hence, after day 1, there are:
After day 2, there are:
or 
jellybeans. 
, so 
Solution By: armang32324
Solution 2 (answer choices)
Testing the answers choices out, we see that the answer is 
.
Video Solution by Daily Dose of Math
https://youtu.be/qJYI_qOMyTo?si=jFa9C5BObcWaI_r6
~Thesmartgreekmathdude
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. 
