Difference between revisions of "2000 AMC 12 Problems/Problem 3"
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After day 2, there are: | 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> | <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> | ||
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+ | Solution By: armang32324 | ||
== Solution 2 (answer choices) == | == Solution 2 (answer choices) == | ||
Testing the answers choices out, we see that the answer is <math>\boxed{B}</math>. | 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 == |
Latest revision as of 00:36, 15 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.