Difference between revisions of "2000 AMC 12 Problems/Problem 3"
(→Solution 1) |
|||
| (8 intermediate revisions by 5 users not shown) | |||
| Line 5: | Line 5: | ||
<math> \textbf{(A)} \ 40 \qquad \textbf{(B)} \ 50 \qquad \textbf{(C)} \ 55 \qquad \textbf{(D)} \ 60 \qquad \textbf{(E)} \ 75 </math> | <math> \textbf{(A)} \ 40 \qquad \textbf{(B)} \ 50 \qquad \textbf{(C)} \ 55 \qquad \textbf{(D)} \ 60 \qquad \textbf{(E)} \ 75 </math> | ||
| − | == Solution 1 == | + | == 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 | |
| − | < | + | == Solution 2 (answer choices) == |
| + | Testing the answers choices out, we see that the answer is <math>\boxed{B}</math>. | ||
| + | == Solution 3 == | ||
| + | We can work backwards to figure out what the starting amount is. If she ate <math>20\%</math> of the jellybeans, <math>80\% = \frac{4}{5}</math> of the jellybeans were remaining at the end of each day. This means that we can multiply by <math>\frac{1}{80\%} = \frac{5}{4}</math> to get the amount of jellybeans that she had at the end of the previous day. So we have: | ||
| + | |||
| + | <math>32 \cdot \frac{5}{4} \cdot \frac{5}{4} = 40 \cdot \frac{5}{4} = \boxed{(B) 50}</math> | ||
| + | |||
| + | ~anabel.disher | ||
| + | ==Video Solution by Daily Dose of Math== | ||
| − | + | https://youtu.be/qJYI_qOMyTo?si=jFa9C5BObcWaI_r6 | |
| − | + | ~Thesmartgreekmathdude | |
| − | |||
== See also == | == See also == | ||
Latest revision as of 12:40, 11 April 2025
- 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 
.
Solution 3
We can work backwards to figure out what the starting amount is. If she ate of the jellybeans, 
of the jellybeans were remaining at the end of each day. This means that we can multiply by 
to get the amount of jellybeans that she had at the end of the previous day. So we have:
~anabel.disher
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
