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

(Solution 1)
(Solution 1)
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<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 ==
 
Since Jenny eats <math>20\%</math> of her jelly beans per day, <math>80\%=\frac{4}{5}</math> of her jelly beans remain after one day.
 
 
Let <math>x</math> be the number of jelly beans in the jar originally.
 
 
<cmath>\frac{4}{5}\cdot x=32</cmath>, so
 
 
<cmath>\frac{4}{5}\cdot x=32</cmath>, so
 
 
<cmath>x=\frac{5}{4}\cdot32= 40 \Rightarrow \boxed{A} .</cmath>
 
  
 
== Solution 2 (answer choices) ==
 
== Solution 2 (answer choices) ==

Revision as of 20:59, 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 $20\%$ 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?

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

Solution 2 (answer choices)

Testing the answers choices out, we see that the answer is $\boxed{B}$.

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

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