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

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.

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 1 (Algebra)

We can begin by labeling the number of initial jellybeans $x$ . If she ate $20\%$ of the jellybeans, then $80\%$ is remaining. Hence, after day 1, there are: $0.8 * x$

After day 2, there are: $0.8 * 0.8 * x$ or $0.64x$ jellybeans. $0.64x = 32$ , so $x = \boxed{(B) 50}$

Solution By: armang32324

Solution 2 (answer choices)

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

Video Solution by Daily Dose of Math

https://youtu.be/qJYI_qOMyTo?si=jFa9C5BObcWaI_r6

~Thesmartgreekmathdude

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.

@@ Line 1: / Line 1: @@
+{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #3]] and [[2000 AMC 10 Problems|2000 AMC 10 #3]]}}
 == 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> \mathrm{(A) \ 40 } \qquad \mathrm{(B) \ 50 } \qquad \mathrm{(C) \ 55 } \qquad \mathrm{(D) \ 60 } \qquad \mathrm{(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 ==
+== Solution 1 (Algebra)==
-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.
+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>
-Let <math>x</math> be the number of jelly beans in the jar originally.
+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>\frac{4}{5}\cdot\frac{4}{5}\cdot x=32</math>
+Solution By: armang32324
-<math>\frac{16}{25}\cdot x=32</math>
+== Solution 2 (answer choices) ==
+Testing the answers choices out, we see that the answer is <math>\boxed{B}</math>.
-<math>\displaystyle x=\frac{25}{16}\cdot32= 50 \Rightarrow B </math>
+==Video Solution by Daily Dose of Math==
+https://youtu.be/qJYI_qOMyTo?si=jFa9C5BObcWaI_r6
+~Thesmartgreekmathdude
 == See also ==
-* [[2000 AMC 12 Problems]]
+{{AMC12 box|year=2000|num-b=2|num-a=4}}
-*[[2000 AMC 12/Problem 2|Previous Problem]]
+{{AMC10 box|year=2000|num-b=2|num-a=4}}
-*[[2000 AMC 12/Problem 4|Next problem]]
 [[Category:Introductory Algebra Problems]]
+{{MAA Notice}}

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