Difference between revisions of "2021 Fall AMC 12A Problems/Problem 7"

(Solution 2)
(Video Solution by TheBeautyofMath)
 
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<math>\textbf{(A)}\ {-}18.5  \qquad\textbf{(B)}\ {-}13.5 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 13.5 \qquad\textbf{(E)}\ 18.5</math>
 
<math>\textbf{(A)}\ {-}18.5  \qquad\textbf{(B)}\ {-}13.5 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 13.5 \qquad\textbf{(E)}\ 18.5</math>
  
==Solution 1==
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==Solution==
 
The formula for expected values is <cmath>\text{Expected Value}=\sum(\text{Outcome}\cdot\text{Probability}).</cmath>
 
The formula for expected values is <cmath>\text{Expected Value}=\sum(\text{Outcome}\cdot\text{Probability}).</cmath>
 
We have
 
We have
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
t &= \frac15\cdot50 + \frac15\cdot20 + \frac15\cdot20 + \frac15\cdot5 + \frac15\cdot5 \\
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t &= 50\cdot\frac15 + 20\cdot\frac15 + 20\cdot\frac15 + 5\cdot\frac15 + 5\cdot\frac15 \\
&= \frac15\cdot(50+20+20+5+5) \\
+
&= (50+20+20+5+5)\cdot\frac15 \\
&= \frac15\cdot100 \\
+
&= 100\cdot\frac15 \\
 
&= 20, \\
 
&= 20, \\
s &= \frac{50}{100}\cdot50 + \frac{20}{100}\cdot20 + \frac{20}{100}\cdot20 + \frac{5}{100}\cdot5 + \frac{5}{100}\cdot5 \\
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s &= 50\cdot\frac{50}{100} + 20\cdot\frac{20}{100} + 20\cdot\frac{20}{100} + 5\cdot\frac{5}{100} + 5\cdot\frac{5}{100} \\
 
&= 25 + 4 + 4 + 0.25 + 0.25 \\
 
&= 25 + 4 + 4 + 0.25 + 0.25 \\
 
&= 33.5.
 
&= 33.5.
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~MRENTHUSIASM
 
~MRENTHUSIASM
  
== Solution 2 ==
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==Video Solution (Simple and Quick)==
First,
+
https://youtu.be/nZBIoUS0mdk
<cmath>
 
\begin{align*}
 
t & = \frac{50 + 20 + 20 + 5 + 5}{5} \\
 
& = 20 .
 
\end{align*}
 
</cmath>
 
  
 +
~Education, the Study of Everything
  
Second, we compute <math>s</math>.
 
We have
 
<cmath>
 
\begin{align*}
 
s & = \frac{50}{100} \cdot 50 + \frac{20}{100} \cdot 20 + \frac{20}{100} \cdot 20
 
+ \frac{5}{100} \cdot 5 + \frac{5}{100} \cdot 5 \\
 
& = 33.5 .
 
\end{align*}
 
</cmath>
 
  
Therefore, <math>t - s = 20 - 33.5 = - 13.5</math>.
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==Video Solution by TheBeautyofMath==
 +
for AMC 10: https://youtu.be/ycRZHCOKTVk?t=789
 +
 
 +
for AMC 12: https://youtu.be/wlDlByKI7A8?t=157
  
 +
~IceMatrix
  
Therefore, the answer is <math>\boxed{\textbf{(B) }-13.5}</math>.
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==Video Solution by WhyMath==
 +
https://youtu.be/f7vhOCnvl0k
  
~Steven Chen (www.professorchenedu.com)
+
~savannahsolver
  
 
==See Also==
 
==See Also==

Latest revision as of 14:32, 13 June 2023

The following problem is from both the 2021 Fall AMC 10A #10 and 2021 Fall AMC 12A #7, so both problems redirect to this page.

Problem

A school has $100$ students and $5$ teachers. In the first period, each student is taking one class, and each teacher is teaching one class. The enrollments in the classes are $50, 20, 20, 5,$ and $5$. Let $t$ be the average value obtained if a teacher is picked at random and the number of students in their class is noted. Let $s$ be the average value obtained if a student was picked at random and the number of students in their class, including the student, is noted. What is $t-s$?

$\textbf{(A)}\ {-}18.5  \qquad\textbf{(B)}\ {-}13.5 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 13.5 \qquad\textbf{(E)}\ 18.5$

Solution

The formula for expected values is \[\text{Expected Value}=\sum(\text{Outcome}\cdot\text{Probability}).\] We have \begin{align*} t &= 50\cdot\frac15 + 20\cdot\frac15 + 20\cdot\frac15 + 5\cdot\frac15 + 5\cdot\frac15 \\ &= (50+20+20+5+5)\cdot\frac15 \\ &= 100\cdot\frac15 \\ &= 20, \\ s &= 50\cdot\frac{50}{100} + 20\cdot\frac{20}{100} + 20\cdot\frac{20}{100} + 5\cdot\frac{5}{100} + 5\cdot\frac{5}{100} \\ &= 25 + 4 + 4 + 0.25 + 0.25 \\ &= 33.5. \end{align*} Therefore, the answer is $t-s=\boxed{\textbf{(B)}\ {-}13.5}.$

~MRENTHUSIASM

Video Solution (Simple and Quick)

https://youtu.be/nZBIoUS0mdk

~Education, the Study of Everything


Video Solution by TheBeautyofMath

for AMC 10: https://youtu.be/ycRZHCOKTVk?t=789

for AMC 12: https://youtu.be/wlDlByKI7A8?t=157

~IceMatrix

Video Solution by WhyMath

https://youtu.be/f7vhOCnvl0k

~savannahsolver

See Also

2021 Fall AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
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
2021 Fall AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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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