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Difference between revisions of "2021 Fall AMC 10A Problems/Problem 10"

(Created page with "==Problem== A school has <math>100</math> students and <math>5</math> teachers. In the first period, each student is taking one class, and each teacher is teaching one class....")
 
(→‎Solution)
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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
−
<math></math>\begin{align*}
+
<cmath>\begin{align*}
 
t &= \frac15\cdot50 + \frac15\cdot20 + \frac15\cdot20 + \frac15\cdot5 + \frac15\cdot5 \\
 
t &= \frac15\cdot50 + \frac15\cdot20 + \frac15\cdot20 + \frac15\cdot5 + \frac15\cdot5 \\
 
&= \frac15\cdot(50+20+20+5+5) \\
 
&= \frac15\cdot(50+20+20+5+5) \\
Line 16: Line 16:
 
&= 25 + 4 + 4 + 0.25 + 0.25 \\
 
&= 25 + 4 + 4 + 0.25 + 0.25 \\
 
&= 33.5.
 
&= 33.5.
−
\end{align*}
+
\end{align*}</cmath>
 
Therefore, the answer is <math>t-s=\boxed{\textbf{(B)}\  {-}13.5}.</math>
 
Therefore, the answer is <math>t-s=\boxed{\textbf{(B)}\  {-}13.5}.</math>

Revision as of 18:18, 22 November 2021

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 &= \frac15\cdot50 + \frac15\cdot20 + \frac15\cdot20 + \frac15\cdot5 + \frac15\cdot5 \\ &= \frac15\cdot(50+20+20+5+5) \\ &= \frac15\cdot100 \\ &= 20, \\ s &= \frac{50}{100}\cdot50 + \frac{20}{100}\cdot20 + \frac{20}{100}\cdot20 + \frac{5}{100}\cdot5 + \frac{5}{100}\cdot5 \\ &= 25 + 4 + 4 + 0.25 + 0.25 \\ &= 33.5. \end{align*} Therefore, the answer is $t-s=\boxed{\textbf{(B)}\ {-}13.5}.$

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