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

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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==
==Solution 1==
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
Line 21: Line 21:


~MRENTHUSIASM
~MRENTHUSIASM
== Solution 2 ==
First, <math>t = \frac{50 + 20 + 20 + 5 + 5}{5} = 20</math>.
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>.
Therefore, the answer is <math>\boxed{\textbf{(B) }-13.5}</math>.
~Steven Chen (www.professorchenedu.com)


==See Also==
==See Also==

Revision as of 20:05, 25 November 2021

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 1

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 $t-s=\boxed{\textbf{(B)}\ {-}13.5}.$

~MRENTHUSIASM

Solution 2

First, $t = \frac{50 + 20 + 20 + 5 + 5}{5} = 20$.

Second, we compute $s$. We have \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*}

Therefore, $t - s = 20 - 33.5 = - 13.5$.


Therefore, the answer is $\boxed{\textbf{(B) }-13.5}$.

~Steven Chen (www.professorchenedu.com)

See Also

2021 Fall AMC 12A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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