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

Latest revision as of 02:02, 13 July 2024

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 10: / Line 10: @@
 We have
 <cmath>\begin{align*}
-t &= \frac15\cdot50 + \frac15\cdot20 + \frac15\cdot20 + \frac15\cdot5 + \frac15\cdot5 \\
+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, \\
-s &= \frac{50}{100}\cdot50 + \frac{20}{100}\cdot20 + \frac{20}{100}\cdot20 + \frac{5}{100}\cdot5 + \frac{5}{100}\cdot5 \\
+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.
@@ Line 22: / Line 22: @@
 ~MRENTHUSIASM
-== Solution 2 ==
+==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>.
+==Video Solution by TheBeautyofMath==
+for AMC 10: https://youtu.be/ycRZHCOKTVk?t=789
+for AMC 12: https://youtu.be/wlDlByKI7A8?t=157
-Therefore, the answer is <math>\boxed{\textbf{(B) }-13.5}</math>.
+~IceMatrix
-~Steven Chen (www.professorchenedu.com)
+==Video Solution by WhyMath==
+https://youtu.be/f7vhOCnvl0k
+~savannahsolver
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