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Difference between revisions of "2023 AMC 12B Problems/Problem 18"

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<cmath>y_2 = y_1 + 18, \quad z_2 = z_1 + 18</cmath>
 
<cmath>y_2 = y_1 + 18, \quad z_2 = z_1 + 18</cmath>
  
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<cmath> y = \frac{ y_1 n_1 + y_2 n_2 }{ n_1 + n_2} = \frac{ y_1 n_1 + (y_1 + 18) n_2 }{ n_1 + n_2} = y_1 + \frac{18 y_2 }{ n_1 + n_2}</cmath>
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<cmath> y = \frac{ y_1 n_1 + y_2 n_2 }{ n_1 + n_2} = \frac{ y_1 n_1 + (y_1 + 18) n_2 }{ n_1 + n_2} = y_1 + \frac{18 n_2 }{ n_1 + n_2}</cmath>
  
 
<cmath> z = \frac{ z_1 k_1 + z_2 k_2 }{ k_1 + k_2} = \frac{ z_1 k_1 + (z_1 + 18) k_2 }{ k_1 + k_2} = z_1 = \frac{ 18 k_2 }{ k_1 + k_2}</cmath>
 
<cmath> z = \frac{ z_1 k_1 + z_2 k_2 }{ k_1 + k_2} = \frac{ z_1 k_1 + (z_1 + 18) k_2 }{ k_1 + k_2} = z_1 = \frac{ 18 k_2 }{ k_1 + k_2}</cmath>
  
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<cmath>y - z = y_1 + \frac{18 y_2 }{ n_1 + n_2} - z_1  - \frac{ 18 k_2 }{ k_1 + k_2} = y_1 - z_1 + 18 \left( \frac{y_2 }{ n_1 + n_2} - \frac{k_2 }{ k_1 + k_2} \right) = 3 + 18 \left( \frac{y_2 }{ n_1 + n_2} - \frac{k_2 }{ k_1 + k_2} \right)</cmath>
+
<cmath>y - z = y_1 + \frac{18 n_2 }{ n_1 + n_2} - z_1  - \frac{ 18 k_2 }{ k_1 + k_2} = y_1 - z_1 + 18 \left( \frac{n_2 }{ n_1 + n_2} - \frac{k_2 }{ k_1 + k_2} \right) = 3 + 18 \left( \frac{n_2 }{ n_1 + n_2} - \frac{k_2 }{ k_1 + k_2} \right)</cmath>
  
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<math>\frac{y_2 }{ n_1 + n_2}</math> at most is <math>1</math>, <math>\frac{k_2 }{ k_1 + k_2}</math> is at least <math>0</math>, meaning that <math>\frac{y_2 }{ n_1 + n_2} - \frac{k_2 }{ k_1 + k_2}</math> is at most <math>1</math>.
+
<math>\frac{n_2 }{ n_1 + n_2}</math> at most is <math>1</math>, <math>\frac{k_2 }{ k_1 + k_2}</math> is at least <math>0</math>, meaning that <math>\frac{n_2 }{ n_1 + n_2} - \frac{k_2 }{ k_1 + k_2}</math> is at most <math>1</math>.
  
 
Therefore, <cmath>y - z \le 3 + 18 = 21</cmath>
 
Therefore, <cmath>y - z \le 3 + 18 = 21</cmath>
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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
 
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
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==Solution 4==
 
==Solution 4==

Latest revision as of 19:27, 23 May 2024

Problem

Last academic year Yolanda and Zelda took different courses that did not necessarily administer the same number of quizzes during each of the two semesters. Yolanda's average on all the quizzes she took during the first semester was $3$ points higher than Zelda's average on all the quizzes she took during the first semester. Yolanda's average on all the quizzes she took during the second semester was $18$ points higher than her average for the first semester and was again $3$ points higher than Zelda's average on all the quizzes Zelda took during her second semester. Which one of the following statements cannot possibly be true?

$\textbf{(A)}$ Yolanda's quiz average for the academic year was $22$ points higher than Zelda's.

$\textbf{(B)}$ Zelda's quiz average for the academic year was higher than Yolanda's.

$\textbf{(C)}$ Yolanda's quiz average for the academic year was $3$ points higher than Zelda's.

$\textbf{(D)}$ Zelda's quiz average for the academic year equaled Yolanda's.

$\textbf{(E)}$ If Zelda had scored $3$ points higher on each quiz she took, then she would have had the same average for the academic year as Yolanda.

Solution 1

Denote by $A_i$ the average of person with initial $A$ in semester $i \in \left\{1, 2 \right\}$ Thus, $Y_1 = Z_1 + 3$, $Y_2 = Y_1 + 18$, $Y_2 = Z_2 + 3$.

Denote by $A_{12}$ the average of person with initial $A$ in the full year. Thus, $Y_{12}$ can be any number in $\left( Y_1 , Y_2 \right)$ and $Z_{12}$ can be any number in $\left( Z_1 , Z_2 \right)$.

Therefore, the impossible solution is $\boxed{\textbf{(A)}~\text{Yolanda's quiz average for the academic year was 22 points higher than Zelda's.}}$

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Solution 2

We can use process of elimination by finding possible solutions to answer choices. Let $y_1$ and $y_2$ be the number of quizzes Yolanda took in the first and second semesters, respectively. Define $z_1$ and $z_2$ similarly for Zelda.

Answer choice B is satisfied by $(y_1,y_2,z_1,z_2) = (289,1,1,289)$.

Answer choice C and E are both satisfied by $(y_1,y_2,z_1,z_2) = (17,17,17,17)$.

Answer choice D is satisfied by $(y_1,y_2,z_1,z_2) = (7,5,5,7)$.

Therefore the answer is $\boxed{\textbf{(A)}~\text{Yolanda's quiz average for the academic year was 22 points higher than Zelda's.}}$

~cantalon

Solution 3 (Algebra)

Let Yolanda's average for semester $1$ be $y_1$, the number of quizzes Yolanda took in semester $1$ be $n_1$, Zelda's average for semester $1$ be $z_1$, the number of quizzes Zelda took in semester $1$ be $k_1$, Yolanda's average for semester $2$ be $y_2$, the number of quizzes Yolanda took in semester $2$ be $n_2$, Zelda's average for semester $2$ be $z_2$, the number of quizzes Zelda took in semester $2$ be $k_2$, Yolanda's average for the entire year be $y$, Zelda's average for the entire year be $z$.

From the problem we know that

\[y_1 = z_1 + 3, \quad y_2 = z_2 + 3\]

\[y_2 = y_1 + 18, \quad z_2 = z_1 + 18\]

\[y = \frac{ y_1 n_1 + y_2 n_2 }{ n_1 + n_2} = \frac{ y_1 n_1 + (y_1 + 18) n_2 }{ n_1 + n_2} = y_1 + \frac{18 n_2 }{ n_1 + n_2}\]

\[z = \frac{ z_1 k_1 + z_2 k_2 }{ k_1 + k_2} = \frac{ z_1 k_1 + (z_1 + 18) k_2 }{ k_1 + k_2} = z_1 = \frac{ 18 k_2 }{ k_1 + k_2}\]

\[y - z = y_1 + \frac{18 n_2 }{ n_1 + n_2} - z_1 - \frac{ 18 k_2 }{ k_1 + k_2} = y_1 - z_1 + 18 \left( \frac{n_2 }{ n_1 + n_2} - \frac{k_2 }{ k_1 + k_2} \right) = 3 + 18 \left( \frac{n_2 }{ n_1 + n_2} - \frac{k_2 }{ k_1 + k_2} \right)\]

$\frac{n_2 }{ n_1 + n_2}$ at most is $1$, $\frac{k_2 }{ k_1 + k_2}$ is at least $0$, meaning that $\frac{n_2 }{ n_1 + n_2} - \frac{k_2 }{ k_1 + k_2}$ is at most $1$.

Therefore, \[y - z \le 3 + 18 = 21\]

Hence, $\boxed{\textbf{(A)}~\text{Yolanda's quiz average for the academic year was 22 points higher than Zelda's}}$ is not possible.

~isabelchen

Solution 4

Denote $y_1$ and $y_2$ as the quiz averages of Yolanda in the $1$st and $2$nd semesters, respectively. Similarly, denote $z_1$ and $z_2$ as the quiz averages of Zelda in the $1$st and $2$nd semesters.

We have $y_1 = z_1 + 3$, so $y_1 - 3 = z_1$. We also know that $y_2 = 18 +y_1 = 3 + z_2$, implying $z_2 = 15 + y_1$.

The average quiz scores for both students must lie between the averages of each semester, i.e $\mathrm{Avg_{y}}\in[y_1, 18+y_1],$ and $\mathrm{Avg_{z}}\in[y_1-3, y_1+15].$ Since $z_2>z_1$, $y_2 > y_1$, and $y_1>z_1$, we have $\min\{\mathrm{Avg}\} = \min\{\mathrm{Avg}_z\} = z_1$ and $\max\{\mathrm{Avg}\} = \max\{\mathrm{Avg}_y\} = y_2$. Therefore the maximum difference between the two yearly averages is

\[|y_2 - z_1| = |(y_1 + 18)-(y_1-3)| = 21 < 22.\]

Therefore, $\boxed{\textbf{(A)}~\text{Yolanda's quiz average for the academic year was 22 points higher than Zelda's}}$ is not possible.

-Benedict T (countmath1)

Video Solution 1 by OmegaLearn

https://youtu.be/BlJ-ArZ6JEw

Video Solution 2

https://youtu.be/A9ywJSZgD6c

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

See Also

2023 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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

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