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2023 AMC 12B Problems/Problem 18: Difference between revisions

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Professorchenedu (talk | contribs)
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Created page with "==Solution== Denote by <math>A_i</math> the average of person with initial <math>A</math> in semester <math>i \in \left\{1, 2 \right\}</math> Thus, <math>Y_1 = Z_1 + 3</math>..."
 
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Therefore, the impossible solution is
Therefore, the impossible solution is
\boxed{\textbf{(A) Yolanda's quiz average for the academic year was 22 points higher than Zelda's.}}
<math>\boxed{\textbf{(A) Yolanda's quiz average for the academic year was 22 points higher than Zelda's.}}</math>


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

Revision as of 19:03, 15 November 2023

Solution

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

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

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