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Difference between revisions of "2022 AMC 10B Problems/Problem 11"

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==Solution 2 (Elimination)==
 
==Solution 2 (Elimination)==
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Suppose we have five schools: Euclid HS with <math>50</math> students and <math>10</math> T-shirts sold, school <math>A</math> with <math>51</math> students and <math>10</math> T-shirts sold, school <math>B</math> with <math>49</math> students and <math>10</math> T-shirts sold, school <math>C</math> with <math>49</math> students and <math>9</math> T-shirts sold, and school <math>D</math> with <math>51</math> students and <math>9</math> T-shirts sold (this configuration is legal). Then, school <math>B</math> rules out <math>A</math>, school <math>A</math> rules out <math>C</math>, school <math>D</math> rules out <math>D</math>, and school <math>C</math> rules out <math>E</math>, leaving us with <math>B</math> as the correct answer.
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Suppose we have five schools: Euclid HS with <math>50</math> students and <math>10</math> T-shirts sold, school <math>A</math> with <math>51</math> students and <math>10</math> T-shirts sold, school <math>B</math> with <math>49</math> students and <math>10</math> T-shirts sold, school <math>C</math> with <math>49</math> students and <math>9</math> T-shirts sold, and school <math>D</math> with <math>51</math> students and <math>9</math> T-shirts sold (this configuration is legal). Then, school <math>B</math> rules out <math>\textbf{(A)}</math>, school <math>A</math> rules out <math>\textbf{(C)}</math>, school <math>D</math> rules out <math>\textbf{(D)}</math>, and school <math>C</math> rules out <math>\textbf{(E)}</math>, leaving us with <math>\boxed{\textbf{(B)}}</math> as the correct answer.
  
 
~mathboy100
 
~mathboy100

Revision as of 04:56, 18 November 2022

Problem

All the high schools in a large school district are involved in a fundraiser selling T-shirts. Which of the choices below is logically equivalent to the statement "No school bigger than Euclid HS sold more T-shirts than Euclid HS"?

$\textbf{(A) }$ All schools smaller than Euclid HS sold fewer T-shirts than Euclid HS.

$\textbf{(B) }$ No school that sold more T-shirts than Euclid HS is bigger than Euclid HS.

$\textbf{(C) }$ All schools bigger than Euclid HS sold fewer T-shirts than Euclid HS.

$\textbf{(D) }$ All schools that sold fewer T-shirts than Euclid HS are smaller than Euclid HS.

$\textbf{(E) }$ All schools smaller than Euclid HS sold more T-shirts than Euclid HS.

Solution

Let $B$ denote a school that is bigger than Euclid HS, and $M$ denote a school that sold more T-shirts than Euclid HS.

It follows that $\neg B$ denotes a school that is not bigger than Euclid HS, and $\neg M$ denotes a school that did not sell more T-shirts than Euclid HS.

Converting everything to conditional statements (if-then form), the given statement becomes \[B\implies\neg M.\] Its contrapositive is $M\implies\neg B,$ which is $\boxed{\textbf{(B)}}.$

Note that "not bigger than" does not mean "smaller than", and "not selling more" does not mean "selling less". There is an equality case. Therefore, none of the other answer choices is equivalent to $B\implies\neg M.$

~MRENTHUSIASM

Solution 2 (Elimination)

Suppose we have five schools: Euclid HS with $50$ students and $10$ T-shirts sold, school $A$ with $51$ students and $10$ T-shirts sold, school $B$ with $49$ students and $10$ T-shirts sold, school $C$ with $49$ students and $9$ T-shirts sold, and school $D$ with $51$ students and $9$ T-shirts sold (this configuration is legal). Then, school $B$ rules out $\textbf{(A)}$, school $A$ rules out $\textbf{(C)}$, school $D$ rules out $\textbf{(D)}$, and school $C$ rules out $\textbf{(E)}$, leaving us with $\boxed{\textbf{(B)}}$ as the correct answer.

~mathboy100

See Also

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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

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