Difference between revisions of "1951 AHSME Problems/Problem 21"

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== Problem ==
 
== Problem ==
Given: <math> x > 0, y > 0, x > y</math> and <math> z\not \equal{} 0</math>. The inequality which is not always correct is:
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Given: <math> x > 0, y > 0, x > y</math> and <math> z\ne 0</math>. The inequality which is not always correct is:
  
<math> \textbf{(A)}\ x \plus{} z > y \plus{} z \qquad\textbf{(B)}\ x \minus{} z > y \minus{} z \qquad\textbf{(C)}\ xz > yz</math>
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<math> \textbf{(A)}\ x + z > y + z \qquad\textbf{(B)}\ x - z > y - z \qquad\textbf{(C)}\ xz > yz</math>
 
<math> \textbf{(D)}\ \frac {x}{z^2} > \frac {y}{z^2} \qquad\textbf{(E)}\ xz^2 > yz^2</math>
 
<math> \textbf{(D)}\ \frac {x}{z^2} > \frac {y}{z^2} \qquad\textbf{(E)}\ xz^2 > yz^2</math>
  
 
== Solution ==  
 
== Solution ==  
<math>\textbf{(A)}\ x \plus{} z > y \plus{} z\implies x>y</math>, just subtract <math>z</math> from both sides
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<math>\textbf{(A)}\ x + z > y + z\implies x>y</math>, just subtract <math>z</math> from both sides
  
<math>\textbf{(B)}\ x \minus{} z > y \minus{} z\implies x>y</math>, just add <math>z</math> to both sides
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<math>\textbf{(B)}\ x - z > y - z\implies x>y</math>, just add <math>z</math> to both sides
  
 
<math>\textbf{(C)}\ xz > yz\implies x>y\text{ iff }x>0</math>, so that means that our desired answer is <math>\boxed{\textbf{(C)}\ xz > yz}</math>.  
 
<math>\textbf{(C)}\ xz > yz\implies x>y\text{ iff }x>0</math>, so that means that our desired answer is <math>\boxed{\textbf{(C)}\ xz > yz}</math>.  

Revision as of 23:01, 13 March 2015

Problem

Given: $x > 0, y > 0, x > y$ and $z\ne 0$. The inequality which is not always correct is:

$\textbf{(A)}\ x + z > y + z \qquad\textbf{(B)}\ x - z > y - z \qquad\textbf{(C)}\ xz > yz$ $\textbf{(D)}\ \frac {x}{z^2} > \frac {y}{z^2} \qquad\textbf{(E)}\ xz^2 > yz^2$

Solution

$\textbf{(A)}\ x + z > y + z\implies x>y$, just subtract $z$ from both sides

$\textbf{(B)}\ x - z > y - z\implies x>y$, just add $z$ to both sides

$\textbf{(C)}\ xz > yz\implies x>y\text{ iff }x>0$, so that means that our desired answer is $\boxed{\textbf{(C)}\ xz > yz}$.

As a check:

$\textbf{(D)}\ \frac {x}{z^2} > \frac {y}{z^2}\implies x>y$, we can divide $z^2$ safely and without worry because $z^2>0$.

$\textbf{(E)}\ xz^2 > yz^2\implies x>y$, similar reasoning as above but instead, multiply $z^2$.

See Also

1951 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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All AHSME Problems and Solutions

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