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

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== Solution ==  
 
== Solution ==  
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<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{(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{(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>.
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As a check:
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<math>\textbf{(D)}\ \frac {x}{z^2} > \frac {y}{z^2}\implies x>y</math>, we can divide <math>z^2</math> safely and without worry because <math>z^2>0</math>.
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<math>\textbf{(E)}\ xz^2 > yz^2\implies x>y</math>, similar reasoning as above but instead, multiply <math>z^2</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 21:31, 10 April 2013

Problem

Given: $x > 0, y > 0, x > y$ and $z\not \equal{} 0$ (Error compiling LaTeX. Unknown error_msg). The inequality which is not always correct is:

$\textbf{(A)}\ x \plus{} z > y \plus{} z \qquad\textbf{(B)}\ x \minus{} z > y \minus{} z \qquad\textbf{(C)}\ xz > yz$ (Error compiling LaTeX. Unknown error_msg) $\textbf{(D)}\ \frac {x}{z^2} > \frac {y}{z^2} \qquad\textbf{(E)}\ xz^2 > yz^2$

Solution

$\textbf{(A)}\ x \plus{} z > y \plus{} z\implies x>y$ (Error compiling LaTeX. Unknown error_msg), just subtract $z$ from both sides

$\textbf{(B)}\ x \minus{} z > y \minus{} z\implies x>y$ (Error compiling LaTeX. Unknown error_msg), 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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