Difference between revisions of "1951 AHSME Problems/Problem 21"
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== 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{(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 20:31, 10 April 2013
Problem
Given: 
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)
Solution
$\textbf{(A)}\ x \plus{} z > y \plus{} z\implies x>y$ (Error compiling LaTeX. Unknown error_msg), just subtract 
from both sides
$\textbf{(B)}\ x \minus{} z > y \minus{} z\implies x>y$ (Error compiling LaTeX. Unknown error_msg), just add to both sides
, so that means that our desired answer is 
.
As a check:
, we can divide 
safely and without worry because 
.
, similar reasoning as above but instead, multiply .
See Also
| 1951 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
