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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>
+
<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
+
<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
+
<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 22:01, 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 (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

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