1951 AHSME Problems/Problem 21

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

$\textbf{(C)}\ xz > yz\implies x>y\text{ if }x>0$ , 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 $z^2$ .

1951 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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1951 AHSME Problems/Problem 21

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