1970 AHSME Problems/Problem 17

Problem

If $r>0$ , then for all $p$ and $q$ such that $pq\ne 0$ and $pr>qr$ , we have

$\text{(A) } -p>-q\quad \text{(B) } -p>q\quad \text{(C) } 1>-q/p\quad \text{(D) } 1<q/p\quad \text{(E) None of these}$

Solution

If $r>0$ and $pr>qr$ , we can divide by the positive number $r$ and not change the inequality direction to get $p>q$ . Multiplying by $-1$ (and flipping the inequality sign because we're multiplying by a negative number) leads to $-p < -q$ , which directly contradicts $A$ .

If $p < 0$ (which is possible but not guaranteed), we can divide by $p$ to get $1 > \frac{q}{p}$ . This contradicts $D$ .

If $(p, q, r) = (3, 2, 1)$ , then the condition that $pr > qr$ is satisfied. However, $-p = -3$ and $q = 2$ , so $-p > q$ is false, eliminating $B$ .

If $(p, q, r) = (2, -3, 1)$ , then $pr > qr$ is also satisfied. However, $-\frac{q}{p} = 1.5$ < so $1 > -\frac{q}{p}$ is false, eliminating $C$ .

All four options do not follow from the premises, leading to $\fbox{E}$ as the correct answer.

1970 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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1970 AHSME Problems/Problem 17

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