Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1970 AHSME Problems/Problem 17"

m (→‎See also)
(Undo revision 167905 by Pilar888 (talk) Vandalism act detected ...)
(Tag: Undo)
 
(5 intermediate revisions by 2 users not shown)
Line 10: Line 10:
  
 
== Solution ==
 
== Solution ==
−
<math>\fbox{E}</math>
+
If <math>r>0</math> and <math>pr>qr</math>, we can divide by the positive number <math>r</math> and not change the inequality direction to get <math>p>q</math>.  Multiplying by <math>-1</math> (and flipping the inequality sign because we're multiplying by a negative number) leads to <math>-p < -q</math>, which directly contradicts <math>A</math>.  Thus, <math>A</math> is always false.
 +
 
 +
If <math>p < 0</math> (which is possible but not guaranteed), we can divide the true statement <math>p>q</math> by <math>p</math> to get <math>1 > \frac{q}{p}</math>.  This contradicts <math>D</math>.  Thus, <math>D</math> is sometimes false, which is bad enough to be eliminated.
 +
 
 +
If <math>(p, q, r) = (3, 2, 1)</math>, then the condition that <math>pr > qr</math> is satisfied.  However, <math>-p = -3</math> and <math>q = 2</math>, so <math>-p > q</math> is false for at least this case, eliminating <math>B</math>.
 +
 
 +
If <math>(p, q, r) = (2, -3, 1)</math>, then <math>pr > qr</math> is also satisfied.  However, <math>-\frac{q}{p} = 1.5</math>, so <math>1 > -\frac{q}{p}</math> is false, eliminating <math>C</math>.
 +
 
 +
All four options do not follow from the premises, leading to <math>\fbox{E}</math> as the correct answer.
  
 
== See also ==
 
== See also ==

Latest revision as of 01:37, 17 December 2021

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 $A$. Thus, $A$ is always false.

If $p < 0$ (which is possible but not guaranteed), we can divide the true statement $p>q$ by $p$ to get $1 > \frac{q}{p}$. This contradicts $D$. Thus, $D$ is sometimes false, which is bad enough to be eliminated.

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 for at least this case, 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.

See also

1970 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
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
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1970_AHSME_Problems/Problem_17&oldid=167906"