Difference between revisions of "1970 AHSME Problems/Problem 17"
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| − | + | == Problem == | |
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| − | + | If <math>r>0</math>, then for all <math>p</math> and <math>q</math> such that <math>pq\ne 0</math> and <math>pr>qr</math>, we have | |
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| − | + | <math>\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}</math> | |
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== Solution == | == Solution == | ||
Latest revision as of 02:37, 17 December 2021
Problem
If 
, then for all 
and 
such that 
and 
, we have
Solution
If and , we can divide by the positive number 
and not change the inequality direction to get 
. Multiplying by 
(and flipping the inequality sign because we're multiplying by a negative number) leads to 
, which directly contradicts 
. Thus, is always false.
If 
(which is possible but not guaranteed), we can divide the true statement by to get 
. This contradicts 
. Thus, is sometimes false, which is bad enough to be eliminated.
If 
, then the condition that 
is satisfied. However, 
and 
, so 
is false for at least this case, eliminating 
.
If 
, then is also satisfied. However, 
, so 
is false, eliminating 
.
All four options do not follow from the premises, leading to 
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. 
