Difference between revisions of "1970 AHSME Problems/Problem 17"
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== 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 == | ||
| − | {{AHSME box|year=1970|num-b=16|num-a=18}} | + | {{AHSME 35p box|year=1970|num-b=16|num-a=18}} |
[[Category: Introductory Algebra Problems]] | [[Category: Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 01:37, 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. 
