Difference between revisions of "1951 AHSME Problems/Problem 21"
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== Problem == | == Problem == | ||
− | Given: <math> x > 0, y > 0, x > y</math> and <math> z\ | + | Given: <math> x > 0, y > 0, x > y</math> and <math> z\ne 0</math>. The inequality which is not always correct is: |
− | <math> \textbf{(A)}\ x | + | <math> \textbf{(A)}\ x + z > y + z \qquad\textbf{(B)}\ x - z > y - z \qquad\textbf{(C)}\ xz > yz</math> |
<math> \textbf{(D)}\ \frac {x}{z^2} > \frac {y}{z^2} \qquad\textbf{(E)}\ xz^2 > yz^2</math> | <math> \textbf{(D)}\ \frac {x}{z^2} > \frac {y}{z^2} \qquad\textbf{(E)}\ xz^2 > yz^2</math> | ||
== Solution == | == Solution == | ||
− | <math>\textbf{(A)}\ x | + | <math>\textbf{(A)}\ x + z > y + z\implies x>y</math>, just subtract <math>z</math> from both sides |
− | <math>\textbf{(B)}\ x | + | <math>\textbf{(B)}\ x - z > y - z\implies x>y</math>, just add <math>z</math> to both sides |
<math>\textbf{(C)}\ xz > yz\implies x>y\text{ iff }x>0</math>, so that means that our desired answer is <math>\boxed{\textbf{(C)}\ xz > yz}</math>. | <math>\textbf{(C)}\ xz > yz\implies x>y\text{ iff }x>0</math>, so that means that our desired answer is <math>\boxed{\textbf{(C)}\ xz > yz}</math>. |
Revision as of 23:01, 13 March 2015
Problem
Given: and . The inequality which is not always correct is:
Solution
, just subtract from both sides
, just add to both sides
, so that means that our desired answer is .
As a check:
, we can divide safely and without worry because .
, similar reasoning as above but instead, multiply .
See Also
1951 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.