Difference between revisions of "2007 AMC 8 Problems/Problem 15"
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Therefore, the answer is <math>\boxed{\textbf{(A)}\ a+c<b}</math> | Therefore, the answer is <math>\boxed{\textbf{(A)}\ a+c<b}</math> | ||
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| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/UdzJetT-XOY | ||
| + | |||
| + | ~savannahsolver | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2007|num-b=14|num-a=16}} | {{AMC8 box|year=2007|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 21:46, 20 April 2021
Problem
Let 
and 
be numbers with 
. Which of the following is
impossible?
Solution
According to the given rules, every number needs to be positive. Since is always greater than 
, adding a positive number (
) to will always make it greater than .
Therefore, the answer is 
Video Solution by WhyMath
~savannahsolver
See Also
| 2007 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
