Difference between revisions of "2007 AMC 8 Problems/Problem 15"
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==Solution 2== | ==Solution 2== | ||
| − | We can test numbers into the inequality we’re given. The simplest is <math>0<1<2<3</math>. We can see that <math>3+1>2</math>, so <math>\boxed{\textbf{(A)} }</math> is correct. | + | We can test numbers into the inequality we’re given. The simplest is <math>0<1<2<3</math>. We can see that <math>3+1>2</math>, so <math>\boxed{\textbf{(A) }a+c<b}</math> is correct. |
—jason.ca | —jason.ca | ||
Revision as of 16:18, 3 November 2023
Contents
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 
Solution 2
We can test numbers into the inequality we’re given. The simplest is 
. We can see that 
, so 
is correct.
—jason.ca
Video Solution by WhyMath
~savannahsolver
Video Solution
https://www.youtube.com/watch?v=_ZHS4M7kpnE
Video Solution 2
https://youtu.be/GxR1giTQeD0 Soo, DRMS, NM
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. 
