Difference between revisions of "1985 AJHSME Problems/Problem 8"
5849206328x (talk | contribs) (New page: ==Problem== If <math>a = - 2</math>, the largest number in the set <math>\{ - 3a, 4a, \frac {24}{a}, a^2, 1\}</math> is <math>\text{(A)}\ -3a \qquad \text{(B)}\ 4a \qquad \text{(C)}\ \fr...) |
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==Solution== | ==Solution== | ||
| − | {{ | + | Since all the numbers are small, we can just evaluate the set to be <cmath>\{ (-3)(-2), 4(-2), \frac{24}{-2}, (-2)^2, 1 \}= \{ 6, -8, -12, 4, 1 \}</cmath> |
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| + | The largest number is <math>6</math>, which corresponds to <math>-3a</math>. | ||
| + | |||
| + | <math>\boxed{\text{A}}</math> | ||
==See Also== | ==See Also== | ||
[[1985 AJHSME Problems]] | [[1985 AJHSME Problems]] | ||
Revision as of 21:50, 13 January 2009
Problem
If 
, the largest number in the set 
is
Solution
Since all the numbers are small, we can just evaluate the set to be ![\[\{ (-3)(-2), 4(-2), \frac{24}{-2}, (-2)^2, 1 \}= \{ 6, -8, -12, 4, 1 \}\]](http://latex.artofproblemsolving.com/5/8/2/5825f2d4723d8be7827adb71aa391982d7276465.png)
The largest number is 
, which corresponds to 
.
