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> |
+ | |||
+ | The largest number is <math>6</math>, which corresponds to <math>-3a</math>. | ||
+ | |||
+ | <math>\boxed{\text{A}}</math> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/_rGBskmj29M | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
− | [[ | + | {{AJHSME box|year=1985|num-b=7|num-a=9}} |
+ | [[Category:Introductory Algebra Problems]] | ||
+ | |||
+ | |||
+ | {{MAA Notice}} |
Revision as of 09:23, 10 January 2023
Contents
Problem
If , the largest number in the set
is
Solution
Since all the numbers are small, we can just evaluate the set to be
The largest number is , which corresponds to
.
Video Solution
~savannahsolver
See Also
1985 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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.