Difference between revisions of "1985 AJHSME Problems/Problem 8"
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<math>\text{(A)}\ -3a \qquad \text{(B)}\ 4a \qquad \text{(C)}\ \frac {24}{a} \qquad \text{(D)}\ a^2 \qquad \text{(E)}\ 1</math> | <math>\text{(A)}\ -3a \qquad \text{(B)}\ 4a \qquad \text{(C)}\ \frac {24}{a} \qquad \text{(D)}\ a^2 \qquad \text{(E)}\ 1</math> | ||
− | ==Solution== | + | ==Solution 1== |
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> | 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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<math>\boxed{\text{A}}</math> | <math>\boxed{\text{A}}</math> | ||
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+ | ==Solution 2== | ||
+ | |||
+ | Since a is negative, some numbers will be positive and some will be negative. The positive numbers are <math>-3a, a^2, 1.</math> We know that <math>a=-2</math> so <math>1<a^2<-3a.</math> Thus our answer is <math>\boxed{\text{A}}.</math> | ||
+ | |||
+ | ~sanaops9 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/_rGBskmj29M | ||
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== | ||
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{{AJHSME box|year=1985|num-b=7|num-a=9}} | {{AJHSME box|year=1985|num-b=7|num-a=9}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
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+ | |||
+ | {{MAA Notice}} |
Latest revision as of 13:32, 1 August 2024
Problem
If , the largest number in the set
is
Solution 1
Since all the numbers are small, we can just evaluate the set to be
The largest number is , which corresponds to
.
Solution 2
Since a is negative, some numbers will be positive and some will be negative. The positive numbers are We know that
so
Thus our answer is
~sanaops9
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.