Difference between revisions of "1985 AJHSME Problems/Problem 8"

Latest revision as of 12:32, 1 August 2024

Problem

If $a = - 2$ , the largest number in the set $\{ - 3a, 4a, \frac {24}{a}, a^2, 1\}$ is

$\text{(A)}\ -3a \qquad \text{(B)}\ 4a \qquad \text{(C)}\ \frac {24}{a} \qquad \text{(D)}\ a^2 \qquad \text{(E)}\ 1$

Solution 1

Since all the numbers are small, we can just evaluate the set to be $\{ (-3)(-2), 4(-2), \frac{24}{-2}, (-2)^2, 1 \}= \{ 6, -8, -12, 4, 1 \}$

The largest number is $6$ , which corresponds to $-3a$ .

$\boxed{\text{A}}$

Solution 2

Since a is negative, some numbers will be positive and some will be negative. The positive numbers are $-3a, a^2, 1.$ We know that $a=-2$ so $1<a^2<-3a.$ Thus our answer is $\boxed{\text{A}}.$

~sanaops9

Video Solution

https://youtu.be/_rGBskmj29M

~savannahsolver

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.

@@ Line 16: / Line 16: @@
 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==

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