1985 AJHSME Problems/Problem 8

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 $\boxed{\text{A}}.$

Video Solution

https://youtu.be/_rGBskmj29M

~savannahsolver

1985 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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1985 AJHSME Problems/Problem 8

Contents

Problem

Solution 1

Solution 2

Video Solution

See Also