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1985 AJHSME Problems/Problem 8

Revision as of 21:50, 13 January 2009 by 5849206328x (talk | contribs)

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

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 $-3a$.

$\boxed{\text{A}}$

See Also

1985 AJHSME Problems

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