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1985 AJHSME Problems/Problem 8

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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 • 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
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