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1985 AJHSME Problem 8

Revision as of 23:02, 29 June 2023 by Leonidastheconquerer (talk | contribs) (→‎In-depth Solution by Boundless Brain!)

Problem

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

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

In-depth Solution by BoundlessBrain!

https://youtu.be/jo3HHDTqbZQ

Solution

Evaluate each number in the set: \[-3a = -3(-2) = 6\] \[4a = 4(-2) = -8\] \[\frac{24}{a} = \frac{24}{-2} = -12\] \[a^2 = (-2)^2 = 4\]

The largest number in this set is $\boxed{\text{(A) -3a}}.$

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