Difference between revisions of "1985 AJHSME Problem 8"

(Solution)
(In-depth Solution by Boundless Brain!)
 
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<math>\text{(A)}\ -3a \qquad \text{(B)}\ 4a \qquad \text{(C)}\ \frac {24}{a} \qquad \text{(D)}\ a^2 \qquad \text{(E)}\ 1</math>
 
<math>\text{(A)}\ -3a \qquad \text{(B)}\ 4a \qquad \text{(C)}\ \frac {24}{a} \qquad \text{(D)}\ a^2 \qquad \text{(E)}\ 1</math>
  
== In-depth Solution by Boundless Brain!==
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== In-depth Solution by BoundlessBrain!==
 
https://youtu.be/jo3HHDTqbZQ
 
https://youtu.be/jo3HHDTqbZQ
  

Latest revision as of 23:02, 29 June 2023

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 $\boxed{\text{(A) -3a}}.$