Difference between revisions of "1985 AJHSME Problem 8"
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== Problem == | == Problem == | ||
| − | If <math>a = - 2</math>, the largest number in the set <math> - 3a, 4a, \frac {24}{a}, a^2, 1</math> is | + | If <math>a = - 2</math>, the largest number in the set <math>\{ - 3a, 4a, \frac {24}{a}, a^2, 1 \}</math> is |
<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 | + | == In-depth Solution by BoundlessBrain!== |
https://youtu.be/jo3HHDTqbZQ | https://youtu.be/jo3HHDTqbZQ | ||
Latest revision as of 23:12, 15 June 2025
Problem
If 
, the largest number in the set 
is
In-depth Solution by BoundlessBrain!
Solution
Evaluate each number in the set:
![\[-3a = -3(-2) = 6\]](http://latex.artofproblemsolving.com/7/5/9/7595b1ed75a2b420cccc17ddf2882ba521ae48f7.png)
![\[4a = 4(-2) = -8\]](http://latex.artofproblemsolving.com/4/b/3/4b32955e5211a53280b1bf580dccf9ece43f70d6.png)
![\[\frac{24}{a} = \frac{24}{-2} = -12\]](http://latex.artofproblemsolving.com/4/3/7/4378dbd0db714c1a8232dd36fad386efe52f15af.png)
![\[a^2 = (-2)^2 = 4\]](http://latex.artofproblemsolving.com/3/7/9/37985bc2198b84808cdeab772ff864b3700e9b94.png)
The largest number in this set is 
