Difference between revisions of "1985 AJHSME Problems/Problem 8"

m
m (See Also)
Line 17: Line 17:
 
{{AJHSME box|year=1985|num-b=7|num-a=9}}
 
{{AJHSME box|year=1985|num-b=7|num-a=9}}
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 +
 +
 +
{{MAA Notice}}

Revision as of 16:04, 3 July 2013

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 (ProblemsAnswer KeyResources)
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
All AJHSME/AMC 8 Problems and Solutions


The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png