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

Latest revision as of 09:23, 10 January 2023

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

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

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 $-3a$ .

$\boxed{\text{A}}$

~savannahsolver

1985 AJHSME (Problems • Answer Key • 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
All AJHSME/AMC 8 Problems and Solutions

@@ Line 7: / Line 7: @@
 ==Solution==
-{{Solution}}
+Since all the numbers are small, we can just evaluate the set to be <cmath>\{ (-3)(-2), 4(-2), \frac{24}{-2}, (-2)^2, 1 \}= \{ 6, -8, -12, 4, 1 \}</cmath>
+The largest number is <math>6</math>, which corresponds to <math>-3a</math>.
+<math>\boxed{\text{A}}</math>
+==Video Solution==
+https://youtu.be/_rGBskmj29M
+~savannahsolver
 ==See Also==
-[[1985 AJHSME Problems]]
+{{AJHSME box|year=1985|num-b=7|num-a=9}}
+[[Category:Introductory Algebra Problems]]
+{{MAA Notice}}