Difference between revisions of "1986 AJHSME Problems/Problem 14"

Latest revision as of 20:17, 3 July 2013

If $200\leq a \leq 400$ and $600\leq b\leq 1200$ , then the largest value of the quotient $\frac{b}{a}$ is

$\text{(A)}\ \frac{3}{2} \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 6 \qquad \text{(D)}\ 300 \qquad \text{(E)}\ 600$

$\frac{b}{a}$ will be largest if $b$ is the largest it can be, and $a$ is the smallest it can be.

Since $b$ can be no larger than $1200$ , $b = 1200$ . Since $a$ can be no less than $200$ , $a = 200$ . $\frac{1200}{200} = 6$

$6$ is $\boxed{\text{C}}$

1986 AJHSME (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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==
-Obviously, <math>\frac{b}{a}</math> will be largest if <math>b</math> is the largest it can be, and <math>a</math> is the smallest it can be.
+<math>\frac{b}{a}</math> will be largest if <math>b</math> is the largest it can be, and <math>a</math> is the smallest it can be.
 Since <math>b</math> can be no larger than <math>1200</math>, <math>b = 1200</math>. Since <math>a</math> can be no less than <math>200</math>, <math>a = 200</math>. <math>\frac{1200}{200} = 6</math>
-<math>6</math> is <math>\boxed{\text{C}}</math>.
+<math>6</math> is <math>\boxed{\text{C}}</math>
 ==See Also==
-[[1986 AJHSME Problems]]
+{{AJHSME box|year=1986|num-b=13|num-a=15}}
+[[Category:Introductory Algebra Problems]]
+{{MAA Notice}}