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

m
m (Solution: removed the word "obviously" where unnecessary.)
Line 7: Line 7:
 
==Solution==
 
==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>
 
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>

Revision as of 20:16, 8 October 2012

Problem

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

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

Solution

$\frac{b}{a}$ 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 $\boxed{\text{C}}$

See Also

1986 AJHSME (ProblemsAnswer KeyResources)
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