Difference between revisions of "2009 AMC 10B Problems/Problem 2"

(New page: == Problem == Which of the following is equal to <math>\dfrac{\frac{1}{3}-\frac{1}{4}}{\frac{1}{2}-\frac{1}{3}}</math>? <math> \text{(A) } \frac 14 \qquad \text{(B) } \frac 13 \qquad \te...)
 
m (Solution)
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We can multiply both by <math>12</math>, getting <math>\dfrac{4-3}{6-4} = \boxed{\dfrac 12}</math>.
 
We can multiply both by <math>12</math>, getting <math>\dfrac{4-3}{6-4} = \boxed{\dfrac 12}</math>.
  
Alternately, we can directly compute that the numerator is <math>\dfrac 1{12}</math>, the denumerator is <math>\dfrac 16</math>, and hence their ratio is <math>\dfrac 12</math>.
+
Alternately, we can directly compute that the numerator is <math>\dfrac 1{12}</math>, the denominator is <math>\dfrac 16</math>, and hence their ratio is <math>\dfrac 12</math>.
  
 
== See Also ==
 
== See Also ==
  
 
{{AMC10 box|year=2009|ab=B|num-b=1|num-a=3}}
 
{{AMC10 box|year=2009|ab=B|num-b=1|num-a=3}}

Revision as of 15:02, 21 February 2012

Problem

Which of the following is equal to $\dfrac{\frac{1}{3}-\frac{1}{4}}{\frac{1}{2}-\frac{1}{3}}$?

$\text{(A) } \frac 14 \qquad \text{(B) } \frac 13 \qquad \text{(C) } \frac 12 \qquad \text{(D) } \frac 23 \qquad \text{(E) } \frac 34$

Solution

Multiplying the numerator and the denumerator by the same value does not change the value of the fraction. We can multiply both by $12$, getting $\dfrac{4-3}{6-4} = \boxed{\dfrac 12}$.

Alternately, we can directly compute that the numerator is $\dfrac 1{12}$, the denominator is $\dfrac 16$, and hence their ratio is $\dfrac 12$.

See Also

2009 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AMC 10 Problems and Solutions