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

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== Solution ==
 
== Solution ==
  
Multiplying the numerator and the denumerator by the same value does not change the value of the fraction.
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Multiplying the numerator and the denominator by the same value does not change the value of the fraction.
 
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>.
  

Revision as of 16:03, 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 denominator 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