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>.
  
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>.
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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>\boxed{(C)\frac {1} {2}}</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}}
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{{MAA Notice}}

Latest revision as of 15:27, 8 December 2018

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 $\boxed{(C)\frac {1} {2}}$.

See Also

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

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