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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...)
 
 
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</math>
 
</math>
  
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== Solution ==
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== Solution 1 ==
  
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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>.
  
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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>.
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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>.
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== Solution 2 (Full Solution) ==
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We write both the numerator and denominator with a denominator of <math>12</math> first, since the LCM of <math>2,3,</math> and <math>4</math> is <math>3\cdot4=12</math>.
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Next, we multiply both the numerator and denominator by <math>12</math>. <math>\dfrac{\frac{1}{3}-\frac{1}{4}}{\frac{1}{2}-\frac{1}{3}}=\dfrac{\frac{4}{12}-\frac{3}{12}}{\frac{6}{12}-\frac{4}{12}}=\dfrac{\frac{1}{12}}{\frac{2}{12}}=\boxed{(C)\dfrac{1}{2}}</math>.
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-sosiaops
  
 
== 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 21:08, 27 December 2020

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 1

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

Solution 2 (Full Solution)

We write both the numerator and denominator with a denominator of $12$ first, since the LCM of $2,3,$ and $4$ is $3\cdot4=12$. Next, we multiply both the numerator and denominator by $12$. $\dfrac{\frac{1}{3}-\frac{1}{4}}{\frac{1}{2}-\frac{1}{3}}=\dfrac{\frac{4}{12}-\frac{3}{12}}{\frac{6}{12}-\frac{4}{12}}=\dfrac{\frac{1}{12}}{\frac{2}{12}}=\boxed{(C)\dfrac{1}{2}}$. -sosiaops

See Also

2009 AMC 10B (Problems • Answer Key • Resources)
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Problem 1 		Followed by
Problem 3
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All AMC 10 Problems and Solutions

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