# 2009 AMC 10B Problems/Problem 2

## 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 