2019 AMC 8 Problems/Problem 3

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Problem

3. Which of the following is the correct order of the fractions $\frac{15}{11},\frac{19}{15},$ and $\frac{17}{13},$ from least to greatest?

$\textbf{(A) }\frac{15}{11}< \frac{17}{13}< \frac{19}{15}  \qquad\textbf{(B) }\frac{15}{11}< \frac{19}{15}<\frac{17}{13}    \qquad\textbf{(C) }\frac{17}{13}<\frac{19}{15}<\frac{15}{11}    \qquad\textbf{(D) } \frac{19}{15}<\frac{15}{11}<\frac{17}{13}   \qquad\textbf{(E) }   \frac{19}{15}<\frac{17}{13}<\frac{15}{11}$

Solution

Consider subtracting 1 from each of the fractions. Our new fractions would then be $\frac{4}{11}, \frac{4}{15}, and \frac{4}{13}$. Since $\frac{4}{15}<\frac{4}{13}<\frac{4}{11}$, it follows that the answer is $\boxed{\qquad\textbf{(E) }}   \frac{19}{15}<\frac{17}{13}<\frac{15}{11}$

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