2019 AMC 8

Complete problem set with solutions and individual problem pages

Problem 3 Easy

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?

  • A.

    \frac{15}{11}< \frac{17}{13}< \frac{19}{15}

  • B.

    \frac{15}{11}< \frac{19}{15}<\frac{17}{13}

  • C.

    \frac{17}{13}<\frac{19}{15}<\frac{15}{11}

  • D.

    \frac{19}{15}<\frac{15}{11}<\frac{17}{13}

  • E.

    \frac{19}{15}<\frac{17}{13}<\frac{15}{11}

Answer:E

Solution 1

We take a common denominator:

\frac{15}{11},\frac{19}{15}, \frac{17}{13} = \frac{15\cdot 15 \cdot 13}{11\cdot 15 \cdot 13},\frac{19 \cdot 11 \cdot 13}{15\cdot 11 \cdot 13}, \frac{17 \cdot 11 \cdot 15}{13\cdot 11 \cdot 15} = \frac{2925}{2145},\frac{2717}{2145},\frac{2805}{2145}.

Since 2717<2805<2925 it follows that the answer is \boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}.

 

Solution 2

When \frac{x}{y}>1 and z>0, \frac{x+z}{y+z}<\frac{x}{y}. Hence, the answer is \boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}.

Related Problems
Problem 1 2019 AMC 8
Problem 2 2019 AMC 8
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Problem 5 2019 AMC 8
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