2019 AMC 8 Problems/Problem 3

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 1

each one is 1+4/x so we are really comparing 4/11,4/15, and 4/13 where you can see 4/11>4/13>4/15 so the answer is E

Solution 2

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 $\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$ .

-xMidnightFirex

~ dolphin7 - I took your idea and made it an explanation.

Solution 2

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

This is also similar to Problem 20 on the AMC 2012.

Solution 3

We use our insane mental calculator to find out that $\frac{15}{11} \approx 1.36$ , $\frac{19}{15} \approx 1.27$ , and $\frac{17}{13} \approx 1.31$ . Thus, our answer is $\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$ .

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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All AJHSME/AMC 8 Problems and Solutions

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

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2019 AMC 8 Problems/Problem 3

Contents

Problem 3

Solution 1

Solution 2

Solution 2

Solution 3

See Also