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 (Bashing)

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.

- Clearness by doulai1

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 2012 AMC 8.

Solution 3 (probably won't use this solution)

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

~~ by an insane math guy. ~~ random text that is here to distract you.

Solution 4

Suppose each fraction is expressed with denominator $2145$ : $\frac{2925}{2145}, \frac{2717}{2145}, \frac{2805}{2145}$ . Clearly $2717<2805<2925$ so the answer is $\boxed{\textbf{(E)}}$ .

Note: Duplicate of Solution 1

Solution 5 -SweetMango77

We notice that each of these fraction's numerator $-$ denominator $=4$ . If we take each of the fractions, and subtract $1$ from each, we get $\frac{4}{11}$ , $\frac{4}{15}$ , and $\frac{4}{13}$ . These are easy to order because the numerators are the same, we get $\frac{4}{15}<\frac{4}{13}<\frac{4}{11}$ . Because it is a subtraction by a constant, in order to order them, we keep the inequality signs to get $\boxed{\textbf{(E)}\;\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$ .

Solution 6

Adding on to Solution 5, we can turn each of the fractions $\frac{15}{11}$ , $\frac{17}{13}$ , and $\frac{19}{15}$ into $1$ $\frac{4}{11}$ , $1$ $\frac{4}{13}$ , and $1$ $\frac{4}{15}$ , respectively. We now subtract $1$ from each to get $\frac{4}{11}$ , $\frac{4}{15}$ , and $\frac{4}{13}$ . Since their numerators are all 4, this is easy because we know that $\frac{1}{15}<\frac{1}{13}<\frac{1}{11}$ and therefore $\frac{4}{15}<\frac{4}{13}<\frac{4}{11}$ . Reverting them back to their original fractions, we can now see that the answer is $\boxed{\textbf{(E)}\;\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$ .

~by ChipmunkT

Solution 7

Since each numerator is more than denominator bybthebsame amount ( $N/(N-4$ ), the pairwise cross products comparisons are all $(N)(M-4)$ < $(N-4)(M) \equiv N > M$ . (Intuitively, the fraction is closer to 1 when the numerator and denominator are larger, and closer to infinity when they are smaller.)

Thus, the one with the smaller numerator makes a larger fraction: $\boxed{\textbf{(E)}\;\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$ .

Video Solution

The Learning Royal: https://youtu.be/IiFFDDITE6Q

Video Solution 2

Solution detailing how to solve the problem: https://www.youtube.com/watch?v=q27qEcr7TbQ&list=PLbhMrFqoXXwmwbk2CWeYOYPRbGtmdPUhL&index=4

Video Solution 3

https://youtu.be/xNrhMAbaEcI

~savannahsolver

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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