Difference between revisions of "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 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 $\boxed{\textbf{(E)}\frac{19}{15}<\frac{17}{13}<\frac{15}{11}}$ .

-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 (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.

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 $\boxed{\textbf{(E)}}$ .

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 $\frac{4}{19}$ . These are easy to order because the numerators are the same, $\frac{4}{11}>\frac{4}{13}>\frac{4}{15}$ . Because it is a subtraction by a constant, in order to order them, we keep the inequality signs to get $\boxed{\text{(E)}\;\frac{15}{11}>\frac{17}{13}>\frac{19}{15}}.$

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

The butterfly method is a method when you multiply the denominator of the second fraction and multiply it by the numerator from the first fraction.

@@ Line 27: / Line 27: @@
 ==Solution 4==
 Suppose each fraction is expressed with denominator <math>2145</math>: <math>\frac{2925}{2145}, \frac{2717}{2145}, \frac{2805}{2145}</math>. Clearly <math>2717<2805<2925</math> so the answer is <math>\boxed{\textbf{(E)}}</math>.
+==Solution 5 -SweetMango77==
+We notice that each of these fraction's numerator <math>-</math> denominator <math>=4</math>. If we take each of the fractions, and subtract <math>1</math> from each, we get <math>\frac{4}{11}, \frac{4}{15}</math>, and <math>\frac{4}{19}</math>. These are easy to order because the numerators are the same, <math>\frac{4}{11}>\frac{4}{13}>\frac{4}{15}</math>. Because it is a subtraction by a constant, in order to order them, we keep the inequality signs to get <math>\boxed{\text{(E)}\;\frac{15}{11}>\frac{17}{13}>\frac{19}{15}}.</math>
 ==See also==

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