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Difference between revisions of "2011 CEMC Gauss (Grade 8) Problems/Problem 4"

(Created page with "==Problem== The largest number in the list <math>\{\frac{3}{10}, \, \frac{9}{20}, \, \frac{12}{25}, \frac{27}{50}, \, \frac{49}{100}\}</math> is <math> \text{ (A) }\ \frac{3}...")
 
 
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~anabel.disher
 
~anabel.disher
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{{CEMC box|year=2011|competition=Gauss (Grade 8)|num-b=3|num-a=5}}

Latest revision as of 15:00, 19 October 2025

Problem

The largest number in the list $\{\frac{3}{10}, \, \frac{9}{20}, \, \frac{12}{25}, \frac{27}{50}, \, \frac{49}{100}\}$ is

$\text{ (A) }\ \frac{3}{10} \qquad\text{ (B) }\ \frac{9}{20} \qquad\text{ (C) }\ \frac{12}{25} \qquad\text{ (D) }\ \frac{27}{50} \qquad\text{ (E) }\ \frac{49}{100}$

Solution 1

Expressing the elements of the set as decimals gives:

$\{0.3, \, 0.45, \, 0.48, \, 0.54, \, 0.49\}$

The largest of these is $0.54$, which is equal to $\boxed {\textbf {(D) } \frac{27}{50}}$.

~anabel.disher

Solution 2

To compare the fractions, we can also express the fractions in terms of the same common denominator, and then see which number has the highest numerator. This gives:

$\frac{3}{10} = \frac{3 \times 10}{10 \times 10} = \frac{30}{100}$

$\frac{9}{20} = \frac{9 \times 5}{20 \times 5} = \frac{45}{100}$

$\frac{12}{25} = \frac{12 \times 4}{25 \times 4} = \frac{48}{100}$

$\frac{27}{50} = \frac{27 \times 2}{50 \times 2} = \frac{54}{100}$

$\frac{49}{100} = \frac{49 \times 1}{100 \times 1} = \frac{49}{100}$

Since the denominators are the same, we can look at the fraction with the highest numerator, which is $\frac{54}{100} = \boxed {\textbf {(D) } \frac{27}{50}}$.

~anabel.disher

2011 CEMC Gauss (Grade 8) (Problems • Answer Key • Resources)
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CEMC Gauss (Grade 8)
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