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2013 CEMC Gauss (Grade 8) Problems/Problem 3

The following problem is from both the 2013 CEMC Gauss (Grade 8) #3 and 2013 CEMC Gauss (Grade 7) #7, so both problems redirect to this page.

Problem

The smallest number in the set $\left\{ \frac{1}{2}, \frac{2}{3}, \frac{1}{4}, \frac{5}{6}, \frac{7}{12} \right\}$ is

$\textbf{(A)}\ \frac{1}{2} \qquad\textbf{(B)}\ \frac{2}{3} \qquad\textbf{(C)}\ \frac{1}{4} \qquad\textbf{(D)}\ \frac{5}{6} \qquad\textbf{(E)}\ \frac{7}{12}$

Solution 1

We can convert all of the fractions to a common denominator, and then see which fraction has the smallest denominator.

We notice $12 = 2^2 \times 3$, $6 = 2 \times 3$, $4 = 2^2$, $3 = 3$, and $2 = 2$. Thus, the common denominator is $12$.

$\frac{1}{2} = \frac{1 \times 2 \times 3}{2 \times 2 \times 3} = \frac{6}{12}$

$\frac{2}{3} = \frac{2 \times 2^2}{3 \times 2^2} = \frac{8}{12}$

$\frac{1}{4} = \frac{1 \times 3}{4 \times 3} = \frac{3}{12}$

$\frac{5}{6} = \frac{5 \times 2}{6 \times 2} = \frac{10}{12}$

$\frac{7}{12} = \frac{7 \times 1}{12 \times 1} = \frac{7}{12}$

Of these fractions, the one that had the smallest numerator was $\frac{3}{12}$, which is equivalent to $\boxed {\textbf {(C) } \frac{1}{4}}$.

~anabel.disher

Solution 2

We can notice that except for $\frac{1}{4}$, all of the numerators of the fractions are greater than or equal to $\frac{1}{2}$ of the denominator. $\frac{1}{2} \times 3 = \frac{3}{2} < 2$, $\frac{1}{2} \times 2 = 1 = 1$, $\frac{1}{2} \times 6 = 3 < 5$, and $\frac{1}{2} \times 12 = 6 < 7$

Thus, $\boxed {\textbf {(C) } \frac{1}{4}}$ is the smallest number in the set.

~anabel.disher

Solution 3

$\frac{1}{2} = \frac{2}{2} - \frac{1}{2} = 1 - \frac{1}{2}$

$\frac{2}{3} = \frac{3}{3} - \frac{1}{3} = 1 - \frac{1}{3}$. This must be larger than $\frac{1}{2}$ because $\frac{1}{3}$ has a larger denominator than $\frac{1}{2}$, so a smaller fraction is being subtracted from $1$

$\frac{1}{4}$ has to be smaller than $\frac{1}{2}$ due to having a larger denominator

$\frac{5}{6} = \frac{6}{6} - \frac{1}{6} = 1 - \frac{1}{6}$. This must be larger than $\frac{2}{3}$ using similar logic.

$\frac{7}{12} = \frac{6}{12} + \frac{1}{12} = \frac{1}{2} + \frac{1}{12}$. This must be larger than $\frac{1}{2}$ because $\frac{1}{12} > 0$.

Putting this altogether, we can see that $\boxed {\textbf {(C) } \frac{1}{4}}$ is the smallest number in the set.

~anabel.disher

2013 CEMC Gauss (Grade 8) (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
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CEMC Gauss (Grade 8)
2013 CEMC Gauss (Grade 7) (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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
CEMC Gauss (Grade 7)
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