Question

n7.3, & n7.4 use benchmarks and a number line to order this set of numbers from least to greatest. a) \\(\frac{11}{12}\\), \\(\frac{1}{3}\\), \\(\frac{7}{6}\\), \\(\frac{5}{4}\\) from least to greatest: \\(\frac{1}{3}\\) \\(\frac{11}{12}\\) \\(\frac{7}{6}\\) \\(\frac{5}{4}\\)

Explanation:

Step1: Find a common denominator

The fractions are $\frac{1}{3}$, $\frac{11}{12}$, $\frac{7}{6}$, $\frac{5}{4}$. The least common denominator (LCD) of 3, 12, 6, and 4 is 12.

Step2: Convert each fraction to twelfths

• $\frac{1}{3} = \frac{1\times4}{3\times4} = \frac{4}{12}$
• $\frac{11}{12}$ remains $\frac{11}{12}$
• $\frac{7}{6} = \frac{7\times2}{6\times2} = \frac{14}{12}$
• $\frac{5}{4} = \frac{5\times3}{4\times3} = \frac{15}{12}$

Step3: Compare the numerators

Now we have $\frac{4}{12}$, $\frac{11}{12}$, $\frac{14}{12}$, $\frac{15}{12}$. Since $4 < 11 < 14 < 15$, the original fractions in order from least to greatest are $\frac{1}{3}$, $\frac{11}{12}$, $\frac{7}{6}$, $\frac{5}{4}$.

Answer:

$\frac{1}{3}$, $\frac{11}{12}$, $\frac{7}{6}$, $\frac{5}{4}$