Question

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

Explanation:

Step1: Find common denominator

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

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

Step2: Compare numerators

Now we have $\frac{4}{12}$ (from $\frac{1}{3}$), $\frac{11}{12}$, $\frac{14}{12}$ (from $\frac{7}{6}$), $\frac{15}{12}$ (from $\frac{5}{4}$). Since $4 < 11 < 14 < 15$, the order of the original fractions is $\frac{1}{3} < \frac{11}{12} < \frac{7}{6} < \frac{5}{4}$.

Step3: Verify with number line

The number line has marks from 0 to $1\frac{1}{2}$ (18/12). Plotting the fractions (as twelfths): $\frac{4}{12}$ (close to 0), $\frac{11}{12}$ (near 1), $\frac{14}{12}$ (1 and 2/12), $\frac{15}{12}$ (1 and 3/12). This confirms the order.

Answer:

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