Question

Arrange the fractions given below in order from least to greatest (Ascending order)

$1\frac{6}{18},2\frac{5}{6},2\frac{1}{3},3\frac{2}{6}$

Answer 1

The improper fraction of $1\frac{6}{18},2\frac{5}{6},2\frac{1}{3},3\frac{2}{6}$is

$$\begin{gathered} 1\frac{6}{18}=\frac{18+6}{18}=\frac{24}{18}=\frac{8}{6} \\ 2\frac{5}{6}=\frac{12+5}{6}=\frac{17}{6} \\ 2\frac{1}{3}=\frac{6+1}{3}=\frac{7}{3} \\ 3\frac{2}{6}=\frac{18+2}{6}=\frac{20}{6} \end{gathered}$$

But $2\frac{1}{3}=\frac{6+1}{3}=\frac{7}{3}$, to make equivalent other fraction multiply and divide by $2$.

$\Rightarrow 2\frac{1}{3}=\frac{6+1}{3}=\frac{7}{3}\times \frac{2}{2}=\frac{14}{6}$

Hence, the improper fraction of $1\frac{6}{18},2\frac{5}{6},2\frac{1}{3},3\frac{2}{6}$ is $\frac{8}{6},\frac{17}{6},\frac{14}{6},\frac{20}{6}$

Since, the denominator of given fraction are same

$\therefore$The ascending order of $1\frac{6}{18},2\frac{5}{6},2\frac{1}{3},3\frac{2}{6}$is $\frac{8}{6},\frac{14}{6},\frac{17}{6},\frac{20}{6}$