Sum of three fractions is $$2\frac{{11}}{{24}}.$$  On dividing the largest fraction by the smallest fraction, $$\frac{7}{6}$$ is obtained which is $$\frac{1}{3}$$ greater than the middle fraction. The smallest fraction is

Solution(By Examveda Team)

Let the three fractions be p, q and r, where p < q < r.
According to the question,
$$\eqalign{ & \frac{r}{p} = \frac{7}{6} \cr & \Rightarrow r = \frac{7}{6}p \cr & {\text{Again, middle fraction}} \cr & = q = \frac{7}{6} - \frac{1}{3} = \frac{{7 - 2}}{6} = \frac{5}{6} \cr & \therefore p + q + r = 2\frac{{11}}{{24}} \cr & \Rightarrow p + \frac{5}{6} + \frac{7}{6}p = \frac{{59}}{{24}} \cr & \Rightarrow p + \frac{{7p}}{6} = \frac{{59}}{{24}} - \frac{5}{6} \cr & \Rightarrow \frac{{6p + 7p}}{6} = \frac{{59 - 20}}{{24}} = \frac{{39}}{{24}} \cr & \Rightarrow 13p = \frac{{39}}{{24}} \times 6 = \frac{{39}}{4} \cr & \Rightarrow p = \frac{{39}}{{4 \times 13}} = \frac{3}{4} \cr} $$