a, b and c are three fractions such that a < b < c. If c is divided by a, the result is $$\frac{9}{2}$$, which exceeds b by $$\frac{23}{6}$$. The sum of a, b and c is $$\frac{19}{12}$$ , What is the value of (2a + b - c)?

$$\frac{c}{a} = \frac{9}{2}$$

$$c = \frac{9a}{2}$$

b + $$\frac{23}{6} = \frac{9}{2}$$

b = $$\frac{9}{2} - \frac{23}{6} = \frac{2}{3}$$

a + b + c = 19/12

$$a +  \frac{2}{3} + \frac{9a}{2} = \frac{19}{2}$$

$$\frac{11a}{2} = \frac{19}{2} -  \frac{2}{3}$$

$$\frac{11a}{2} = \frac{11}{12}$$

a = $$\frac{1}{6}$$

$$c = \frac{9}{2} \times \frac{1}{6} = \frac{3}{4}$$

 2a + b - c = $$\frac{2}{6} + \frac{2}{3} - \frac{3}{4} = \frac{3}{12} = \frac{1}{4}$$