Three pipes A, B, C have flow rates of 2 liters, y liters and 3 liters per minute, (2 < y < 3) respectively. The lowest and the highest flow rates of the pipes are decreased by a constant quantity x. If the reciprocals of the flow rates of A, B, C are in arithmetic progression both before and after the change, then x = 

According to 1st condition,

$$\left(\frac{\ 1}{y}-\frac{1\ }{2}\right)=\left(\frac{1\ }{3}-\frac{1\ }{y}\right).$$

or,$$\frac{\ 2}{y}=\frac{1\ }{3}+\frac{1\ }{2}=\frac{5\ }{6}.$$

or,$$y=\frac{12\ }{5}.$$

According to second condition,

$$\frac{1\ }{y}-\frac{1\ }{2-x}=\frac{1\ }{3-x}-\frac{1\ }{y}.$$

or,$$\frac{2\ }{y}=\frac{1\ }{3-x}+\frac{1\ }{2-x}.$$

or,$$\frac{5}{6}=\frac{3-x+2-x}{\left(3-x\right)\left(2-x\right)}=\frac{\left(5-2x\right)}{\left(x^2-5x+6\right)}.$$

or,$$\left(5x^2-25x+30\right)=30-12x.$$

or,$$5x^2=13x.$$

or,$$x=2.6.$$

A is correct choice.