Let the fraction be $\frac{x}{y}$

According to the question

Given x+y=11

= x=11-y ---Eq(1)

$$\begin{gathered} = \frac{x+1}{y-2}=\frac{3}{2} ---Eq\left(2\right) \\ Substitute \text{'}x\text{'} in Eq\left(2\right) we get, \\ =\frac{{\displaystyle (11-y)+1}}{{\displaystyle y-2}}=\frac{{\displaystyle 3}}{{\displaystyle 2}} \\ =\frac{{\displaystyle (11-y)+1}}{{\displaystyle y-2}}=\frac{{\displaystyle 3}}{{\displaystyle 2}} \\ =\frac{{\displaystyle 12-y}}{{\displaystyle y-2}}=\frac{{\displaystyle 3}}{{\displaystyle 2}} \end{gathered}$$

= 2(12-y)=3(y-2)

= 24-2y=3y-6

= -2y-3y=-6-24

= -5y= -30

= y=6

Substitute 'y' in Eq(1) we get,

= x=11-(6)

= x=5

∴ (x,y)=(5,6)

Hence, Option B is correct.