Let, in the mid-semester examination, the total score of girls and boys be $g$ and $b$ respectively.
So, $\frac{g}{30} = \frac{b}{20} + 5 \quad \longrightarrow (1)$
Average score of the class $ = \frac{g+b}{50} $
Let, in the final exam, the total score of girls and boys be ${g}’$ and ${b}’$ respectively.
Now, $ \frac{{g}’}{30} = \frac{g}{30} – 3 \Rightarrow \boxed{ {g}’ = g-90}$
And, $\frac{{g}’+{b}'}{50} = \frac{g+b}{50} + 2 $
$\Rightarrow \frac{{g}’+{b}'}{50} = \frac{g+b+100}{50} $
$\Rightarrow {g}’+{b}' = g+b+100 $
$\Rightarrow g-90+{b}' = g+b+100 $
$\Rightarrow \boxed { {b}' = b+190 } $
So, the new average of boys $= \frac{{b}'}{20} = \frac{b+190}{20} = \frac{b}{20} + \frac{190}{20} = \frac{b}{20} + 9.5 $
$\therefore$ The increase in the average score of the boys is $9.5.$
Correct Answer $ : \text{A}$