Assuming, its usual speed is $x \hspace{0.1cm} km/hr $
∴ New Speed will be $= x+ 33.33\% \hspace{0.1cm} of \hspace{0.1cm} x \\ = x+ \dfrac{1}{3} \text{ of x } \\ = \dfrac{4x}{3} \hspace{0.1cm} km/hr$
In $x \hspace{0.1cm} km/hr$ speed $1500 \hspace{0.1cm} km$ can be covered in $\dfrac{1500}{x} \hspace{0.1cm} hr$ $\qquad \left [ \because Distance = Speed \times Time \\ \qquad \qquad Or, Time = \dfrac{Distance}{Speed}\right ]$
Now In $\dfrac{4x}{3} km/hr$ speed $1500 \hspace{0.1cm} km$ can be covered in $\dfrac{1500}{\dfrac{4x}{3}} = \dfrac{1500 \times 3}{4x} \hspace{0.1cm} hr$
As the plane lates for half an hour, the difference between the usual time taken by the plane and time taken by plane in new speed will be half an hour.
$∴\dfrac{1500}{x} - \dfrac{1500 \times 3}{4x} = \dfrac{1}{2}$
Or, $\dfrac{(1500 \times 4) - (1500 \times 3)}{4x} = \dfrac{1}{2}$
Or. $\dfrac{6000 - 4500}{4x} = \dfrac{1}{2}$
Or, $\dfrac{1500}{4x} = \dfrac{1}{2}$
Or, $4x = 1500 \times 2$
Or, $x = \dfrac{3000}{4} = 750\hspace{0.1cm} km/hr$
So, the plane's usual speed is $750\hspace{0.1cm} km/hr$
∴ New speed will be $\dfrac{4x}{3}$ i.e. $\dfrac{4 \times 750}{3} = 1000 \hspace{0.1cm} km/hr$
∴ The plane's usual speed is $750 \hspace{0.1cm} km/hr$ & New speed is $1000 \hspace{0.1cm} km/hr$ which means the plane increases its speed by $250 \hspace{0.1cm} km/hr$ than its ususal speed to make up the timing .