An aeroplane covers 2500km, 1200km and 500km at the rate of 500km/hr, 400km/hr and 250km/hr respectively. Find its average speed for the whole journey.

Average body speed is defined as the total number of distances  the body has traveled divided by the total time  the body has taken. Assuming that the object travels the total distance of d in the time interval of t, its average velocity is v = dt. However, suppose the plane closes the distance with three different windshields. First travel 2500 km at a speed of 500 km / h,  then travel a distance of 1200 km at a speed of 400 km / h. Finally, drive 500km at  250km / h.  The three notes must be $d_1, d_2, and d_3,$ respectively. And the corresponding three  speeds were $v_1, v_2, v_3$ 
In that case, the total distance traveled by the plane is $d_1 + d_2 + d_3$. Let's calculate the total time. Let's first look at the time it took to cover each class. Let the time required for $d_1, d_2, and d_3,$ be $t_1, t_2, and t_3,$ respectively. Now use the expression t = dv.

This means that $t_1=\frac{d_1}{v_1}$, $t_2=\frac{d_2}{v_2}$ and $t_3=\frac{d_3}{v_3}$.
Hence, the total time taken by the aeroplane is $t_1+t_2+t_3=\frac{d_1}{v_1}+\frac{d_2}{v_2}+\frac{d_3}{v_3}$.
Therefore, the average speed of the aeroplane is $v=\frac{d_1+d_2+d_3}{t_1+t_2+t_3}=\frac{d_1+d_2+d_3}{{\displaystyle \frac{d_1}{v_1}+\frac{d_2}{v_2}+\frac{d_3}{v_3}}}$.
Substitute the values of the distances and the speeds.
$\Rightarrow v=\frac{2500+1200+500}{{\displaystyle \frac{2500}{500}+\frac{1200}{400}+\frac{500}{250}}}=\frac{4200}{5+3+2}=\frac{4200}{10}=420km{h}^{-1}$.
Therefore, the average speed of the aircraft is 420km/h.