An aeroplane left 40 minutes late …

Let the usual speed of the plane be x km/hr. Then,
New speed = (x + 400) km/hr.
Time taken to cover 1600 km when the speed is x km/hr {tex} = \frac{{1600}}{x}{/tex} hrs.
Time taken to cover 1600 km when the speed is (x + 400) km/hr {tex} = \frac{{1600}}{{x + 400}}{/tex} hrs.
{tex}\therefore \frac{{1600}}{x} - \frac{{1600}}{{x + 400}} = \frac{{40}}{{60}}{/tex} {tex}\left[ {\because 40{\text{ minutes}} = \frac{{10}}{{60}}hr} \right]{/tex}
{tex} \Rightarrow \frac{{1600(x + 400) - 1600x}}{{x(x + 400)}} = \frac{2}{3}{/tex}
{tex} \Rightarrow \frac{{1600x + 640000 - 1600x}}{{{x^2} + 400x}} = \frac{2}{3}{/tex}
{tex} \Rightarrow \frac{{640000}}{{{x^2} + 400x}} = \frac{2}{3}{/tex}
{tex}\Rightarrow{/tex} 1920000 = 2(x² + 400x)
{tex} \Rightarrow \frac{{1920000}}{2} = {x^2} + 400x{/tex}
{tex}\Rightarrow{/tex} 960000 = x² + 400x
{tex}\Rightarrow{/tex} x² + 400x - 960000 = 0
{tex}\Rightarrow{/tex} x² + 1200x - 800x - 960000 = 0
{tex}\Rightarrow{/tex} x(x + 1200) - 800(x + 1200) = 0
{tex}\Rightarrow{/tex} (x + 1200)(x - 800) = 0
{tex}\Rightarrow{/tex} x - 800 = 0 [{tex}\because{/tex} Speed can not be negative {tex}\therefore{/tex} x + 1200 {tex}\ne{/tex} 0]
{tex}\Rightarrow{/tex} x = 800
Hence, the usual speed of the plane is 800 km/hr.