Question

a plane left 30 minutes later than the scheduled time and in order to reach its destination 1500 km away in time it had to increase its speed by 250 km per hour from its usual speed find its usual speed

Answer 1

$\textbf{Answer}$

Suppose usual speed of plane = x km/hour

Total distance = 1500 km

$\textbf{Time = Distance / Speed}$

=> Usual time = (1500/x) hours -----(1)

Since plane left 30 minutes later,in that case -

$\textbf{Distance remains same}$

$\textbf{Speed is increased by 250 km/hour}$

$\textbf{Time to travel is 30 minutes less}$

30 minutes = 30/60 hour = 1/2 hour

$\textbf{New speed = (x + 250) km/hour}$

$\textbf{New time = 1500/(x+250) hour}$

$\textbf{According to the question,}$

(1500/x) = 1500/(x+250) + 1/2

=> (1500/x) - 1500/(x + 250) = 1/2

=> 1500 ( 1/x - 1/(x + 250) = 1/2

=> 3000 (x + 250 - x) = (x) (x + 250)

=> 750000 = x^2 + 250x

=> x^2 + 250x - 750000 = 0

=> x^2 + 1000x - 750x - 750000 = 0

=> x(x + 1000) - 750(x + 1000) = 0

=> (x - 750) (x + 1000) = 0

=> x = 750 or x = - 1000

Since x is the usual speed of the plane,
which can not be a negative number.

$\textbf{Usual speed of plane is 750 km/hour}$

$\textbf{Hope My Answer Helped}$

$\textbf{Thanks}$

Answer 2

Let the usual speed of the aeroplane be x km/h.



=> Time taken to cover the journey at usual speed = ( 1500/x ) hours





=> New speed of the aeroplane = ( x + 250 ) km/h



=> Time taken to cover the journey at new speed = ( 1500 ÷ { x + 250 } ) hours





According to the given information,


Difference in two times = 30 minutes = 1/2 hours



$=> \frac{1500}{x} - \frac{1500 }{x + 250} = \frac{1}{2} \\ \\ \\ = > 1500(x + 250) - 1500x = \frac{x(x + 250)}{2} \\ \\ \\ => {x }^{2} + 250 - 750000 = 0 \\ \\ \\ = > (x - 750)(x + 1000) = 0$





=> x = 750 or -1000 but speed cant be negative, so x = 750



Hence, usual speed of aeroplane is 750 km/hr