Question

a plane started late by 30 minutes. I to reach a destination 1500 km away in time the piolet increased the speed by 100km/hr. the find the original speed of the plane

Answer 1

Let the usual speed of the plane = x km/hr.

Increased speed of the plane = (x +100) km/hr

Time taken to reach the destination at usual speed, t1=
$\frac{1500}{x}$


Time taken to reach the destination at increased speed, t2=
$\frac{1500}{x + 100}$


Given:  

 t 1 – t 2 = 30 min

$= > \: \frac{1500}{x} - \frac{1500}{x + 100} = \frac{30}{60} \\ \\ \frac{1500x + 150000 - 1500x}{x(x + 100)} = \frac{1}{2} \\ \\ {x}^{2} + 100x - 300000 = 0 \\ \\ {x}^{2} + 600x - 500x - 300000 = 0 \\ \\ x = 500 \: or \: - 600 \\ x \: = 500$

therefore original speed is 500 km/h

Answer 2

Solution :-

Let the original speed of train be x km/hr

New speed = (x + 100) km/hr

We know that,

Time = Distance / Speed

Given : A plane left 30 minutes or 1/2 hours later than the scheduled time.

According to the question,

=> 1500/x - 1500/(x + 100) = 1/2

=> (1500x + 15000 - 1500x)/x(x + 100) = 1/2

=> 2(15000) = x(x + 100)

=> 30000 = x² + 100x

=> x² + 100x - 30000 = 0

=> x² + 600x - 500x - 30000 = 0

=> x(x + 600) - 500(x + 600) = 0

=> (x - 500) (x + 600) = 0

=> x = 500 or x = - 600

∴ x ≠ - 600 (Because speed can't be negative)

Hence,

Its usual speed = 500 km/hr