Question

A plane 30 minutes late than its schedule time and in order to reach tha destination 1500km away in time it had to increase its speed by 100 km/h form the usual speed find its usual speed

Answer 1

Let the usual speed of the plane be 'x'km/hour
Distance    Speed               Time    
1500 km   x km/h            1500/x hours 
1500 km   x + 100 km/h  1500/x+10 hours
Now, the delay time = 30 minutes = 1/2 hour
Also , according to the given equation
The total time taken by the plane at its original speed = Delay time + Time taken after increasing the speed
∴ 1500/x = 1500/x+100 + 1/2
∴ 1500/x - 1500/ x +100 = 1/2
∴ 1500 ( 1/x - 1/x+100) = 1/2
∴ 1/x - 1/x+100 = 1/3000
∴ x+100-x / x²+100x = 1/3000
∴ 10 / x² + 100x = 1/3000
∴ x² + 100x = 3000*10
∴ x² + 100x - 30000 = 0
∴ x² + 600x - 500x - 30000 = 0
∴ x(x+600) - 500(x+600) = 0
∴ (x+600)(x-500) = 0
∴ x = 500 or x = - 600
Now, clearly speed can never be negative.
∴ x = Usual speed = 500 km/hour.
∴ The usual speed of the plane is 500 kilometers/hour.
Hope this helps you !
# Dhruvsh 

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