A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500km away in time, it had to increase its speed by 100 km/h from the usual speed. Find its usual speed.
Guys plzz answer it with correct explanation...
Answers
Answer:
Let the usual time taken by the aeroplane = x km/hr
Distance to the destination = 1500 km
Case (i)
Speed = Distance / Time = (1500 / x) Hrs
Case (iI)
Time taken by the aeroplane = (x - 1/2) Hrs
Distance to the destination = 1500 km
Speed = Distance / Time = 1500 / (x - 1/2) Hrs
Increased speed = 250 km/hr
⇒ [1500 / (x - 1/2)] - [1500 / x] = 250
⇒ 1/(2x2 - x) = 1/6
⇒ 2x2 - x = 6
⇒ (x - 2)(2x + 3) = 0
⇒ x = 2 or -3/2
Since, the time can not be negative,
The usual time taken by the aeroplane = 2 hrs
and the usual speed = (1500 / 2) = 750 km/hr.
Answer:
500 km/hr
Step-by-step explanation:
Let the usual speed be 'x' km/hr.
Time(t₁) = Distance/Speed = 1500/x.
Given that it had do increase its speed by 100 km/hr.
So, Increased speed = (x + 100) km/hr.
Time(t₂) = 1500/x + 100.
∴ Given that A plane left 30 minutes late.
⇒ (1500/x) - (1500/x + 100) = 30/60
⇒ (1500x + 1500 * 100 - 1500x)/x(x + 100) = 1/2
⇒ 1500 * 100 * 2 = x(x + 100)
⇒ 300000 = x² + 100x
⇒ x² + 100x - 300000 = 0
⇒ x² + 600x - 500x - 300000 = 0
⇒ x(x + 600) - 500(x + 600) = 0
⇒ (x - 500)(x + 600) = 0
⇒ x = 500, -600{Ignore negative values as speed cannot be negative}
⇒ x = 500.
Therefore, usual speed of the plane = 500 km/hr.
Hope it helps!