A plane left 30 min. late than from its schedule time in order to reach the destination 1500 km away in the time, it had increase its speed by 100km/ hr from usual speed .
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Your question seems to be incomplete. We have to find the usual speed.
Now,
Given that Plane left 30 minutes late = (1/2) hours.
Let the usual speed of the plane be 'x'.
(i)
Given Distance = 1500 km.
We know that Time = Distance/Speed
= > t1 = (1500/x) ------ (1)
(ii)
Given that it had increase its speed by 100km/hr.
So, New Speed = x + 100 km/hr
We know that Time = Distance/Speed
Time t2 = (1500/x + 100) ----- (2)
-------------------------------------------------------------------------------------------------------------
Now,
= > (1500/x) - (1500/x + 100) = (1/2)
= > 2 * 1500(x + 100) - 2x * 1500 = x(x + 100)
= > 3000x + 300000 - 3000x = x^2 + 100x
= > x^2 + 100x - 300000 = 0
= > x^2 + 600x - 500x - 300000 = 0
= > x(x + 600) - 500(x + 600) = 0
= > (x - 500)(x + 600) = 0
= > x = 500,-600[Neglect negative value]
= > x = 500.
Therefore, the usual speed of the plane = 500 km/hr.
hope this helps!
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