Question

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 .

Answer 1

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!

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