Question

an aeroplane left 30 minutes later than it schedule time and in order to reach is destination 1500 km away in time,it had to increase its speed by 250 km/hr from its usual speed.determine its usual speed.

Answer 1

Let the earlier speed be X km / hr

Increased speed = (X + 250) km / hr

1500 / X - 1500 / ( X +250) = 1/2

MULTIPLY BY 2X*( X +250)

3000 ( X +250) - 3000X = X( X +250)

3000X + 750000 - 3000X = X^2 + 250X

X^2 + 250X - 750,000 = 0

X^2 +1000X -750X - 750,000 = 0

X ( X + 1000) -750 ( X + 1000) = 0

( X - 750) ( X + 1000) = 0

X = 750

X + 250 = 1000 km / hr ANSWER
CHECK
1500/1000 = 1.5 hours
1500/750 = 2 hours
2 - 1.5 = 1/2 hour

Answer 2

Solution :-

Let the original speed of train be x km/hr
New speed = (x + 250) 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 + 250) = 1/2
=> (1500x + 37500 - 1500x)/x(x + 250) = 1/2
=> 2(37500) = x(x + 250)
=> 75000 = x² + 250x
=> x² + 250x - 75000 = 0
=> x² + 1000x - 750x - 75000 = 0
=> x(x + 1000) - 750(x + 1000) = 0
=> (x - 750) (x + 1000) = 0
=> x = 750 or x = - 1000

∴ x ≠ - 1000 (Because speed can't be negative)


Hence,
Its usual speed = 750 km/hr