Question

A train left 30 minutes later than the scheduled time and in order to reach its destination 1500 km away in time it has to increase its speed by 250 km per hour from its usual speed

Answer 1

Hey

Here is your answer,

Let the usual speed of the plane = x km/h

Time taken to cover 1500 km = (1500/x) hours

New speed = (100 + x) km/h

Time taken = 1500/(100 + x) - This time is 30 minutes (0.5 hours) less than the first time

So;

1500/x - 1500/(100 + x) = 0.5
150000 = 0.5(x² + 100x)
x² + 100x - 300000 = 0

This is a quadratic equation

x = [-200+/-√(100)² - 4(1)(-300000)]/2(1)
x = 500 or - 600

So, the usual speed of the plane = 500 km/h

Hope it helps you!

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