An aeroplane left 50 minutes later than its schedule time, and in order to reach the destination, 1250 km away, in time, it had to increase its speed b 250 km / hr from its usual speed. Find its usual speed.

Given: Increment in speed = 250 km/h

Distance = 1250 km

To find: Its usual speed

Method Used:

To solve the quadratic equation by factorisation method, follow the steps:

1) Multiply the coefficient of x² and constant term.

2) factorise the result obtained in step 1.

3) Now choose the pair of factors in such a way that after adding or subtracting(splitting) them

You get coefficient of x.

Explanation:

Given, aeroplane left 50 minutes later than its schedule time,

and in order to reach the destination, 1250 km away, in time,

it had to increase its speed to 250 km / hr. from its usual speed.

Let the usual speed be ‘a’.

Increased speed = a + 25

As,

So usual time is

When speed is increased time will be

As speed increases time decreases

So according to question,

⇒ 6 × 1250 × (a + 250 –a) = 5(a² + 250a)

⇒ a² + 250a – 375000 = 0

⇒ a² + 750a – 500a – 375000 = 0

⇒ a(a + 750) – 500(a + 750) = 0

⇒ (a + 750) (a – 500) = 0

⇒ a = 500 km/hr

Hence the usual speed is 500 km/hr.