Answer :
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.
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