An aeroplan

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 or a = -750

∵ speed can’t be negative ∴ a = 500

Hence the usual speed is 500 km/hr.

OR

Let the numbers are ‘a’ and ‘b’

According to given conditions:

a + b = 15

⇒ b = 15 – a …. (1)

Also,

From (1),

⇒ 15 × 10 = 3(15a – a²)

⇒ 15 × 10 = 45a – 3a²

⇒ 3a² – 45a + 150 = 0

⇒ a² – 15a + 50 = 0

⇒ a² – 15a – 5a + 50 = 0

⇒ a (a – 10) – 5(a – 10) = 0

⇒ (a – 5) (a – 10) = 0

⇒ a = 5, 10

If a = 5, b = 15 – 5 = 10

If a = 10, b = 15 – 10 = 5

Hence, Numbers are 5,10 or 10, 5.