Answer :

45km/hr

Let, original speed of train = s

Current speed = s’

Original time taken = t

Current time taken = t’

Given –

Distance d = 360 km

s’ = s + 5

t’ = t – 48 min

To find – original speed s

Formula - distance = speed × time

Answer –

We know that, 1 hour = 60 minutes

Therefore, 48 min hrs

∴ t’ = t - 0.8

distance = speed × time

∴ d = s × t = s’ × t’

∴ 360 = st ………(1)

∴ s’t’ = 360

∴ (s + 5) (t – 0.8) = 360

∴ st – 0.8s + 5t – 4 = 360

∴ 360 – 0.8s + 5t – 4 = 360 ………from (1)

∴ – 0.8s + 5t – 4 = 0

………from (1)

Multiplying above equation throughout by -s,

∴ 0.8s^{2} - 1800 + 4s = 0

Dividing above equation by 0.8,

∴ s^{2} - 2250 + 5s = 0

∴ s^{2} + 5s - 2250 = 0

∴ s^{2} + 50s – 45s - 2250 = 0

∴ s(s + 50) – 45(s + 50) = 0

∴ (s + 50) (s – 45) = 0

∴ s = -50 or s = 45

But speed can never be negative.

∴ s = 45

Therefore, original speed of the train is 45 km/hr.

**OR**

Given equation –

∴ 3x^{2} – 6x = -2

∴ 3x^{2} – 6x + 2 = 0

Comparing above equation with ax^{2} + bx + c = 0, we get

a = 3 , b = -6 , c = 2

therefore, roots of above equation are

Hence, the solution for given equation is .

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