Q. 373.8( 36 Votes )

# A train covers a

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,

0.8s2 - 1800 + 4s = 0

Dividing above equation by 0.8,

s2 - 2250 + 5s = 0

s2 + 5s - 2250 = 0

s2 + 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 –

3x2 – 6x = -2

3x2 – 6x + 2 = 0

Comparing above equation with ax2 + 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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