Q37 of 40 Page 1

A train covers a distance of 360 km at a uniform speed. Had the speed been 5km/hour more, it would have taken 48 minutes less for the journey. Find the original speed of the train.

OR


Solve the following equation:



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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