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Let the average speed of the passenger train be 'x' km/hr and the average speed of the express train be (x + 11) km/hr
Distance between Mysore and Bangalore = 132 km
It is given that the time taken by the express train to cover the distance of 132 km is 1 hour less than the passenger train to cover the same distance.
So, time taken by passenger train = 132/x hr
The time taken by the express train = {(132)/x+11} hr
Now, according to the question
{132/(x+11)} = 132/x + 1
After taking L.C.M. of 132/x +1 and then solving it we get (132+x)/x.
Now,
{132/(x+11)} = (132+x)/x
By cross multiplying, we get
132x = x²+132x+11x+1452
x² + 11x - 1452 = 0
x² + 44x - 33x - 1452 = 0
x(x+44) - 33(x+44) = 0
(x+33) (x+44) = 0
x - 44 or x = 33
As the speed cannot be in negative therefore, x = 33 or the speed of the passenger train = 33 km/hr and the speed is 33 + 11 = 44 km/hr.
Let the speed of the passenger train be x km/h & the speed of the express train be (x+11)km/h.
Time taken by the passenger train, T = distance/speed = (132/x)hr & time taken by the express train, T`= (132/x+11)hr.
ATQ, T - T` = 1 or 132/x - (132/x+11)=1 or 132x + 1452 - 132x /x(x+11) = 1 or x² + 11x = 1452 or x² + 11x - 1452 = 0 Discriminant = b² - 4ac or
D = 121- 4 X -1452 X 1 = 121 + 5808 = 5929
x = -11 + √5929/2 or x = -11 + 77/2 = 66/2 = 33
or x = -11-77/2 = -88/2 = -44. Now, since the speed can not be negative, hence x = 33km/h.
The speed of the passenger train = 33km/h.
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Train travels = 132/x hr
The time taken by the express train = {(132)/x+11} hr
According to the question
{132/(x+11)} = 132/x + 1
After taking L.C.M. of 132/x +1 and then solving it we get (132+x)/x.
Now,
{132/(x+11)} = (132+x)/x
By cross multiplying, we get
132x = x²+132x+11x+1452
x² + 11x - 1452 = 0
x² + 44x - 33x - 1452 = 0
x(x+44) - 33(x+44) = 0
(x-33) (x+44) = 0
x = -44 or x = 33
As the speed cannot be in negative
therefore, x = 33
or
the speed of the passenger train = 33 km/hr and the speed is 33 + 11 = 44 km/hr
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