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An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11km/h more than that of the passenger train, find the average speed of the two trains.

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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/hrDistance between Mysore and Bangalore = 132 kmIt 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...
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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.

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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 ...
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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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Passenger train speed is 33 km/hr And Express train speed is 44km/hr.
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Train travels = 132/x hrThe time taken by the express train = {(132)/x+11} hrAccording to the question{132/(x+11)} = 132/x + 1After taking L.C.M. of 132/x +1 and then solving it we get (132+x)/x.Now,{132/(x+11)} = (132+x)/xBy cross multiplying, we get132x = x²+132x+11x+1452x² + 11x - 1452 =...
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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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Express Train 44 kmph, Passenger Train 33 kmph
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