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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 trains is 11 km / hr more than that of the passenger train, find the average speeds of the two trains.
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 trains is 11 km / hr more than that of the passenger train, find the average speeds of the two trains.
Time = distance/speed
Now, express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore. The average speed of the express trains is 11 km / hr more than that of the passenger train.
Let the average speed of passenger train be ‘a’.
\(\frac{132}{a}-\frac{132}{a+11}=1\)
⇒ 132 ×(a + 11 –a) = a2 + 11a
⇒ a2 + 11a – 1452 = 0
⇒ a2 + 44a – 33a – 1452 = 0
⇒ a(a + 44) – 33(a + 44) = 0
⇒ (a + 44)(a – 33) = 0
⇒ a = 33 km/hr
Speed of express train = 44 km/hr
speed of passenger train`= x`km/hr
speed of express train `=(x+11)`km/hr
time taken by passenger train`= 132/x = 132/x`
time taken by express train`=132/(x+11) = 132/(x+11)`
acc to question
`132/(x+11) =132/x – 1 `
`132[1/x – 1/(x+11)] = 1`
`132[(x+11-x)/(x(x+11))]=1`
`132*11 = x(x+11)`
`x^2+11x-1452=0`
`x^2 + 44x-3x- 1452=0 `
`x(x+44) – 33(x+44)=0`
`(x-33)(x+44)=0`
`x=33 , -44`
x=-44 cant be possible as x cant be zero
so, x=33 & `x+11 = 44` answer