#### Question: 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 11 km/hr more than that of the passenger train, find the average speed of the two trains.

#### Video Explanation

#### Explanatory Answer

Step 1: Frame equations based on the information about the speed and time taken by the express and passenger trains.

Let the speed of the express train be s km/hr.

Statement: Speed of the express train is 11km/hr more than that of passenger train.

∴ speed of the passenger train = (s – 11)km/hr

Time taken by express train to travel 132 km = \\frac{132}{s}) hours

Time taken by passenger train to travel 132 km = \\frac{132}{s-11}) hours

Statement: Express train takes 1 hour lesser time than the passenger train

∴ \\frac{132}{s}) = \\frac{132}{s-11}) - 1

\\frac{132}{s-11}) - \\frac{132}{s}) = 1

Because s ≠ 0 and s ≠ 11 (if s = 11, speed of passenger train will be 0. Then the passenger train would not have travelled at all), multiply the equation by s(s-11)

132s – 132(s – 11) = s(s – 11)

132s – 132s + 1452 = s^{2} - 11s

s^{2} – 11s - 1452 = 0

##### Step 2: Factorize the equation to find the value of ‘s’

s^{2} – 44s + 33s – 1452 = 0

s(s – 44) + 33(s – 44) = 0

(s – 44) (s + 33) = 0

s = 44 or s = -33

Speed cannot be negative

∴ speed of the express train = 44km/hr

Speed of the passenger train = s - 11 = 44 - 11 = 33 km/hr