Question

travel a distance of 196 km by train and return in car which Travels at a speed of 21 km per hour more than the train if the total journey takes 11 hours square on the average speed of the train and car respectively

Answer 1

Let the speed of the train is x km/hr
Given, the car travels at a speed of 21 km/hr faster than the train,
So, the speed of the car = x + 21 km/hr
Now, Time = distance/speed
Again, the man travels a distance of 196km by train and returns in a car and total journey takes 11 hours,
So, 196/x + 196/(x + 21) = 11
=> 196{1/x + 1/(x + 21)} = 11
=> 196{(x + 21 + x)/x*(x + 21)} = 11
=> 196{(2x + 21)/x*(x + 21)} = 11
=> 196(2x + 21) = 11x*(x + 21)
=> 392x + 4116 = 11x2 + 231x
=> 11x2 + 231x - 392x - 4116 = 0
=> 11x2 - 161 x - 4116 = 0
=> 11x2 - 308 x + 147x - 4116 = 0
=> 11x(x - 28) + 147(x - 28) = 0
=> (x - 28)*(11x + 147) = 0
=> x = 28, -147/11
Since speed can not be negative,
So, x = 28
Now, the speed of the car = x + 21 = 28 + 21 = 49 km/hr