Question

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

Answer 1

Let the speed of the train is x km/h
speed of car = x + 21 km/h
time = distance/speed

the man travels a distance of 196 km by train and returns in a car and total journey takes 11 hours.

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 × -4116 = 0
11x² - 308x + 147x - 4116 = 0
11x(x - 28) + 147(x - 28) = 0
(x - 28) × ( 11x + 147) = 0
x = 28
speed of the car = x + 21
                          = 28 + 221 = 49 km/h

Answer 2

let the speed of train be x

let the speed of car is equal to x + 21

196 / x + 196 / x + 21 = 11

multiply by x, x + 21

196 into x + 21 + 196 x =11x^2+ 231 x

so 196 x + 4116+196x=11 x^2 + 231 x

11 x ^2 - 4 x 61 x - 4116 =0

11x^2- 308 x + 147 x -4860 =0

11 x (x -48) + 147 ( x - 28 )=0

(11 x + 147)=0 or( x - 28) =0

therefore, x - 147 /11 or

x = 28

therefore the average speed of train is 28km/hr
average speed of car=
x+21=28+21=49km/hr.