Pls solve:A man travels a distance of 196km by train and returns in a car which travels at a speed of 21km/hr faster than the train.If the total journey takes 11hours.Find the speed of the train & car. 
 

Dear Student,

Please find below the solution to the asked query:

Let ,  Speed of train = x km / hr

Given : Car which travels at a speed of 21 km / hr faster than the train , So

Speed of car = ( x + 21 ) km/ hour

We know :  Time  = $\frac{\mathrm{Distance}}{\mathrm{Speed}}$

Also given : A man travels a distance of 196 km by train and returns in a car and the total journey takes 11 hours., So

$$\begin{gathered} \frac{196}{x} + \frac{196}{x +\hspace{0.17em}21} = 11 \\ \Rightarrow \frac{196 \left( x +\hspace{0.17em}21 \right) + 196 x}{x \left( x +\hspace{0.17em}21 \right)}= 11 \\ \Rightarrow \frac{196 x + 4116 + 196 x}{x { }^2 +\hspace{0.17em}21 x }= 11 \\ \Rightarrow \frac{392 x + 4116}{x { }^2 +\hspace{0.17em}21 x }= 11 \\ \Rightarrow 392 x + 4116 = 11x { }^2 +\hspace{0.17em}231 x \\ \Rightarrow 11x { }^2 - 161 x - 4116 = 0 \\ \Rightarrow 11x { }^2 - 308 x + 147 x- 4116 = 0 \\ \Rightarrow 11 x \left( x - 28 \right) + 147 \left( x - 28 \right) = 0 \\ \mathbf{\Rightarrow }\left(\mathbf{ }\mathbf{11}\mathbf{ }\mathit{x}\mathbf{ }\mathbf{+}\mathbf{ }\mathbf{147}\mathbf{ }\right)\left(\mathbf{ }\mathit{x}\mathbf{ }\mathbf{-}\mathbf{ }\mathbf{ }\mathbf{28}\mathbf{ }\right)\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{0} \end{gathered}$$

So,

x  = $- \frac{147}{11}$  and 28  ,  But we assume " x "  as speed of train so that can't be negative , So we neglect x  = $- \frac{147}{11}$ and get

x  = 28

Therefore,

Speed of train = 28 km / hr

And

Speed of car = (  28+ 21 ) km/ hour = 49 km / hr                                            ( Ans )


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