the ratio between the rates of walking of A and B is 2:3.and therefore A takes 10 mins more than the time taken by B to reach the destination.If A had walked at double the speed ,then in what time would have covvered the distance?

Dear Student,

Please find below the solution to the asked query:

We know :  Time  = $\frac{\mathrm{Distance}}{\mathrm{Speed}}$
Given : The ratio between the rates of walking of A and B is 2 : 3 , Let ratio coefficient =  x  , So

Speed of A  = 2 x km / hr  and Speed of B  =  3 x  km  / hr 

And

Let distance from starting point to destination =  d  km , So

From given conditions we get :

$$\begin{gathered} \frac{d}{2 x} - \frac{10}{60} = \frac{d}{3 x} \\ \Rightarrow \frac{d}{2 x} - \frac{1}{6} = \frac{d}{3 x} \\ \Rightarrow \frac{\left( 3 d - x \right)}{6 x}= \frac{d}{3 x} \\ \Rightarrow \left( 3 d - x \right)= \frac{d}{3 x}\times 6 x \\ \Rightarrow 3 d - x = 2 d \\ \mathbf{\Rightarrow }\mathit{d}\mathbf{ }\mathbf{=}\mathbf{ }\mathit{x} \end{gathered}$$
Therefore,

If A had walked at double the speed ,then in what time would have covered the distance = $\frac{d}{2 \times 2 x} = \frac{x}{4 x} = \frac{1}{4} \mathrm{hr} = \frac{1}{4}\times 60 =$15 min                                                     ( Ans )

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