A car travels from a place X to place Y at an average speed of $v$ km/hr, from Y to X at an average speed of…
A car travels from a place X to place Y at an average speed of $v$ km/hr, from Y to X at an average speed of $2v$ km/hr, again from X to Y at an average speed of $3v$ km/hr and again from Y to X at an average speed of $4v$ km/hr. Then the average speed of the car for the entire journey
is less than $v$ km/hr
lies between $v$ and $2v$ km/hr
lies between $2v$ and $3v$ km/hr
lies between $3v$ and $4v$ km/hr
Solution
Let the distance XY $= d$. Total distance $= 4d$. Total time $= \dfrac{d}{v} + \dfrac{d}{2v} + \dfrac{d}{3v} + \dfrac{d}{4v} = \dfrac{d}{v}\left(1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4}\right) = \dfrac{d}{v} \times \dfrac{25}{12}$. Average speed $= \dfrac{4d}{(d/v)(25/12)} = \dfrac{48v}{25} = 1.92v$. This lies between $v$ and $2v$.