A car covers a distance at speed of 60   km   h - 1 . It returns and comes back to the original…

Let car covers a distance of $s$ in straight line from point $A$ to $B$ & then comes back to point $A$.

Average speed $=\frac{\text{total distance covered}}{\text{total time taken}}$.

Now, time taken from $A$ to $B$ will be, $\frac{s}{60}$. Time taken from $B$ to $A$ will be, $\frac{s}{v}$.

Therefore,

$48=\frac{s+s}{{\displaystyle \frac{s}{60}}+{\displaystyle \frac{s}{v}}}$

$\Rightarrow \frac{1}{60}+\frac{1}{v}=\frac{2}{{\displaystyle 48}}$

$\Rightarrow \frac{1}{v}=\frac{1}{24}-\frac{1}{60}$

$$\begin{gathered} \Rightarrow \frac{1}{v}=\frac{60-24}{24\times 60} \\ \Rightarrow v=40 \mathrm{km} {\mathrm{h}}^{-1} \end{gathered}$$