Roohi travels 300 km to her home, partly by train and partly by bus. She takes 4 hours when she travels 60 km by train and the remaining distance by bus. If she travels 100 km by train and the remaining distance by bus, she takes 10 minutes longer. Calculate the speeds of the train and the bus, respectively.

Given: Total distance = 300 km

Let the speed of the train be x km/hr and the speed of the bus is y km/hr.
We know that:  $\mathrm{Time}=\frac{\mathrm{Distance}}{\mathrm{Speed}}$

Given, Roohi travels 60 km by train and 300−60 = 240 km by bus in 4 hours.

So, $\frac{60}{\mathrm{x}}+\frac{240}{\mathrm{y}}=4 ...\left(1\right)$

Also, if Roohi travels 100 km by train and 300−100 = 200 km by bus, and takes 10 minutes more.

Therefore, $\frac{100}{\mathrm{x}}+\frac{200}{\mathrm{y}}=4+\frac{10}{60}$

$\Rightarrow \frac{100}{\mathrm{x}}+\frac{200}{\mathrm{y}}=\frac{25}{6} ...\left(2\right)$

Let $\frac{1}{\mathrm{x}}=\mathrm{u} \mathrm{and} \frac{1}{\mathrm{y}}=\mathrm{v}$

Then, from eq(1) and eq(2),

$60\mathrm{u}+240\mathrm{v}=4 ...\left(3\right)$

$100\mathrm{u}+200\mathrm{v}=\frac{25}{6}$

$\Rightarrow 4\mathrm{u}+8\mathrm{v}=\frac{1}{6} ...\left(4\right)$

Multiplying eq(4) by 15, we get

$60\mathrm{u}+120\mathrm{v}=\frac{15}{6} ...\left(5\right)$

Subtracting eq(5) from eq(3), we get:

$240\mathrm{v}-120\mathrm{v}=4-\frac{15}{6}$

$\Rightarrow 120\mathrm{v}=\frac{9}{6}$

$\Rightarrow 120\mathrm{v}=\frac{3}{2}$

$\Rightarrow \mathrm{v}=\frac{3}{240}$

$\therefore \mathrm{y}=\frac{1}{\mathrm{v}}=\frac{240}{3}=\mathbf{80}\mathbf{ }\mathbf{km}\mathbf{/}\mathbf{h}$

From eq(4),

$4\mathrm{u}+8\times \frac{3}{240}=\frac{1}{6}$

$\Rightarrow 4\mathrm{u}+\frac{1}{10}=\frac{1}{6}$

$\Rightarrow 4\mathrm{u}=\frac{1}{15}$

$\Rightarrow \mathrm{u}=\frac{1}{60}$

$\therefore \mathrm{x}=\frac{1}{\mathrm{u}}=\frac{60}{1}=\mathbf{60}\mathbf{ }\mathbf{km}\mathbf{/}\mathbf{h}$

Hence, the speed of the train is 60 km/hr and the speed of the bus is 80 km/hr.