Two cities X and Y are connected by a regular bus service with a bus leaving in either direction every T min. A girl is driving scooty with a speed of 60 km/h in the direction X to Y notices that a bus goes past her every 30 minutes in the direction of her motion, and every 10 minutes in the opposite direction. Choose the correct option for the period T of the bus service and the speed (assumed constant) of the buses

Let the speed of the girl on the scooty be $\mathrm{v\_g}=\; 60$ km/h.

Let the (constant) speed of each bus be $\mathrm{v\_b}$ km/h, and let the interval between successive bus departures in either direction be $T$ minutes.

Because a bus leaves every $T$ minutes, the distance between two consecutive buses traveling in the same direction is

$d=\; v\_b\; \backslash times\; \backslash frac\{T\}\{60\}$ km ( since 60 min = 1 h ).

When a bus travels in the same direction as the girl, the relative speed between the bus and the girl is $\mathrm{v\_b\; -\; v\_g}$. The time between two successive overtakes is therefore

$\mathrm{t\_\{\backslash text\{same\}\}}=\; \backslash frac\{d\}\{v\_b\; -\; v\_g\}$ hours.

We are told this time equals 30 minutes = 0.5 h, so

$\backslash frac\{v\_b\; \backslash times\; \backslash frac\{T\}\{60\}\}\{v\_b\; -\; 60\}\; =\; 0.5$

or equivalently

$v\_b\; T\; =\; 0.5\; \backslash times\; 60\; \backslash times\; (v\_b\; -\; 60)$  (1)

When a bus comes from the opposite direction, the relative speed is $\mathrm{v\_b\; +\; v\_g}$. The interval between such meetings is

$\mathrm{t\_\{\backslash text\{opp\}\}}=\; \backslash frac\{d\}\{v\_b\; +\; v\_g\}$ hours,

which is given as 10 minutes = 1/6 h. Hence

$\backslash frac\{v\_b\; \backslash times\; \backslash frac\{T\}\{60\}\}\{v\_b\; +\; 60\}\; =\; \backslash frac\{1\}\{6\}$

or

$v\_b\; T\; =\; \backslash frac\{1\}\{6\}\; \backslash times\; 60\; \backslash times\; (v\_b\; +\; 60)$  (2)

Equating the right‑hand sides of (1) and (2) eliminates the product $v\_b\; T$:

$0.5\; (v\_b\; -\; 60)\; =\; \backslash frac\{1\}\{6\}\; (v\_b\; +\; 60)$

Multiply by 6 to clear fractions:

$3\; (v\_b\; -\; 60)\; =\; v\_b\; +\; 60$

Expand and simplify:

$3v\_b\; -\; 180\; =\; v\_b\; +\; 60$

$2v\_b\; =\; 240$

$v\_b\; =\; 120$ km/h.

Now substitute $v\_b\; =\; 120$ into (1) to find $T$:

$120\; \backslash ,\; T\; =\; 0.5\; \backslash times\; 60\; \backslash times\; (120\; -\; 60)$

$120\; \backslash ,\; T\; =\; 0.5\; \backslash times\; 60\; \backslash times\; 60\; =\; 30$

$T\; =\; \backslash frac\{30\}\{120\}\; =\; 0.25$ h = 15 minutes.

Thus the bus service departs every **15 minutes** and each bus travels at **120 km/h**.