Question

Two towns $A$ and $B$ are connected by a regular bus service with a bus leaving in either direction every $T$ minutes. A man cycling with a speed of $20km{h}^{-1}$ in the direction $A$ to $B$ notices that a bus goes past him every $18min$ in the direction of his motion, and every $6$ min in the opposite direction. What is the period $T$ of the bus service and with what speed (assumed constant) do buses ply on the road?

• A. 9 min, 40 Kmph
• B. 12 min, 10 Kmph
• C. 12 min, 40 Kmph
• D. 9 min, 60 Kmph

Answer 1

The correct option is A 9 min, 40 Kmph

As both the bus and cyclist are moving in the same direction their relative velocity will be −

Relative velocity of bus $=V_B-V_C=V_B$ - (+20)

Distance covered $=V_T$

$Speed\times Time=Distance$

$(V_B-20)\times 18=V_{BT}$ ... ... ... ... (1)

Case 2 − Bus moving from B to A

As the bus and cyclist are moving in the opposite direction their relative velocity will be −

Relative velocity of bus $=V_B-V_C=V_B$ -(-20)

Distance covered $=V_T$

$Speed\times Time=Distance$

$(V_B+20)\times 6=V_{BT}$ ... ... ... ... (2)

Solving equations (1) and (2) −

$(V_B-20)\times 18$

$V_B=40km/hr$

Substituting value of $V_B$ in equation (2) −

$(40+20)\times 6=40T$

$T=9minutes$

Speed of the bus is $40km/h$ and time interval after which each bus leaves is $9$ minutes.