Question: 21. 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 \mathrm{~km} / \mathrm{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.
(1) 9 \mathrm{~min}, 40 \mathrm{~km} / \mathrm{h}
(2) 25 \mathrm{~min}, 100 \mathrm{~km} / \mathrm{h}
(3) 10 \mathrm{~min}, 90 \mathrm{~km} / \mathrm{h}
(4) 15 \mathrm{~min}, 120 \mathrm{~km} / \mathrm{h}
Answer: Option (4)
Explanation:
Let the speed of each bus be v km/h and buses leave every T minutes. In one hour, number of buses starting from either city is \frac{60}{T}. Thus the spacing between successive buses is
\text{Distance between buses}=\frac{vT}{60}.For buses moving in the same direction as the girl, the relative speed is
v-60.She meets such a bus every 30 minutes, so
(v-60)\cdot \frac{30}{60}=\frac{vT}{60}.Simplifying:
\frac{v-60}{2}=\frac{vT}{60} \quad \Rightarrow \quad T=\frac{30(v-60)}{v}.For buses approaching from the opposite direction, relative speed is
v+60.She meets one every 10 minutes, so
(v+60)\cdot \frac{10}{60}=\frac{vT}{60}.Simplifying:
\frac{v+60}{6}=\frac{vT}{60} \quad \Rightarrow \quad T=\frac{10(v+60)}{v}.Equate the two expressions for T:
\frac{30(v-60)}{v}=\frac{10(v+60)}{v}.Multiply both sides by v:
30(v-60)=10(v+60).Expand and solve:
30v-1800=10v+600 20v=2400 v=120 \text{ km/h}Now substitute value of v into either expression for T:
T=\frac{30(v-60)}{v}=\frac{30(120-60)}{120}=15\text{ min}.Thus the bus speed is 120 \text{ km/h}
and the time interval is 15\text{ min}.