Question

A bus leaves the bus stand at 9 am and travels at a constant speed of 60 km/h. It reaches its destination 3.5 hours later than its original time. The next day, it travels \( \frac{2}{3} \) of its route within \( \frac{1}{3} \) of its initial time and travels the rest of the route within 40 minutes. At what time does it reach its destination normally?

• A. 1:15 PM
• B. 1:17 PM
• C. 3:12 PM
• D. 1:25 PM

Answer 1

The Correct Option is A

Let the total distance of the route be \(D\) kilometers, and the normal time to reach the destination be \(T\) hours.
Step 1: Time taken on the first day
On the first day, the bus travels at a speed of 60 km/h and reaches the destination 3.5 hours later than the normal time. The time taken on the first day is:
\[\text{Time taken on day 1} = T + 3.5.\]
The distance traveled is \(D\), and the speed is 60 km/h, so:
\[\text{Time taken on day 1} = \frac{D}{60}.\]
Equating the two expressions for time taken on day 1:
\[\frac{D}{60} = T + 3.5. \quad \text{(1)}\]
Step 2: Time taken on the second day
On the second day, the bus travels \(\frac{2}{3}\) of its route within \(\frac{1}{3}\) of its initial time. The time taken for this part of the journey is:
\[\frac{1}{3}T.\]
The remaining \(\frac{1}{3}\) of the route is traveled in 40 minutes, which is \(\frac{2}{3}\) hours.
So, the time taken on the second day is:
\[\text{Time taken on day 2} = \frac{1}{3}T + \frac{2}{3}.\]
The total distance traveled on the second day is also \(D\), and the speed is again 60 km/h:
\[\text{Time taken on day 2} = \frac{D}{60}.\]
Equating the two expressions for time taken on day 2:
\[\frac{D}{60} = \frac{1}{3}T + \frac{2}{3}. \quad \text{(2)}\]
Step 3: Solve the system of equations
Now, solve equations (1) and (2) simultaneously.
From equation (1):
\[\frac{D}{60} = T + 3.5,\]
\[D = 60(T + 3.5). \quad \text{(3)}\]
From equation (2):
\[\frac{D}{60} = \frac{1}{3}T + \frac{2}{3},\]
\[D = 60\left(\frac{1}{3}T + \frac{2}{3}\right),\]
\[D = 20T + 40. \quad \text{(4)}\]
Step 4: Set equations (3) and (4) equal
Now, equate equations (3) and (4):
\[60(T + 3.5) = 20T + 40.\]
Simplifying:
\[60T + 210 = 20T + 40,\]
\[60T - 20T = 40 - 210,\]
\[40T = -170,\]
\[T = \frac{-170}{40} = 4.25 \, \text{hours}.\]
Step 5: Find the normal time
The normal time \(T = 4.25\) hours, which is 4 hours and 15 minutes. Since the bus leaves at 9 am, the normal time to reach the destination is:
\[9:00 \, \text{am} + 4 \, \text{hours} \, 15 \, \text{minutes} = 1:15 \, \text{pm}.\]
Final Answer
The bus reaches its destination normally at:
\[\boxed{1:15 \, \text{pm}}.\]