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

Answer: (A)

1:15 PM

Answer 2

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:
$\boxed{1:15 \, \text{pm}}.$