A bus starts at 9 am and follows a fixed route every day. One day, it traveled at a constant speed of 60 km per hour and reached its destination 3.5 hours later than its scheduled arrival time. Next day, it traveled two-thirds of its route in one-third of its total scheduled travel time, and the remaining part of the route at 40 km per hour to reach just on time. The scheduled arrival time of the bus is

Question

Accepted Answer

#### Step 1: Define variables
Let the scheduled time taken by the bus to complete its journey be $$ T $$ hours.  
The total distance traveled by the bus is given as $$ S = 60(T + 3.5) $$, based on the first day's travel.

#### Step 2: Analyze the second day's travel
On the second day:
- The bus covered two-thirds of the distance ($$ \frac{2}{3}S $$) in one-third of its scheduled time ($$ \frac{1}{3}T $$).
- The remaining one-third of the distance ($$ \frac{1}{3}S $$) was covered at a speed of 40 km/h in two-thirds of the scheduled time ($$ \frac{2}{3}T $$).

#### Step 3: Relate speed and time
From the second day's travel:
- For the first part:  
$$
	ext{Speed for } \frac{2}{3}S = \frac{\frac{2}{3}S}{\frac{1}{3}T} = \frac{2S}{T}.
$$  
This speed is given as 40 km/h. Thus:  
$$
\frac{2S}{T} = 40.
$$

Simplify to find $$ S/T $$:  
$$
S/T = 20.
$$

The usual speed of the bus is $$ S/T = 20 	imes 4 = 80 $$ km/h.

#### Step 4: Solve for $$ T $$
From the first day's travel:  
$$
60(T + 3.5) = 80T.
$$  
Simplify:  
$$
60T + 210 = 80T,
$$  
$$
20T = 210,
$$  
$$
T = 10.5.
$$

#### Step 5: Calculate scheduled arrival time
The bus usually takes $$ T = 10.5 $$ hours for its journey. Starting at **9:00 AM**, it will complete its journey at:  
$$
9:00 + 10.5 	ext{ hours} = 7:30 PM.
$$

Thus, the scheduled arrival time of the bus is **7:30 PM**.

Answer

Answer

Answer