Rahul starts on his journey at 5 pm at a constant speed so that he reaches his destination at 11 pm the same day. However, on his way, he stops for 20 minutes, and after that, increases his speed by 3 km per hour to reach on time. If he had stopped for 10 minutes more, he would have had to increase his speed by 5 km per hour to reach on time. His initial speed, in km per hour, was

The correct answer is Option 15.

Let us break down the problem step-by-step to determine Rahul's initial speed.

Step 1: Total time of journey
Rahul starts his journey at 5 pm and reaches his destination at 11 pm.
Total time available to complete the journey, T = 11 pm - 5 pm = 6 hours.

Let the total distance of the journey be D km, and Rahul's initial constant speed be v km/hr.

Under normal circumstances without any stops, the total distance is given by:

$D=v\times 6$

Step 2: Case 1 - Stopping for 20 minutes
Rahul stops for 20 minutes. Expressing 20 minutes in hours:
20 minutes = 20 / 60 hours = 1 / 3 hour.

Let t be the time (in hours) Rahul traveled at his initial speed v before stopping.
The distance covered in time t is $D_1=v·t$.

The remaining time to reach on time is:
$t_{\text{remaining, 1}}=6-t-\frac{1}{3}=\frac{17}{3}-t$ hours.

For the remaining distance, his speed is increased by 3 km/hr, so the new speed is (v + 3) km/hr.
The remaining distance covered is $D_2=(v+3)\times (\frac{17}{3}-t)$.

Since total distance $D=D_1+D_2$:

$6v=vt+(v+3)(\frac{17}{3}-t)$

Expanding the right-hand side:

Simplifying the equation:

Multiplying by 3:

$v=51-9t$     --- (Equation 1)

Step 3: Case 2 - Stopping for 10 minutes more (30 minutes total)
If he stopped for 10 minutes more, his total stoppage time is 20 + 10 = 30 minutes = 1/2 hour.

The remaining time to reach on time is:
$t_{\text{remaining, 2}}=6-t-\frac{1}{2}=\frac{11}{2}-t$ hours.

In this case, his speed for the remaining distance is increased by 5 km/hr, making it (v + 5) km/hr.
Equating the total distance again:

$6v=vt+(v+5)(\frac{11}{2}-t)$

Expanding the right-hand side:

Simplifying the equation:

Multiplying by 2:

$v=55-10t$     --- (Equation 2)

Step 4: Solving for initial speed v
Equating Equation 1 and Equation 2:

$51-9t=55-10t$

$10t-9t=55-51$

$t=4\text{hours}$

Substitute t = 4 into Equation 1:

$v=51-9\left(4\right)$

$v=51-36=15\text{km/hr}$

Thus, Rahul's initial speed was 15 km per hour.