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 option is 15.

Let us understand the problem and solve it step-by-step to see why the initial speed is 15 km/h.

Rahul's journey starts at 5 pm and is planned to end at 11 pm. The total scheduled duration for the journey is:

$$11\text{ pm}-5\text{ pm}=6\text{ hours}$$

Let his initial speed be $v$ km/h. Thus, the total distance of the journey, $d$ (in km), is given by:

$$d=6v$$

Suppose he travels at his initial speed $v$ for a duration of $t$ hours before stopping. The distance covered in this time is:

$$d_1=vt$$

The remaining distance to be covered is:

$$d_2=d-d_1=6v-vt=v(6-t)$$

Case 1: He stops for 20 minutes.
20 minutes is equal to $\frac{20}{60}=\frac{1}{3}$ hours.
To reach on time, the time remaining for the rest of the journey must be:

$$t_{\text{rem1}}=6-t-\frac{1}{3}$$

During this remaining time, he increases his speed by 3 km/h, making his new speed $v+3$ km/h. The equation for the remaining distance is:

$$(v+3)(6-t-\frac{1}{3})=v(6-t)$$

Let $y=6-t$ represent the remaining scheduled hours before he stopped. Substituting this, we get:

$$(v+3)(y-\frac{1}{3})=vy$$

Expanding the left side:

$$vy-\frac{v}{3}+3y-1=vy$$

Subtracting $vy$ from both sides gives:

$$3y-\frac{v}{3}-1=0\Rightarrow 3y=\frac{v}{3}+1$$

Multiplying by 3, we obtain our first relation:

$$9y=v+3\text{ — (Equation 1)}$$

Case 2: He stops for 10 minutes more (total 30 minutes of stop time).
30 minutes is equal to $\frac{30}{60}=\frac{1}{2}$ hours.
To reach on time in this case, his new speed must be increased by 5 km/h, making it $v+5$ km/h. The equation is:

$$(v+5)(y-\frac{1}{2})=vy$$

Expanding the left side:

$$vy-\frac{v}{2}+5y-\frac{5}{2}=vy$$

Subtracting $vy$ from both sides gives:

$$5y-\frac{v}{2}-\frac{5}{2}=0\Rightarrow 5y=\frac{v}{2}+\frac{5}{2}$$

Multiplying by 2, we obtain our second relation:

$$10y=v+5\text{ — (Equation 2)}$$

Solving the System of Equations:
We have two linear equations in $v$ and $y$:

1) $9y-v=3$

2) $10y-v=5$

Subtracting Equation 1 from Equation 2:

$$(10y-v)-(9y-v)=5-3$$

$$y=2$$

Substitute $y=2$ back into Equation 1 to find the initial speed $v$:

$$9(2)=v+3$$

$$18=v+3$$

$$v=18-3=15$$

Thus, his initial speed was 15 km/h.