⚡ AZUCATION SHORTCUT:
Let remaining time be $T$ and initial speed be $v$.
Case 1: $vT = (v+3)(T - 1/3) \implies 3T - v/3 = 1$
Case 2: $vT = (v+5)(T - 1/2) \implies 5T - v/2 = 2.5$
Solving these two linear equations:
$9T - v = 3$ and $10T - v = 5$.
Subtracting gives $T = 2$. Substituting $T=2$ gives $v = 9(2) - 3 = \mathbf{15}$.

Step 1: Set up the variables
Let Rahul's initial speed be $v$ km/hr.
Suppose he travels for $x$ hours before stopping. The remaining time left to reach on time is $T$ hours (where $T = 6 - x$).
The remaining distance to be covered is $D = v \times T$.

Step 2: Scenario 1 (20-minute stop)
He stops for 20 minutes ($\frac{1}{3}$ hr), so he must cover distance $D$ in $(T - \frac{1}{3})$ hours at speed $(v + 3)$.
$$vT = (v + 3)(T - \frac{1}{3})$$ $$vT = vT - \frac{v}{3} + 3T - 1$$ $$3T - \frac{v}{3} = 1 \implies 9T - v = 3 \quad \dots(1)$$

Step 3: Scenario 2 (30-minute stop)
He stops for 30 minutes ($\frac{1}{2}$ hr), so he must cover distance $D$ in $(T - \frac{1}{2})$ hours at speed $(v + 5)$.
$$vT = (v + 5)(T - \frac{1}{2})$$ $$vT = vT - \frac{v}{2} + 5T - 2.5$$ $$5T - \frac{v}{2} = 2.5 \implies 10T - v = 5 \quad \dots(2)$$

Step 4: Final Calculation
Subtract equation (1) from equation (2):
$$(10T - v) - (9T - v) = 5 - 3$$ $$\mathbf{T = 2 \text{ hours}}$$ Substitute $T = 2$ in equation (1):
$$9(2) - v = 3 \implies 18 - v = 3 \implies \mathbf{v = 15 \text{ km/hr}}.$$ The initial speed was **15 km/hr**.

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