A takes 30 minutes more than B to cover a distance of 15 km at a certain speed. But if A doubles his speed, he takes one hour less than B to cover the same distance. What is the speed (in km/h) of B?

Let the speed of A is $$V_1$$ and speed of B is $$V_2$$.

Let B takes t hour to cover 15Km distance.

So, time taken by B to cover 15Km, $$t=\dfrac{15}{V_2}$$

time taken by A to cover 15Km $$t+0.5=\dfrac{15}{V_1}$$ --------(i)

As per the question, when $$V_1=2 \times$$ the initial velocity

Then time taken by A to cover 15Km, $$t-1=\dfrac{15}{2V_1}$$ --------(ii)

From equation (i) and (ii),

$$\Rightarrow \dfrac{t+0.5}{t-1}=\dfrac{\dfrac{15}{V_1}}{\dfrac{15}{2V_1}}$$

$$\Rightarrow  t+0.5=2t-2$$

$$\Rightarrow 2t-t=2+0.5$$hour

$$\Rightarrow t=2.5=2\dfrac{1}{2}$$hour

Hence the speed of the car B$$V_2=\dfrac{15}{2.5}=6 km/h$$