X takes 3 hours more than Y to walk a distance of 30 km, but if X doubles his race, he is able to be ahead of Y by $1\dfrac{1}{2}$ hours, then the speed of their walking will be :

1. X’s speed = $\dfrac{10}{3}$ km/hr, Y’s speed = 5 km/hr

2. X’s speed = 5 km/hr, Y’s speed = $\dfrac{10}{3}$ km/hr

3. X’s speed = 10 km/hr, Y’s speed = $\dfrac{5}{3}$ km/hr

4. X’s speed = $\dfrac{5}{3}$ km/hr, Y’s speed = 10 km/hr

Let X's speed and Y's speed be x km/hr and y km/hr respectively.

Time = $\dfrac{\text{Distance}}{\text{Speed}}$

Given,

X takes 3 hours more than Y to walk 30 km.

⇒ $\dfrac{30}{x} = \dfrac{30}{y} + 3$

⇒ $\dfrac{30}{y} = \dfrac{30}{x} - 3$ ………(1)

Given,

If X doubles his race, he is able to be ahead of Y by $1\dfrac{1}{2}$ hours.

⇒ $\dfrac{30}{2x} = \dfrac{30}{y} - \dfrac{3}{2}$

⇒ $\dfrac{30}{y} = \dfrac{30}{2x} + \dfrac{3}{2}$ ………(2)

From equation (1) and (2), we get :

$\Rightarrow \dfrac{30}{x} - 3 = \dfrac{30}{2x} + \dfrac{3}{2} \\[1em] \Rightarrow \dfrac{30}{x} - \dfrac{30}{2x} = \dfrac{3}{2} + 3 \\[1em] \Rightarrow \dfrac{60 - 30}{2x} = \dfrac{3 + 6}{2} \\[1em] \Rightarrow \dfrac{30}{2x} = \dfrac{9}{2} \\[1em] \Rightarrow x = \dfrac{30}{9} = \dfrac{10}{3} \text{ km/hr}.$

Substituting value of x in equation (1), we get :

$\Rightarrow \dfrac{30}{y} = \dfrac{30}{\dfrac{10}{3}} - 3 \\[1em] \Rightarrow \dfrac{30}{y} = \dfrac{90}{10} - 3 \\[1em] \Rightarrow \dfrac{30}{y} = 9 - 3 \\[1em] \Rightarrow \dfrac{30}{y} = 6 \\[1em] \Rightarrow y = \dfrac{30}{6} = 5 \text{ km/hr}.$

Hence, option 1 is the correct option.