In the given diagram (not draw to scale), railway stations A, B, C, P and Q are connected by straight tracks. Track PQ is parallel to BC. The time taken by a train travelling at 90 km/hr to reach B from A by the shortest route is :

1. 8 minutes

2. 12 minutes

3. 16.8 minutes

4. 20 minutes

In △ APQ and △ ABC,

⇒ ∠PAQ = ∠BAC (Common angle)

⇒ ∠APQ = ∠ABC (Corresponding angles are equal)

∴ △ APQ ~ △ ABC (By A.A. axiom)

From figure,

Let AP = x km.

We know that,

Corresponding sides of similar triangle are proportional.

$\Rightarrow \dfrac{AP}{AB} = \dfrac{PQ}{BC} \\[1em] \Rightarrow \dfrac{x}{AP + PB} = \dfrac{20}{50} \\[1em] \Rightarrow \dfrac{x}{x + 18} = \dfrac{2}{5} \\[1em] \Rightarrow 5x = 2(x + 18) \\[1em] \Rightarrow 5x = 2x + 36 \\[1em] \Rightarrow 5x - 2x = 36 \\[1em] \Rightarrow 3x = 36 \\[1em] \Rightarrow x = \dfrac{36}{3} \\[1em] \Rightarrow x = 12.$

AB = AP + BP = 12 + 18 = 30 km.

Time = $\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{AB}{90} = \dfrac{30}{90} = \dfrac{1}{3} \text{ hr} = \dfrac{1}{3} \times 60$ = 20 minutes.

Hence, Option 4 is the correct option.