A goods train leaves a station at 6 p.m., followed by an express train which leaves at 8 p.m. and travels 20 km/hour faster than the goods train. The express train arrives at a station, 1040 km away, 36 minutes before the goods train. Assuming that speed of both the trains remain constant between the two stations; calculate their speeds.

Let speed of goods train be x km/hr

Speed of express train will be (x + 20) km/hr

Time taken to cover 1040 km by,

Goods train = $\dfrac{1040}{x}$ hours

Express train = $\dfrac{1040}{x + 20}$ hours

Since express train leaves 2 hours after the goods train and arrives 36 minutes before the goods train,

∴ It takes 2 hours 36 minutes less i.e., (2 x 60) + 36 = 156 minutes

$\therefore \dfrac{1040}{x} - \dfrac{1040}{x + 20} = \dfrac{156}{60} \\[1em] \Rightarrow \dfrac{1040(x + 20) - 1040x}{x(x + 20)} = \dfrac{13}{5} \\[1em] \Rightarrow \dfrac{1040x + 20800 - 1040x}{x^2 + 20x} = \dfrac{13}{5} \\[1em] \Rightarrow \dfrac{20800}{x^2 + 20x} = \dfrac{13}{5} \\[1em] \Rightarrow 20800 \times 5 = 13(x^2 + 20x) \\[1em] \Rightarrow 104000 = 13x^2 + 260x \\[1em] \Rightarrow 13x^2 + 260x - 104000 = 0 \\[1em] \Rightarrow 13(x^2 + 20x - 8000) = 0 \\[1em] \Rightarrow x^2 + 20x - 8000 = 0 \\[1em] \Rightarrow x^2 + 100x - 80x - 8000 = 0 \\[1em] \Rightarrow x(x + 100) - 80(x + 100) = 0 \\[1em] \Rightarrow (x - 80)(x + 100) = 0 \\[1em] \Rightarrow x - 80 = 0 \text{ or } x + 100 = 0 \\[1em] \Rightarrow x = 80 \text{ or } x = -100.$

Since speed cannot be negative,

∴ x = 80, (x + 20) = 100.

Hence, speed of goods train = 80 km/hr and express train = 100 km/hr.