QA train passes a 264-meter-long post and platform in 8 seconds and 20 seconds, respectively. What is the train's speed?

Time taken to move the post = 8 s 
=> x/v = 8 
=> x = 8v --- (1)

Time taken to cross the platform 264 m long = 20 s
(x + 264)/v = 20
=> x + 264 = 20v ---(2)

Equation (1) represents the time it takes for the train to move past the post, which is 8 seconds. The distance the train travels in this time is given by x, which is the length of the train. The speed of the train is given by v, so x/v represents the time it takes for the train to move a distance x at a speed v. Setting this equal to 8 seconds allows you to solve for the speed of the train.

Equation (2) represents the time it takes for the train to cross the platform, which is 20 seconds. The distance the train travels in this time is given by x + 264, where x is the length of the train and 264 is the length of the platform. Setting this equal to 20v allows you to solve for the speed of the train.

You can then use these two equations to solve for the speed of the train by substituting the first equation into the second and solving for v. This gives you the final result of v = 22 m/s, or 79.2 km/hr.