The cities of Kalsdale and Lamsville are connected by a railway line whose length in kilometers is divisible by 264.
A train started from Kalsdale towards Lamsville and, 6 hours after departing, developed an engine snag which forced it to continue at 5/7 velocity and under an hour late.
If the engine snag had occurred 98 kilometers later, it would have arrived 24 minutes sooner at its destination.
How far apart are Kalsdale and Lamsville, what was the initial speed of the train, and how many minutes late was the train?
Let d1 be the distance (km) traveled in the first 6 hours and v be the initial velocity (km/hr).
Since 98km at reduced speed takes 24 minutes more than it would at full speed,
98*7/(5*v) = 98/v + 24/60
which solves as v = 98 km/hr
which indicates d1 = 588.
So the distance is at least 588+98 = 686 km, which is 158 km over a multiple of 264. The next multiple of 264 is 792 km, which is therefore the distance between K and L.
At normal speed, 792 km would be covered in 792/98 = 8 + 8/98 hours.
As it happened, the first 588 miles was covered in 6 hours, and the remaining 204 miles in 204*7/(98*5) = 8 + 448/490 hours.
448/490  8/98 = 32/35  4/49 = 204/245 hours or about 49.959... minutes that the train arrived late.

Posted by Charlie
on 20060619 15:01:34 