After traveling exactly half of its journey from station A to station B, a train was held up for 10 minutes. In order to arrive
at City B on schedule, the engine driver had to increase the speed of the train by 12 km/hour. Find the original speed of the train before it was held up, if it is known that the distance between the two stations is 120 km.
Let's define variables: s = original speed s + 12 = faster speed The time for the half of the route is: 60 / s The time for the second half of the route is: 60 / (s + 12) The equation for the time of the trip is: 60 / s + 60 / (s + 12) + 1/6 = 120 / s Where, 1/6: held up for 10 minutes (in hours). Rewriting the equation we have: 6s (60) + s (s + 12) = 60 * 6 (s + 12) 360s + s ^ 2 + 12s = 360s + 4320 s ^ 2 + 12s = 4320 s ^ 2 + 12s - 4320 = 0 We factor the equation: (s + 72) (s-60) = 0 We take the positive root so that the problem makes physical sense. s = 60 Km / h Answer: The original speed of the train before it was held up is: s = 60 Km / h
The nearest million is 46 million since if the number to the right of "nearest so and so" is less than 5, you round down but if it is 5 or more then you round up. Hope this helps!
