15 POINTS: 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.

Answer:So let's just mke things easier by imagining that x (km/h) was the original speed of the train.From the information, we can conclude that:- The train's speed when it went from station A to the stopping point was x (km/h) => The time it took to finish the first half of its journey was 60/x (h)- The speed of the train when it went from the stopping point to station B was x + 12 (km/h) => The time it took to finish the other half of its journey was 60/x + 12 (h)- The original time the train was supposed to take to finish the whole journey was 120/x (h)Since the train still arrived at city B on schedule, we have:    120/x = 60/x + 60/x + 12 + 1/6⇔ 6 × 120 × (x + 12) = 6 × 60 × (x+12) + 6 × 60x + x(x + 12)⇔ 720x + 8640 = 360x + 4320 + 360x + x² + 12x⇔ 720x - 360x - 360x - x² - 12x = -8640 + 4320⇔ - x² - 12x = -4320⇔ -x² - 12x + 4320 = 0⇔ -x² + 60x - 72x + 4320 = 0⇔ -x(x - 60) - 72(x - 60) = 0⇔ (-x - 72)(x - 60) = 0⇔ -x - 72 = 0 or x - 60 = 0⇔ -x = 72 or x = 60⇔ x = -72 or x = 60Since we know that the time can't be a negative number, the only available option left is 60 km/h.