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.

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 = 60

Since we know that the time can't be a negative number, the only available option left is 60 km/h.

hi,

i'm 75% sure the answer is 1,1 to both lines.

have a great day

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