# From points A and B, the distance between which is 120 km, two buses left at the same time towards each other

From points A and B, the distance between which is 120 km, two buses left at the same time towards each other. On the way, the first made a stop for 10 minutes, the second – for 5 minutes. The first bus arrived at B 25 minutes earlier than the second arrived at A. It can be assumed that the bus speeds were constant, with the speed of the first bus exceeding the speed of the second bus by 20 km / h. How long did it take for the passengers of each of these buses to travel between points and B?

Let’s take the speed of the 1st bus for x, then:

x – 20 – speed of the 2nd;

120 / x – time of movement of the 1st;

120 / (x – 20) – 2nd;

120 / x + 1/6 – travel time of the 1st;

120 / (x – 20) + 1/12 – 2nd.

Since the 1st arrived 5/12 hours earlier, we get the equation:

120 / (x – 20) + 1/12 – (120 / x + 1/6) = 5/12

120 / (x – 20) – 120 / x = 1/6

120x – 120x + 2400 = 1/6 * (x – 20) * x

x ^ 2 – 20x – 14400 = 0

x12 = (20 + -√400 – 4 * 1 * (144000)) / 2 = (20 + -140) / 2 = 80.

The negative root is meaningless.

80 – 20 = 60.

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