# The distance between stations A and B is 120 km. After the train that left A to B, in 3 hours the second

**The distance between stations A and B is 120 km. After the train that left A to B, in 3 hours the second train departed in the same direction, the speed of which is 10 km / h more than the speed of the first train. It is known that the first train arrived at station B 2 hours earlier than the second. How many hours will the second train take from A to B?**

Let the speed of the first train be x km / h, then the speed of the second train is (x + 10) km / h. The first train covered the distance between stations in 120 / x hours, and the second train in 120 / (x + 10) hours. The second train left 3 hours later than the first and arrived at the station 2 hours later, which means it was on the way 3 – 2 = 1 hour less than the first or (120 / x – 120 / (x + 10)) hours. Let’s make an equation and solve it.

120 / x – 120 / (x + 10) = 1;

O.D.Z. x ≠ 0, x ≠ -10;

(120 (x + 10) – 120x) / (x (x + 10)) = (x (x + 10)) / (x (x + 10));

120 (x + 10) – 120x = x (x + 10);

120x + 1200 – 120x = x ^ 2 + 10x;

x ^ 2 + 10x – 1200 = 0;

D = b ^ 2 – 4ac;

D = 10 ^ 2 – 4 * 1 * (-1200) = 100 + 4800 = 4900; √D = 70;

x = (-b ± √D) / (2a);

x1 = (-10 + 70) / 2 = 60/2 = 30 (km / h) – the speed of the first;

x2 = (-10 – 70) / 2 = -80/2 = -40 – speed cannot be negative;

Let’s find the travel time of the second train:

120 / (x + 10) = 120 / (30 + 10) = 120/40 = 3 (h).

Answer. The second train traveled from A to B in 3 hours.