# A bus departed from point M to point N. After 30 minutes, a passenger car left point

A bus departed from point M to point N. After 30 minutes, a passenger car left point N for point M at a speed that exceeded the speed of the bus by 18 km. After 1 hour and 20 minutes, he met a bus, and traveled 3 km more than the bus. Find the distance between points M and N.

1. Take for x (km / h) the speed of the bus, the speed of the car (x + 18) km / h.

2. We translate minutes into hours:

30 minutes = 1/2 hour.

3. The bus left its settlement earlier than the car left and was on the way to the meeting point:

1/2 + 4/3 = 11/6 hours.

4. Considering that the car has traveled 3 kilometers more than the bus to the meeting point, we will compose the equation:

4 (x + 18) / 3 – 11x / 6 = 3;

8x + 144 – 11x = 18;

3x = 126;

x = 42 km / h – bus speed.

1 hour 20 minutes = 1 1/3 hours = 34/3 hours.

Vehicle speed 42 + 18 = 60 km / h.

5. We calculate the distance between settlements: 60 x 4/3 + 42 x 11/6 = 80 + 77 =

157 kilometers.