From city A to city B, the distance between which is 60 km, 2 cyclists left at the same time. The speed of one of them

From city A to city B, the distance between which is 60 km, 2 cyclists left at the same time. The speed of one of them is 4 km / h less than the speed of the other, so he arrived in city B 1h15 min later than the other cyclist. How much time was spent on the first 12 km of the journey by the cyclist who was traveling at a lower speed?

Let the speed of the first cyclist be x km / h, then the speed of the second cyclist is (x – 4) km / h. The first cyclist spent 60 / x hours on the way from city A to city B, and the second cyclist – (60 / (x – 4)) hours. By the condition of the problem, it is known that the second cyclist was on the way longer than the first one by (60 / (x – 4) – 60 / x) hours or by 1 hour 15 minutes = 1 15/60 hours = 1 1/4 hours = 1.25 h. Let’s make an equation and solve it.

60 / (x – 4) – 60 / x = 1.25;

O.D.Z. x ≠ 4; x ≠ 0;

60x – 60 (x – 4) = 1.25x (x – 4);

60x – 60x + 240 = 1.25x ^ 2 – 5x;

1.25x ^ 2 – 5x – 240 = 0;

x ^ 2 – 4x – 192 = 0;

D = b ^ 2 – 4ac;

D = (-4) ^ 2 – 4 * 1 * (-192) = 16 + 768 = 784; √D = 28;

x = (- b ± √D) / (2a);

x1 = (4 + 28) / 2 = 32/2 = 16 (km / h) – speed of the 1st;

x2 = (4 – 28) / 2 = -24/2 = -12 – speed is never negative;

x – 4 = 16 – 4 = 12 (km / h) – speed of the 2nd.

Let’s find how many hours the second cyclist spent on the first 12 km of the journey.

12: 12 = 1 (h).

Answer. 1 hour.