From point A to point B, the cyclist drove along one road 27 km long, and returned back along another road

From point A to point B, the cyclist drove along one road 27 km long, and returned back along another road, which was 7 km shorter than the first. Although on the way back the cyclist reduced his speed by 3 km / h, he still spent 10 minutes less time on the way back than on the way from A to B. How fast did the cyclist go from A to B?

Let’s find the length of the road on the way back:

27 – 7 = 20.

We take the speed of the cyclist from A to B for x, then:

x – 3 – speed on the way back:

27 / x – time of movement from A to B.

20 / (x – 3) – time from B to A.

Since the time difference was 10 minutes = 1/6 hour, we get the equation:

27 / x – 20 / (x – 3) = 1/6

27x – 81 – 20x = 1/6 * x * (x -1)

42 * x – 486 = x ^ 2 – x

x ^ 2 – 43x + 486 = 0

x12 = (43 + -√ (1849 -4 * 486)) / 2 = 26.5 km / h.