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 the 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 x be the speed of the cyclist. Let us express the time that he spent on the way from A to B: 27 / x hours.
On the way back, he reduced the speed by 3 km / h, the distance is 27 – 7 = 20 km, we express the time on the way back: 20 / (x – 3) hours.
On the way from A to B, he spent more time for 10 minutes = 1/6 hour: 27 / x – 20 / (x – 3) = 1/6.
(27x – 81 – 20x) / x (x – 3) = 1/6.
(7x – 81) / (x² – 3x) = 1/6.
According to the rule of proportion:
x² – 3x = 6 (7x – 81);
x² – 3x – 42x + 486 = 0;
x² – 45x + 486 = 0;
D = 2025 – 1944 = 81 (√D = 9);
x1 = (45 – 9) / 2 = 18 (km / h).
x2 = (45 + 9) / 2 = 27 (km / h).
Answer: from A to B the cyclist rode at a speed of 18 km / h or 27 km / h.