# From point A to point B, the cyclist traveled along a 48 km long road, and he returned back along another road

**From point A to point B, the cyclist traveled along a 48 km long road, and he returned back along another road, which is 8 km shorter than the first. Having increased the speed by 4 km / h on the way back, the cyclist spent 1 hour less than on the path from A to B. How fast did the cyclist go from point A to point B?**

Let x be the speed of the cyclist. Let us express the time that he spent on the way from A to B: 48 / x hours.

On the way back, he increased the speed by 4 km / h, the distance is 48 – 8 = 40 km, we express the time on the way back: 40 / (x + 4) hours.

On the way from A to B, he spent more time for 1 hour: 48 / x – 40 / (x + 4) = 1.

(48x + 192 – 40x) / x (x + 4) = 1;

(8x + 192) / (x² + 4x) = 1.

According to the rule of proportion:

x² + 4x = 8x + 192;

x² + 4x – 8x – 192 = 0;

x² – 4x – 192 = 0.

D = 16 + 768 = 784 (√D = 28);

x1 = (4 – 28) / 2 = -17 (not suitable).

x2 = (4 + 28) / 2 = 16 (km / h).

Answer: from A to B the cyclist rode at a speed of 16 km / h.