Analysis of Motion:

Let the speed of the cyclists in still air be $v = 20 \text{ km/h}$. Let the speed of the wind be $w$. The distance between towns A and B is $L = 10 \text{ km}$.

The wind blows from A to B. Therefore:

• Velocity from A to B (with wind): $v_1 = v + w$
• Velocity from B to A (against wind): $v_2 = v – w$

First Meeting:

The cyclists start simultaneously. Let $C_A$ start from A and $C_B$ start from B. They meet at a distance $x_1 = 2 \text{ km}$ from B.

Distance traveled by $C_A$ (downwind): $L – x_1 = 10 – 2 = 8 \text{ km}$.
Distance traveled by $C_B$ (upwind): $x_1 = 2 \text{ km}$.

Since time is equal:

$$ \frac{8}{v+w} = \frac{2}{v-w} \implies 4(v-w) = v+w \implies 3v = 5w $$ $$ w = \frac{3}{5}v = \frac{3}{5}(20) = 12 \text{ km/h} $$

So, $v_1 = 32 \text{ km/h}$ and $v_2 = 8 \text{ km/h}$.

Second Meeting:

After the first meeting, they continue to the towns, turn around (after resting), and meet again 6 km from B.

Let’s check the position “6 km from B”. This is $4 \text{ km}$ from A.

• $C_A$ goes $A \to B$, rests, then $B \to \text{Meet}_2$.
  Path: $10 \text{ km}$ (at $v_1$) + Rest + $6 \text{ km}$ (at $v_2$).
• $C_B$ goes $B \to A$, rests, then $A \to \text{Meet}_2$.
  Path: $10 \text{ km}$ (at $v_2$) + Rest + $4 \text{ km}$ (at $v_1$).

Let the rest time be $\Delta t$. The problem states “a cyclist takes some rest”. This implies only one might rest, or they rest differently. However, the symmetry suggests we must find “in which town and for what period”. This implies the rest occurs at the destination town before the return trip.

Time for $C_A$ to reach $B$: $t_{A \to B} = \frac{10}{32} \text{ h}$.
Time for $C_B$ to reach $A$: $t_{B \to A} = \frac{10}{8} = 1.25 \text{ h}$.

Clearly $C_A$ arrives much earlier. Let’s assume the one who arrives first rests or the question implies a specific rest scenario.

Let’s equate the total time to the second meeting point.

Case 1: Rest at B (by cyclist starting from A).

$$ T_1 = \frac{10}{32} + \Delta t + \frac{6}{8} $$ $$ T_2 = \frac{10}{8} + 0 + \frac{4}{32} \quad (\text{Assuming cyclist starting from B doesn’t rest, or rests 0}) $$ $$ T_2 = 1.25 + 0.125 = 1.375 \text{ h} $$ $$ T_1 = 0.3125 + 0.75 + \Delta t = 1.0625 + \Delta t $$

Equating times: $1.375 = 1.0625 + \Delta t \implies \Delta t = 0.3125 \text{ h}$.

Converting to minutes: $0.3125 \times 60 = 18.75 \text{ minutes}$.

Conclusion:

The cyclist who started from town A rests at town B for 18.75 minutes.