11. Mahmoud runs to a marker and back in 15 minutes. His speed on the way to the marker is 5 m/s and his speed on the way back is \( 4 \mathrm{~m} / \mathrm{s} \). Find the distance to the marker.

To find the distance to the marker, we can use the relationship between speed, distance, and time. Let's denote the distance to the marker as \( d \). When Mahmoud runs to the marker, the time taken is: \[ t_1 = \frac{d}{5 \text{ m/s}} \] When he comes back, the time taken is: \[ t_2 = \frac{d}{4 \text{ m/s}} \] The total time for the round trip is given as 15 minutes, or \( 15 \times 60 = 900 \) seconds. So we can set up the equation: \[ t_1 + t_2 = 900 \text{ seconds} \] Substituting for \( t_1 \) and \( t_2 \): \[ \frac{d}{5} + \frac{d}{4} = 900 \] To solve this equation, we first find a common denominator, which is 20: \[ \frac{4d}{20} + \frac{5d}{20} = 900 \] Combining the terms gives: \[ \frac{9d}{20} = 900 \] Now, multiply both sides by 20: \[ 9d = 18000 \] Finally, dividing by 9 gives: \[ d = 2000 \text{ meters} \] So, the distance to the marker is **2000 meters**.