We have two problems to solve. Problem 6: A man is 32 years older than his son. Ten years ago, the man was three times as old as his son was then. We need to find the present age of the man and his son. Problem 7: Mahmoud runs to a marker and back in 15 minutes. His speed to the marker is 5 m/s and his speed on the way back is 4 m/s. We need to find the distance to the marker.

1. Problem Description

We have two problems to solve.

Problem 6: A man is 32 years older than his son. Ten years ago, the man was three times as old as his son was then. We need to find the present age of the man and his son.

Problem 7: Mahmoud runs to a marker and back in 15 minutes. His speed to the marker is 5 m/s and his speed on the way back is 4 m/s. We need to find the distance to the marker.

2. Solution Steps

Problem 6:

Let $m$ be the man's current age and $s$ be the son's current age.

From the first sentence, we have:

$m = s + 32$

Ten years ago, the man's age was $m - 10$ and the son's age was $s - 10$. From the second sentence, we have:

$m - 10 = 3(s - 10)$

Now we have a system of two equations with two variables:

1. $m = s + 32$

2. $m - 10 = 3(s - 10)$

Substitute equation (1) into equation (2):

$(s + 32) - 10 = 3(s - 10)$

$s + 22 = 3s - 30$

$2s = 52$

$s = 26$

Now substitute $s = 26$ into equation (1):

$m = 26 + 32$

$m = 58$

Problem 7:

Let $d$ be the distance to the marker in meters.

Let $t_1$ be the time it takes to run to the marker, and $t_2$ be the time it takes to run back.

We know that total time is 15 minutes, which is $15 \times 60 = 900$ seconds.

So, $t_1 + t_2 = 900$

We also know that distance = speed × time.

So, $d = 5t_1$ and $d = 4t_2$.

Therefore, $t_1 = \frac{d}{5}$ and $t_2 = \frac{d}{4}$.

Substitute these into the equation $t_1 + t_2 = 900$:

$\frac{d}{5} + \frac{d}{4} = 900$

Multiply by 20 to clear the fractions:

$4d + 5d = 18000$

$9d = 18000$

$d = 2000$

3. Final Answer

Problem 6: The man's present age is 58 years, and the son's present age is 26 years.

Problem 7: The distance to the marker is 2000 meters.