We are given 8 problems. Let's solve them one by one. Problem 1: A man receives a total of $17800$ over four years, and each year he is paid $500$ more than the previous year. We need to find his salary in the first year. Problem 2: Samir buys $x$ cans of soda at $30$ cents each and $(x+4)$ cans of soda at $35$ cents each. The total cost is $3.35$. We need to find $x$. Problem 3: The length of a straight line $ABC$ is $5$ m. The ratio $AB:BC = 2:5$. We need to find the length of $AB$. Problem 4: The opposite angles of a cyclic quadrilateral are $(3x+10)^\circ$ and $(2x+20)^\circ$. We need to find the angles. Problem 5: The interior angles of a hexagon are in the ratio $1:2:3:4:5:9$. We need to find the angles. Problem 6: A man is 32 years older than his son. Ten years ago he was three times as old as his son was then. We need to find the present age of each. Problem 7: 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 m/s. We need to find the distance to the marker. Problem 8: A car completes a journey in 10 minutes. For the first half of the distance, the speed was 60 km/h, and for the second half, the speed was 40 km/h. We need to find the total distance of the journey.

1. Problem Description

We are given 8 problems. Let's solve them one by one.

Problem 1: A man receives a total of $17800$ over four years, and each year he is paid $500$ more than the previous year. We need to find his salary in the first year.

Problem 2: Samir buys $x$ cans of soda at $30$ cents each and $(x+4)$ cans of soda at $35$ cents each. The total cost is $3.35$. We need to find $x$.

Problem 3: The length of a straight line $ABC$ is $5$ m. The ratio $AB:BC = 2:5$. We need to find the length of $AB$.

Problem 4: The opposite angles of a cyclic quadrilateral are $(3x+10)^\circ$ and $(2x+20)^\circ$. We need to find the angles.

Problem 5: The interior angles of a hexagon are in the ratio $1:2:3:4:5:9$. We need to find the angles.

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

Problem 7: 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 m/s. We need to find the distance to the marker.

Problem 8: A car completes a journey in 10 minutes. For the first half of the distance, the speed was 60 km/h, and for the second half, the speed was 40 km/h. We need to find the total distance of the journey.

2. Solution Steps

Problem 1:

Let $x$ be the salary in the first year. Then the salaries for the next three years are $x+500$, $x+1000$, and $x+1500$.

The total salary is $x + (x+500) + (x+1000) + (x+1500) = 17800$.

$4x + 3000 = 17800$

$4x = 14800$

$x = 3700$

Problem 2:

The cost of $x$ cans is $0.30x$.

The cost of $(x+4)$ cans is $0.35(x+4)$.

The total cost is $0.30x + 0.35(x+4) = 3.35$.

$0.30x + 0.35x + 1.40 = 3.35$

$0.65x = 1.95$

$x = \frac{1.95}{0.65} = 3$

Problem 3:

$AB:BC = 2:5$. Let $AB = 2x$ and $BC = 5x$.

$AB + BC = 5$

$2x + 5x = 5$

$7x = 5$

$x = \frac{5}{7}$

$AB = 2x = 2 \times \frac{5}{7} = \frac{10}{7}$

Problem 4:

The opposite angles of a cyclic quadrilateral add up to $180^\circ$.

$(3x+10) + (2x+20) = 180$

$5x + 30 = 180$

$5x = 150$

$x = 30$

The angles are $3(30) + 10 = 100^\circ$ and $2(30) + 20 = 80^\circ$.

Problem 5:

The interior angles of a hexagon add up to $720^\circ$.

The ratio of the angles is $1:2:3:4:5:9$. Let the angles be $x, 2x, 3x, 4x, 5x, 9x$.

$x + 2x + 3x + 4x + 5x + 9x = 720$

$24x = 720$

$x = 30$

The angles are $30^\circ, 60^\circ, 90^\circ, 120^\circ, 150^\circ, 270^\circ$.

Problem 6:

Let the man's present age be $M$ and the son's present age be $S$.

$M = S + 32$

Ten years ago, the man's age was $M - 10$ and the son's age was $S - 10$.

$M - 10 = 3(S - 10)$

$M - 10 = 3S - 30$

$M = 3S - 20$

$S + 32 = 3S - 20$

$2S = 52$

$S = 26$

$M = 26 + 32 = 58$

Problem 7:

Let $d$ be the distance to the marker.

Time taken to run to the marker is $t_1 = \frac{d}{5}$.

Time taken to run back is $t_2 = \frac{d}{4}$.

Total time is $t_1 + t_2 = 15$ minutes $= 15 \times 60 = 900$ seconds.

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

$\frac{4d + 5d}{20} = 900$

$9d = 18000$

$d = 2000$ meters.

Problem 8:

Let $d$ be the total distance. Let $d/2$ be the first half of the distance and $d/2$ the second half of the distance.

Time taken for the first half is $t_1 = \frac{d/2}{60} = \frac{d}{120}$.

Time taken for the second half is $t_2 = \frac{d/2}{40} = \frac{d}{80}$.

Total time is $t_1 + t_2 = 10$ minutes $= \frac{10}{60} = \frac{1}{6}$ hours.

$\frac{d}{120} + \frac{d}{80} = \frac{1}{6}$

$\frac{2d + 3d}{240} = \frac{1}{6}$

$5d = 40$

$d = 8$ km.

3. Final Answer

1. $3700

2. $3

3. $\frac{10}{7}$ m

4. $100^\circ, 80^\circ$

5. $30^\circ, 60^\circ, 90^\circ, 120^\circ, 150^\circ, 270^\circ$

6. Man: 58 years, Son: 26 years

7. 2000 meters

8. 8 km