Let the time Andrew takes to run the 200-meter race be $t_A$ seconds.

We can break this problem into two parts based on the information provided:

1. Andrew beats Benzie by 8 seconds

Let $v_A$ be Andrew's speed and $v_B$ be Benzie's speed. We know that:

$v_A = \frac{200}{t_A}$

Since Andrew beats Benzie by 8 seconds, Benzie's time to finish the 200-meter race is:

$t_B = t_A + 8$

Thus, Benzie's speed can be expressed as:

$v_B = \frac{200}{t_A + 8}$

2. Head start of 32 meters

In the second scenario, Andrew gives Benzie a head start of 32 meters, meaning Benzie only needs to run 168 meters. Both finish the race at the same time, so their times are equal in this case:

$\frac{168}{v_B} = \frac{200}{v_A}$

Now, substitute $v_A$ and $v_B$ in terms of $t_A$ from the earlier expressions:

$\frac{168}{\frac{200}{t_A + 8}} = \frac{200}{\frac{200}{t_A}}$

Simplifying this equation:

$168 \cdot \frac{t_A + 8}{200} = t_A$

$168(t_A + 8) = 200 t_A$

$168 t_A + 1344 = 200 t_A$

$200 t_A - 168 t_A = 1344$

$32 t_A = 1344$

$t_A = \frac{1344}{32} = 42$

Thus, Andrew takes 42 seconds to run the 200-meter race.

Would you like more details or have any questions? Here are five related questions to explore:

1. What is Benzie's time to finish the 200-meter race?
2. What are Andrew's and Benzie's speeds during the race?
3. How much of a head start would Benzie need to beat Andrew by 2 seconds?
4. If Andrew's speed increased by 10%, how long would he take to finish the race?
5. What if the race were 400 meters instead—how long would each take then?

Tip: Always ensure units are consistent (time in seconds, distance in meters) when solving rate problems.