We label Mathville as $M$, Algebratown as $A$, and the other intersection points of roads as shown.

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-012.jpg?height=274&width=458&top_left_y=1701&top_left_x=888)

There is 1 route from $M$ to each of $C$ and $B: M \rightarrow C$ and $M \rightarrow B$.

There are 3 routes to $D: M \rightarrow D$ and $M \rightarrow C \rightarrow D$ and $M \rightarrow B \rightarrow D$.

This means that there are 4 routes to $F$ :

$$
M \rightarrow C \rightarrow F \quad M \rightarrow D \rightarrow F \quad M \rightarrow C \rightarrow D \rightarrow F \quad M \rightarrow B \rightarrow D \rightarrow F
$$

Similarly, there are 4 routes to $E$ :

$$
M \rightarrow B \rightarrow E \quad M \rightarrow D \rightarrow E \quad M \rightarrow C \rightarrow D \rightarrow E \quad M \rightarrow B \rightarrow D \rightarrow E
$$

Finally, there are $4+4=8$ routes to $A$, since every route comes through either $E$ or $F$, no route goes through both $E$ and $F$, and there are 4 routes to each of $E$ and $F$.

ANsWer: $(\mathrm{C})$
On the map shown, there are a number of routes from Mathville to Algebratown.

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-009.jpg?height=215&width=726&top_left_y=955&top_left_x=664)

Each route must travel along the roads in the direction marked by the arrows. The total number of routes from Mathville to Algebratown is
(A) 3
(B) 4
(C) 8
(D) 6
(E) 10

## Part B: Each correct answer is worth 6.