, 2016

When Team A played Team B, if Team B won, then Team B scored more goals than Team A, and if the game ended in a tie, then Team A and Team B scored the same number of goals.

Therefore, if a team has 0 wins, 1 loss, and 2 ties, then it scored fewer goals than its opponent once (the 1 loss) and the same number of goals as its oppponent twice (the 2 ties).

Combining this information, we see that the team must have scored fewer goals than were scored against them.

In other words, it is not possible for a team to have 0 wins, 1 loss, and 2 ties, and to have scored more goals than were scored against them.

We can also examine choices (A), (B), (D), (E) to see that, in each case, it is possible that the team scored more goals than it allowed.

This will eliminate each of these choices, and allow us to conclude that (C) must be correct.

(A): If the team won 2-0 and 3-0 and tied 1-1, then it scored 6 goals and allowed 1 goal.

(B): If the team won 4-0 and lost 1-2 and 2-3, then it scored 7 goals and allowed 5 goals.

(D): If the team won 4-0, lost 1-2, and tied 1-1, then it scored 6 goals and allowed 3 goals.

(E): If the team won 2-0, and tied 1-1 and 2-2, then it scored 5 goals and allowed 3 goals.

Therefore, it is only the case of 0 wins, 1 loss, and 2 ties where it is not possible for the team to score more goals than it allows.

Solution 1

In the given diagram, we can see 3 of the 6 faces, or $\frac{1}{2}$ of the cube.

The remaining 3 faces (also $\frac{1}{2}$ of the cube) is unshaded.

Of the visible faces, $\frac{1}{2}$ of the area is shaded.

Therefore, the fraction of the total surface area that is shaded is $\frac{1}{2}\times \frac{1}{2} = \frac{1}{4}$.

Solution 2

Since the cube is $2 \times 2 \times 2$, the area of each face is $2\times 2 = 4$.

Since a cube has six faces, the total surface area of the cube is $6 \times 4 = 24$.

Each of the three faces that is partially shaded is one-half shaded, since each face is cut into two identical pieces by its diagonal.

Thus, the shaded area on each of these three faces is $2$, and so the total shaded area is $3 \times 2 = 6$.

Therefore, the fraction of the total surface area that is shaded is $\frac{6}{24}=\frac{1}{4}$.