Solution(s):

A line and circle can only intersect each other at at most two points. Two lines can only intersect at one point.

Therefore, the maximum number of intersections is \[ 2 \cdot 2 + 1 = 5. \]

Thus, D is the correct answer.

Solution(s):

We cannot use \(4\) or more dollar \($5\) bills since that would go over the total.

We can use three \($5\) bills and one \($2\) bill. We cannot use two \($5\) bills since we cannot make an odd amount of money with \($2\) bills.

Finally, we can use one \($5\) bill and six \($2\) bills. As above, we cannot use zero \($5\) bills. Therefore, there are \(2\) ways to get the total.

Thus, A is the correct answer.

Solution(s):

To get the smallest possible average, we want to use the smallest \(4\) positive even integers.

This can be achieved as follows: \[ \dfrac{2 + 4 + 6 + 8}{4} = \dfrac{20}{4}\]\[ = 5. \]

Thus, C is the correct answer.

Solution(s):

We don't want to increase the thousands digit, so we can keep that as \(2.\)

This means that we have to increase the tens and hundreds digits to \(1,\) to yield the next palindrome of \(2112.\) The product of its digits is \(4.\)

Thus, B is the correct answer.

Solution(s):

The days of the week cycle every \(7\) days. After \(700\) days, the day of the week will still be Saturday.

After \(6\) more days, the day of the week will be Friday.

Thus, C is the correct answer.

Solution(s):

The birdbath gains \(20 - 18 = 2\) milliliters of water per minute.

The birdbath will continue to gain water until it is fill, when the volume will remain constant.

Thus, A is the correct answer.

Solution(s):

There are a total of \[ 6 + 8 + 4 + 2 + 5 = 25 \] students in the class. The percent that chose \(E\) is \[ 100 \cdot \dfrac{5}{25} = \dfrac{100}{5} = 20 \%. \]

Thus, E is the correct answer.

Solution(s):

Note that France and Spain are the European countries. The number of ‘\(80\)s stamps from these countries respectively is \(15\) and \(9\) for a total of \[ 15 + 9 = 24 \] stamps.

Thus, D is the correct answer.

Solution(s):

Note that Brazil and Peru are the South American countries.

Brazil's ‘\(50\) s and ‘\(60\) s total \(11\) stamps with a cost of \(11 \cdot 6 = 66¢.\)

Peru's ‘\(50\) s and ‘\(60\) s total \(10\) stamps with a cost of \(10 \cdot 4 = 40¢.\)

Therefore, the total cost is \[ 66¢ + 40¢ = 106¢ \]\[= $1.06. \]

Thus, B is the correct answer.

Solution(s):

The total price of all the ‘\(70\) s is \[ 12 \cdot 6 + 12 \cdot 6 + 6 \cdot 4 + 13 \cdot 5 \] \[ = 72 + 72 + 24 + 65 \]\[= 233¢. \] The total number of stamps is \[ 12 + 12 + 6 +13 = 43. \] Therefore, the average is \[ 233 \div 43 \approx 5.4¢. \]

Thus, E is the correct answer.

Solution(s):

Note that the number of tiles the \(n\) th square requires is \(n^2\) since each side has \(n\) tiles.

Therefore, the \(6\) th square will need \(6^2 = 36\) tiles and \(7\) th will need \(7^2 = 49.\) The difference is \(49 - 36 = 13.\)

Thus, C is the correct answer.

Solution(s):

The total probability is \(1.\) We need to subtract the probability of the spinner landing on \(B\) and \(A\) to get \(C,\) which is \[ 1 - \dfrac{1}{3} - \dfrac{1}{2} = \dfrac{1}{6}. \]

Thus, B is the correct answer.

Solution(s):

The larger box will have approximately \[ 125 \cdot 2 \cdot 2 \cdot 2 = 1000 \] jellybeans.

Thus, E is the correct answer.

Solution(s):

let \(x\) be the original price of the items. \(30\%\) is \(.7x.\) Another \(20\%\) off is \(.8 \cdot .7x = .56x.\)

This means that the price is \(44\%\) less than the original price.

Thus, B is the correct answer.

Solution(s):

The number of boxes enclosed by each polygon can be obtained by dividing the polygon into unit squares and right triangles with sidelength \(1\) and adding up their values.

The unit squares count as \(1\) and the triangles count as \(.5.\)

\(A\) has a total area of \(5,\) \(B\) has \(5,\) \(C\) has \(5,\) \(D\) has \(4.5,\) and \(E\) has \(5.5.\)

Thus, E is the correct answer.

Solution(s):

We can find the area of all the triangles since we know both legs.

\[ \begin{gather*} W = \dfrac{3 \cdot 4}{2} = 6 \\ X = \dfrac{3 \cdot 3}{2} = 4.5 \\ Y = \dfrac{4 \cdot 4}{2} = 8 \\ Z = \dfrac{5 \cdot 5}{2} = 12.5 \end{gather*} \]

Plugging these values into the answer choices, we see that \(X+Y=Z\) is the only that is true.

Thus, E is the correct answer.

Solution(s):

Let \(x\) be the number of correct answers. Then she answered \(10 - x\) questions incorrectly.

This gives her a total score of \[ 5x - 2(10 - x) = 7x - 20. \]

We know that this equals \(29,\) and solving yields \[ 7x - 20 = 29 \]\[ x = 7. \]

Thus, C is the correct answer.

Solution(s):

Gage has skated a total of \[ 5 \cdot 75 + 3 \cdot 90 = 645 \text{ min.} \]

For an average of \(85\) minutes over \(9\) days, Gage must have skated a total of \[ 85 \cdot 9 = 765 \text{ min.} \]

This means that Gage must skate \[ 765 - 645 = 120 \text{ min.} \] on the last day. Note that \(120\) minutes is the same as \(2\) hours.

Thus, E is the correct answer.

Solution(s):

Note that the \(0\) digit can either be the tens or the units digit. This gives us \(2\) options for this.

There are \(9\) options for each of the other digits for a total of \[ 2 \cdot 9 \cdot 9 = 162 \] numbers.

Thus, D is the correct answer.

Solution(s):

Note that the area of \(\triangle XYC\) is \(8 \div 2 = 4\) since \(\overline{XC}\) splits \(\triangle XYZ\) into two congruent triangles.

We also know that the unshaded region is \(\frac{1}{4}\) the area of the whole triangle, since all the sides are \(\frac{1}{2}\) the length of the larger triangle.

This means that the shaded region is \(\frac{3}{4}\) the area of the whole triangle, which is \(4 \cdot \frac{3}{4} = 3.\)

Thus, D is the correct answer.

Solution(s):

The probability that there are at least as many head as tails is the same as the probability that there are at least as many tails as heads.

The only overlap between these two scenarios is when the number of heads and the number of tails is equal.

Let \(p\) be the desired probability. Then from our analysis we get that \[ p + p - q = 1, \] where \(q\) is the probability of getting the same number of heads and tails.

To find \(q,\) there are \(\binom{4}{2} = 6\) ways to choose which coins are heads, and there are a total of \(2^4 = 16\) possibilities.

Therefore, \[ q = \dfrac{6}{16} = \dfrac{3}{8}, \] and \[ 2p - \dfrac{3}{8} = 1 \]\[ p = \dfrac{11}{16}. \]

Thus, E is the correct answer.

Solution(s):

We can count the number of unexposed sides to find how many sides contribute to the surface area.

Three cubes have \(1\) side unexposed, two cubes have \(2\) sides unexposed, and one cube has \(3\) sides unexposed.

This gives us a total of \[ 3 \cdot 1 + 2 \cdot 2 + 1 \cdot 3 = 10 \] unexposed sides, which gives us \[ 6 \cdot 6 - 10 = 36 - 10 \]\[= 26 \] exposed sides.

Each exposed side contributes \(1^2 = 1\) to the surface area, for a total surface area of \(1 \cdot 26 = 26.\)

Thus, C is the correct answer.

Solution(s):

Notice that there are repeating \(3 \times 3\) regions with the same pattern (they might be rotated differently).

In this region, there are three unit squares, and two triangles that combine to form another unit square.

This makes the area of the darker region \(4\) and the whole region \(9.\) The desired fraction is then \(\dfrac{4}{9}.\)

Thus, B is the correct answer.

Solution(s):

We can set up a proportion to find the amount of juice Miki can extract from the fruits: \[ \dfrac{p}{12} = \dfrac{8}{3} \] for the pears and \[ \dfrac{o}{12} = \dfrac{8}{2} \] for the oranges. Solving yields \[ p = 32 \text{ and } o = 48. \]

The percent of the whole that is pear juice is \[ 100 \cdot \dfrac{32}{32 + 48} = 100 \cdot \dfrac{2}{5} \]\[= 40 \%. \]

Thus, B is the correct answer.

Solution(s):

WLOG, assume that everyone gave Ott \($1.\) This means that Moe had \($5,\) Loki had \($4,\) and Nick had \($3\) originally.

Ott now has \($3,\) and the total amount of money is \[ $5 + $4 + $3 = $12. \]

This means that Ott has \[ \dfrac{3}{12} = \dfrac{1}{4} \] of the group's money.

Thus, B is the correct answer.