Answer: D

Small Hint:

Count each type of pair of figures separately.

Big Hint:

Each line can meet the circle at most twice, and the two lines can meet once.

Solution:

A line and a circle can intersect at no more than two points, and two distinct lines can intersect at no more than one point.

Arrange the two lines so each meets the circle twice and the two lines meet at a different point. This gives $$2+2+1=5.$$

Thus, D is the correct answer.

Answer: A

Small Hint:

Because $\$17$ is odd, the number of $\$5$ bills must be odd.

Big Hint:

Try the possible odd numbers of $\$5$ bills before the total goes over $\$17.$

Solution:

Since the total $\$17$ is odd, the number of $\$5$ bills must be odd.

One $\$5$ bill leaves $\$12,$ which is six $\$2$ bills. Three $\$5$ bills leave $\$2,$ which is one $\$2$ bill. Five $\$5$ bills is too much.

Therefore, there are $2$ combinations.

Thus, A is the correct answer.

Answer: C

Small Hint:

To make the average as small as possible, choose the smallest possible integers.

Big Hint:

Use the four smallest distinct positive even integers.

Solution:

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.

Answer: B

Small Hint:

The next palindrome after $2002$ still starts and ends with $2.$

Big Hint:

The middle two digits must increase to the smallest matching pair.

Solution:

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.

Answer: C

Small Hint:

Reduce $706$ modulo $7.$

Big Hint:

$700$ days is an exact number of weeks.

Solution:

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.

Answer: A

Small Hint:

Before overflow, the volume rises at a constant positive rate.

Big Hint:

After the birdbath is full, extra water overflows so the volume stays constant.

Solution:

Before the birdbath overflows, it gains $20-18=2$ milliliters of water per minute, so its volume increases steadily.

Once the birdbath is full, the incoming extra water overflows, so the volume remains constant. Only graph A shows an increasing line followed by a horizontal line.

Thus, A is the correct answer.

Answer: E

Small Hint:

Add the heights of all five bars to get the class size.

Big Hint:

Compare the height of bar $E$ with the total number of students.

Solution:

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.

Answer: D

Small Hint:

European means France and Spain in the table.

Big Hint:

Use only the entries in the ‘$80$s column.

Solution:

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.

Answer: B

Small Hint:

South American means Brazil and Peru.

Big Hint:

Before the ‘$70$s means the ‘$50$s and ‘$60$s columns.

Solution:

Brazil and Peru are the South American countries.

Brazil has $4+7=11$ stamps before the ‘$70$s, costing $11\cdot6=66$ cents. Peru has $6+4=10$ such stamps, costing $10\cdot4=40$ cents.

The total cost is $66+40=106$ cents, or $\$1.06.$

Thus, B is the correct answer.

Answer: E

Small Hint:

Compute the total cost of all ‘$70$s stamps.

Big Hint:

Divide the total cost by the number of ‘$70$s stamps, then choose the closest option.

Solution:

The ‘$70$s stamps cost $$\begin{aligned} &12\cdot6+12\cdot6 \\ &\quad {}+6\cdot4+13\cdot5 \\ &\quad {}=72+72+24+65 \\ &\quad {}=233 \end{aligned}$$ cents.

There are $12+12+6+13=43$ stamps, so the average cost is $233\div43,$ a little more than $5$ cents and closest to $5.5$ cents.

Thus, E is the correct answer.

Answer: C

Small Hint:

The sixth and seventh squares have side lengths $6$ and $7.$

Big Hint:

Compare $7^2$ and $6^2.$

Solution:

The sixth and seventh squares have side lengths $6$ and $7$ tile lengths.

They use $6^2=36$ and $7^2=49$ tiles, respectively, so the seventh square requires $49-36=13$ more tiles.

Thus, C is the correct answer.

Answer: B

Small Hint:

The probabilities of $A,$ $B,$ and $C$ add to $1.$

Big Hint:

Subtract the probabilities for $A$ and $B$ from $1.$

Solution:

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.

Answer: E

Small Hint:

Doubling length, width, and height multiplies volume by $2\cdot2\cdot2.$

Big Hint:

Scale the number of jellybeans by the volume factor.

Solution:

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

Thus, E is the correct answer.

Answer: B

Small Hint:

After a $30\%$ discount, $70\%$ of the original price remains.

Big Hint:

The second discount is $20\%$ off the sale price, not the original price.

Solution:

Let the original price be $x.$ After the $30\%$ discount, the price is $0.70x.$

Taking another $20\%$ off that sale price leaves $0.80\cdot0.70x=0.56x.$

The customer pays $56\%$ of the original price, so the total discount is $44\%.$

Thus, B is the correct answer.

Answer: E

Small Hint:

Break each polygon into unit squares and half-unit triangles.

Big Hint:

Compare the five areas by counting full squares plus half-squares.

Solution:

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 $0.5.$

$A$ 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.

Answer: E

Small Hint:

Each right isosceles triangle has area half the square of the side it is built on.

Big Hint:

Compute the areas on the sides $3,$ $4,$ and $5.$

Solution:

For a right isosceles triangle built on a side of length $s,$ the two legs have length $s,$ so its area is $\frac{s^2}{2}.$

$$\begin{aligned} &W=\frac{3\cdot4}{2}=6, \\ &\quad X=\frac{3^2}{2}=4.5, \\ &\quad Y=\frac{4^2}{2}=8, \\ &\quad Z=\frac{5^2}{2}=12.5. \end{aligned}$$

These values satisfy $X+Y=Z,$ and the other listed equations do not.

Thus, E is the correct answer.

Answer: C

Small Hint:

Let $x$ be the number of correct answers.

Big Hint:

Then the number of incorrect answers is $10-x.$

Solution:

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.

Answer: E

Small Hint:

Convert all skating times to minutes.

Big Hint:

Find the total minutes needed for a $9$-day average of $85.$

Solution:

Gage has skated a total of $$5 \cdot 75 + 3 \cdot 90 = 645$$ minutes.

For an average of $85$ minutes over $9$ days, Gage must have skated a total of $$85 \cdot 9 = 765$$ minutes.

This means that Gage must skate $$765 - 645 = 120$$ minutes on the last day. Note that $120$ minutes is the same as $2$ hours.

Thus, E is the correct answer.

Answer: D

Small Hint:

The number is a three-digit number.

Big Hint:

The single $0$ can be in the tens place or the ones place, but not the hundreds place.

Solution:

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.

Answer: D

Small Hint:

The altitude splits the isosceles triangle into two equal halves.

Big Hint:

The segment through the midpoints creates a similar triangle with half the side lengths.

Solution:

Since $XY=XZ$ and $XC$ bisects $YZ,$ altitude $XC$ splits $\triangle XYZ$ into two congruent triangles. The left half $\triangle XYC$ has area $4.$

In $\triangle XYC,$ point $A$ is the midpoint of $XY.$ The horizontal segment through $A$ meets $XC$ halfway up, so the small top triangle is similar to $\triangle XYC$ with scale factor $\frac{1}{2}.$ Its area is therefore $\frac{1}{4}$ of $4,$ or $1.$

The shaded region is the rest of the left half, so its area is $4-1=3.$

Thus, D is the correct answer.

Answer: E

Small Hint:

At least as many heads as tails means $2,$ $3,$ or $4$ heads.

Big Hint:

Count outcomes by choosing which tosses are heads.

Solution:

There are $2^4=16$ equally likely outcomes for four coin tosses.

At least as many heads as tails means getting $2,$ $3,$ or $4$ heads. The number of such outcomes is $$\begin{aligned} &\binom{4}{2}+\binom{4}{3}+\binom{4}{4} \\ &\quad {}=6+4+1=11. \end{aligned}$$

Thus the probability is $\dfrac{11}{16}.$

Thus, E is the correct answer.

Answer: C

Small Hint:

Start with $6$ separate cubes, each with $6$ exposed faces.

Big Hint:

Each glued pair of faces removes two exposed faces.

Solution:

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

Three cubes have $1$ face unexposed, two cubes have $2$ faces unexposed, and one cube has $3$ faces unexposed.

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

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

Thus, C is the correct answer.

Answer: B

Small Hint:

Look for a repeating $3\times3$ block.

Big Hint:

In one block, combine pairs of half dark squares into whole dark squares.

Solution:

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

In this region, there are three dark unit squares and two dark 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.

Answer: B

Small Hint:

Use an equal convenient number of pears and oranges.

Big Hint:

Six pears and six oranges make both juice amounts easy to compute.

Solution:

Use $6$ pears and $6$ oranges, which is an equal number of each fruit.

Since $3$ pears make $8$ ounces, $6$ pears make $16$ ounces. Since $2$ oranges make $8$ ounces, $6$ oranges make $24$ ounces.