Solution:

The logo is made from $5$ congruent tilted squares. Each tilted square has diagonals of length $2$ and $2$, so its area is $\frac{2\cdot2}{2}=2$ square inches.

The total area is $5\cdot2=10$.

Thus, the correct answer is A.

Solution:

Using the definitions, $$\begin{align*} (5\,\blacklozenge\,3) \star 6 &= (5^2-3^2) \star 6 \\ &=16 \star 6 \\ &=(16-6)^2\\ &= 100\end{align*}$$

Thus, the correct answer is D.

Solution:

Since $a< b< c$ and $abc=100$, we have a^3<100, so $a\le4$. Also $a$ must divide $100$, so $a\in\{1,2,4\}$.

If $a=1$, the pairs $(b,c)$ are $(2,50),(4,25),(5,20)$. If $a=2$, the only possible pair is $(5,10)$. If $a=4$, then $bc=25$, and there is no integer $b$ with $4< b< c$.

There are $4$ possible triples.

Thus, the correct answer is E.

Solution:

Reflecting over the diagonal line $q$ changes the orientation of the letter as if the diagonal were a mirror. Reflecting that image over the horizontal line $p$ then places the final letter below $p$, to the right of the vertical line, with the orientation shown in choice E.

Thus, the correct answer is E.

Solution:

If Bella was $6$ five years ago, then she is $11$ right now.

If the kitten was a newborn five years ago, then it is $5$ right now.

Since the sum of all $3$ ages is $30,$ Anna's age is $$30-5-11 = 14$$

Since Anna is $14$ and Bella is $11,$ she is $3$ years older than Bella.

Thus, the correct answer is C.

Solution:

Let the smallest number be $x$. Then the largest is $4x$. Since the three numbers are equally spaced, the middle number is the average of the smallest and largest: $$\frac{x+4x}{2}=15.$$ Thus $5x=30$, so $x=6$.

Thus, the correct answer is C.

Solution:

The song has $4.2\cdot8000=33{,}600$ kilobits. At $56$ kilobits per second, the download takes $$\frac{33{,}600}{56}=600$$ seconds, which is $10$ minutes.

Thus, the correct answer is B.

Solution:

Since every integer from $3$ to $20$ occurs once as a denominator and once as a numerator, they cancel each other out.

After canceling every number out, we have only $1$ and $2$ left as numerators and $21$ and $22$ left as denominators.

The remaining fraction is $\dfrac{1 \cdot 2}{21 \cdot 22} .$ This simplifies to $\dfrac{2}{462} = \dfrac{1}{231}$

Thus, the correct answer is B.

Solution:

The current difference is $212-68 = 144 .$

Since we have $15$ minutes while halving every $5$ minutes, we halve the difference $3$ times.

This means the difference is multiplied by $(\dfrac{1}{2})^3 = \dfrac{1}{8} ,$ so our new difference is $\dfrac{1}{8}\cdot144=18.$ This makes our final temperature $68+18=86.$

Thus, the correct answer is B.

Solution:

She drives $45$ miles per hour for $2$ hours, so the trail is $90$ miles from her house. The graph must rise from $0$ to $90$ miles between $8$ AM and $10$ AM, then stay flat for the $3$-hour hike.

She starts home at $1$ PM. Driving $90$ miles at $60$ miles per hour takes $1.5$ hours, so she gets home at $2{:}30$ PM. This matches choice E.

Thus, the correct answer is E.

Solution:

Since there are $10$ pieces, there were $9$ locations where a bite was made. Since we have $9$ bites and $3$ inches are removed per bite, a total of $27$ inches were removed.

With $27$ inches removed and $17$ inches remaining, we know we started with $27+17 = 44$ inches.

Thus, the correct answer is D.

Solution:

Spinner A gives the tens digit and Spinner B gives the ones digit. There are $4\cdot4=16$ equally likely two-digit numbers.

The possible perfect squares between $51$ and $84$ are $64$ and $81$, and both can occur. Thus the probability is $$\frac{2}{16}=\frac{1}{8}.$$

Thus, the correct answer is B.

Solution:

Let the smaller number be $x$, and let the blank be $c$. Then the other number is $2x+c$, where both $x$ and $c$ are positive integers.

The sum condition gives $x+(2x+c)=28$, so $c=28-3x$. For $c$ to be positive, 28-3x>0, so $x<\frac{28}{3}$.

Thus $x$ can be any integer from $1$ through $9$, giving $9$ possible values of the blank.

Thus, the correct answer is D.

Solution:

The word has $5$ E's and $4$ other letters: B, K, P, and R. To keep no two E's adjacent, the only possible pattern is $$E\_E\_E\_E\_E,$$ with the four non-E letters in the four gaps.

The letters B, K, P, and R can be arranged in those gaps in $4!=24$ ways.

Thus, the correct answer is D.

Solution:

For each weight, only the lowest-price point at that weight can give the lowest price per ounce.

At $1$ ounce, the lowest price is more than $1$ dollar, so the price per ounce is more than $1$.

At $2$ ounces, the lowest price is $2$ dollars, so the price per ounce is $1$.

At $3$ ounces, the lowest price is about $2.5$ dollars, so the price per ounce is about $\frac{2.5}{3}$, less than $1$.

At $4$ ounces, the lowest price is close to $3.9$ dollars, so the price per ounce is close to $\frac{3.9}{4}$, still larger than the $3$-ounce option.

At $5$ ounces, the lowest price is close to $4.5$ dollars, so the price per ounce is close to $\frac{4.5}{5}$, also larger than the $3$-ounce option.

The price per ounce is lowest at $3$ ounces.

Thus, the correct answer is C.

Solution:

Let the numbers be $a,b,c,d$ in order.

Since the average of $a$ and $b$ is $21,$ we know their sum is $21\cdot2 = 42.$
Since the average of $c$ and $d$ is $30,$ we know their sum is $30\cdot2 = 60.$
Since $a+b = 42$ and $c+d = 60 ,$ we know $$\begin{align*} a+b+c+d &= 42 + 60\\ &= 102 \end{align*}$$

Now, with the average of $b$ and $c$ being $26,$ we know their sum is $26\cdot2=52.$ This means $b+c = 52.$

Subtracting this result from the sum of all the terms yields $a+d =50.$
Since $\dfrac{a+d}{2}=25,$ our answer is $25.$

Thus, the correct answer is B.

Solution:

If we take $n!!$ for an even $n$ that is greater or equal to $10,$ then $10$ is one of the numbers we multiply by. Since $10$ is a factor of $n!!,$ we know that the units digit of $10$ is $0,$ which means that it doesn't affect our result.

This means it suffices to compute the units digit of $2!!+4!!+6!!+8!!,$ which is equivalent to: $$\begin{align*} &2+ 2(4)+2(4)(6)\\ &+2(4)(6)(8)\\ &= 2+8+48+384\\ &= 442 \end{align*}$$ The units digit therefore is $2$

Thus, the correct answer is B.

Solution:

The four given midpoints form a parallelogram. Its horizontal base has length $2-(-3)=5$, and its height is $4$, so its area is $5\cdot4=20$.

For any rectangle, the parallelogram formed by joining the side midpoints has half the area of the rectangle. Therefore the rectangle's area is $2\cdot20=40$.

Thus, the correct answer is C.

Solution:

From the dot plot, there are $7$ scores at least $85$: two $85$'s, three $90$'s, one $95$, and one $100$. For the median to be $85$, the $10$th and $11$th scores in increasing order must both be at least $85$, so at least $11$ scores must be at least $85$.

Each regraded student can gain only $5$ points, so the only scores below $85$ that can become at least $85$ are the $80$'s. There are enough $80$'s, and we need $11-7=4$ more scores at least $85$. Regrading four students who originally scored $80$ achieves this.

Thus, the correct answer is C.

Solution:

Adding the numbers in the top row shows that every row and column must have sum $12$.

In the first column, the missing number above $x$ is $14-x$. In the bottom row, the missing number to the right of $x$ is $4-x$. In the middle row, the remaining missing number is $x-1$.

Since $x$ is larger than the other missing numbers, we need x>14-x, x>4-x, and $x> x-1$. The strongest condition is x>7, so the smallest possible integer value is $8$.

Thus, the correct answer is D.

Solution:

Steph made $25$ baskets in $30$ attempts. Candace also took $12+18=30$ attempts, and their overall percentages were the same, so Candace also made $25$ baskets.

Let $f$ be Candace's first-half baskets and $s$ be her second-half baskets. Then $f+s=25$.

Steph's first-half percentage was $\frac{15}{20}=\frac34$, so $\frac{f}{12}<\frac34$, giving f<9. Steph's second-half percentage was $1$, so s<18.

Thus $f\le8$ and $s\le17$. Since $f+s=25$, the only possibility is $f=8$ and $s=17$, so $s-f=9$.

Thus, the correct answer is C.

Solution:

The bus takes $2$ minutes to drive to the next stop and then waits $1$ minute, so it moves through one-stop cycles every $3$ minutes.

Zia makes a decision only when she reaches a stop, every $5$ minutes. Measure stops from the bus's starting stop. Zia starts at stop $3$.

After $5$ minutes, Zia is at stop $4$, while the bus is waiting at stop $2$, so she keeps walking. After $10$ minutes, Zia is at stop $5$, while the bus is between stops $3$ and $4$, so she keeps walking. After $15$ minutes, Zia is at stop $6$, while the bus is at stop $5$, the previous stop, so she waits.

The bus then takes $2$ more minutes to reach her stop, so she boards after $17$ minutes.

Thus, the correct answer is A.

Solution:

Let $k$ be the number of triangles. To have both a triangle line and a circle line, we need $3\le k\le6$.

If $k=3$, the three triangles must form a line. A row or column works, because the remaining six circles contain a full circle row or column. A diagonal does not work, because its complement contains no full line of circles. Thus there are $6$ configurations for $k=3$. By symmetry, there are also $6$ configurations for $k=6$.

If $k=4$, choose the triangle line and then the extra triangle. If the line is a row or column, all $6$ possible extra positions work, because one parallel row or column remains all circles. This gives $6\cdot6=36$ configurations. If the triangle line is a diagonal, no extra position leaves a full circle line. By symmetry, $k=5$ also gives $36$ configurations.

The total number of configurations is $6+36+36+6=84$.

Thus, the correct answer is D.

Solution:

When the net is folded, the matching side lengths give $GF=EF=8$. Since $GFCB$ is a rectangle, $BC=8$. The fold identifies $AB$ with $BC$, so $AB=8$. In rectangle $HJBA$, this gives $HJ=8$, and the opposite side $BJ$ equals $AH=8$.

Since $GH=14$, we have $GJ=14-8=6$. Thus one triangular base of the prism is right triangle $BJG$, with area $$\frac{6\cdot8}{2}=24.$$ The prism length is $GF=8$, so the volume is $24\cdot8=192$.

Thus, the correct answer is C.

Solution:

Let $p_n$ be the probability that the cricket is on its starting leaf after $n$ hops. We have $p_0=1$.

If the cricket is on the starting leaf, the next hop must leave it. If the cricket is not on the starting leaf, exactly one of the $3$ possible hops returns to the start. Therefore $$p_{n+1}=\frac{1-p_n}{3}.$$

Thus $$p_1=0,\quad p_2=\frac13,\quad p_3=\frac29,\quad p_4=\frac{1-\frac29}{3}=\frac{7}{27}.$$

Thus, the correct answer is E.