Solution:

The walls of one bedroom have area $2(12+10)\cdot 8 = 44\cdot 8 = 352$ square feet. Subtracting the $60$ square feet of doorways and windows leaves $352-60=292$ square feet per bedroom.

With $3$ bedrooms, the total is $3\cdot 292 = 876$ square feet.

Thus, the correct answer is E.

Solution:

Since $3\star 5=(3+5)\cdot 5=40$ and $5\star 3=(5+3)\cdot 3=24,$ the difference is $40-24=16.$

Thus, the correct answer is E.

Solution:

The student used $120/30=4$ gallons driving home and $120/20=6$ gallons returning, for $10$ gallons over $240$ miles.

The average is $240/10 = 24$ miles per gallon.

Thus, the correct answer is B.

Solution:

Since $OA=OB=OC,$ triangles $AOB, BOC,$ and $COA$ are isosceles. The base angles give $\angle ABO=\dfrac{180^\circ-140^\circ}{2}=20^\circ$ and $\angle OBC=\dfrac{180^\circ-120^\circ}{2}=30^\circ.$

Therefore $\angle ABC=20^\circ+30^\circ=50^\circ.$

Thus, the correct answer is D.

Solution:

Writing the statements as implications, being a Crup implies being a Dramp, a Dramp implies being an Arog, and an Arog implies being a Braf: $C\Rightarrow D\Rightarrow A\Rightarrow B.$

So every Crup is a Dramp, an Arog, and a Braf. The only listed statement guaranteed true is that all Crups are Arogs and Brafs.

Thus, the correct answer is D.

Solution:

The three blank problems give $3\cdot 1.5 = 4.5$ points, so Sarah needs $100-4.5=95.5$ points from the first $22.$

Since $95.5/6$ lies between $15$ and $16,$ she must answer at least $16$ correctly, which would give a score of $100.5.$

Thus, the correct answer is D.

Solution:

Because $AB=BC=EA$ and $\angle A=\angle B=90^\circ,$ quadrilateral $ABCE$ is a square, so $\angle AEC=90^\circ.$

The remaining sides satisfy $CD=DE=EC,$ so $\triangle CDE$ is equilateral and $\angle CED=60^\circ.$

Therefore $\angle E=\angle AEC+\angle CED=90^\circ+60^\circ=150^\circ.$

Thus, the correct answer is E.

Solution:

Once $a$ and $c$ are chosen, $b=\dfrac{a+c}{2}$ is determined, and $a\lt b\lt c$ holds automatically. For $b$ to be an integer, $a$ and $c$ must share parity.

Choosing two even digits from $\{0,2,4,6,8\}$ gives $\binom{5}{2}=10$ pairs, and choosing two odd digits from $\{1,3,5,7,9\}$ gives another $10.$

This yields $20$ valid numbers.

Thus, the correct answer is D.

Solution:

The final s is the $12$th appearance of the letter s in the message, so it is shifted $1+2+\cdots+12=\dfrac{12\cdot 13}{2}=78$ places to the right.

Since $78=3\cdot 26$ is a multiple of the alphabet length $26,$ the shift returns to the same letter, s.

Thus, the correct answer is D.

Solution:

Taking $BC$ as the base, the area is $\dfrac12(BC)d,$ where $d$ is the distance from $A$ to line $BC.$ The area equals $1$ exactly when $d=\dfrac{2}{BC}.$

The points at this fixed distance from line $BC$ form two lines parallel to $BC,$ one on each side.

Thus, the correct answer is A.

Solution:

The triangle has sides $3,3,2.$ Its area is $\dfrac12\cdot 2\cdot\sqrt{3^2-1^2}=2\sqrt2.$

The circumradius is $R=\dfrac{abc}{4K}=\dfrac{3\cdot 3\cdot 2}{4\cdot 2\sqrt2}=\dfrac{9}{4\sqrt2} =\dfrac{9\sqrt2}{8}.$

The area of the circle is $\pi R^2=\pi\cdot\dfrac{81\cdot 2}{64}=\dfrac{81}{32}\pi.$

Thus, the correct answer is C.

Solution:

$N$ years ago Tom's age was $T-N,$ and the sum of his three children's ages was $T-3N.$

The condition gives $T-N=2(T-3N),$ so $T-N=2T-6N,$ which simplifies to $5N=T.$

Therefore $T/N=5.$

Thus, the correct answer is D.

Solution:

The two circles intersect at $(0,0)$ and $(2,2).$

By symmetry, half the intersection is formed by removing an isosceles right triangle of leg length $2$ from a quarter of one circle. The quarter-circle has area $\dfrac14\pi(2)^2=\pi$ and the triangle has area $\dfrac12(2)^2=2.$

Therefore the whole region has area $2(\pi-2).$

Thus, the correct answer is D.

Solution:

Two girls leave and two boys arrive, so the group size is unchanged. The two girls who left therefore represent $40\%-30\%=10\%$ of the group.

Thus the group has $2\div 0.10=20$ people, and the original number of girls was $40\%$ of $20,$ or $8.$

Thus, the correct answer is C.

Solution:

Let $x=\angle A.$ Then $\angle B=\dfrac{x}{2},$ $\angle C=\dfrac{x}{3},$ and $\angle D=\dfrac{x}{4}.$

The angles sum to $360^\circ,$ so $x+\dfrac{x}{2}+\dfrac{x}{3}+\dfrac{x}{4}=\dfrac{25x}{12}=360.$

Thus $x=\dfrac{12\cdot 360}{25}=172.8\approx 173.$

Thus, the correct answer is D.

Solution:

Suppose the class has $10$ students: one junior and nine seniors. The total of all scores is $10\cdot 84=840.$

The nine seniors total $9\cdot 83=747,$ so the junior's score is $s=840-747=93.$

Thus, the correct answer is C.

Solution:

Let the side length be $s.$ The perpendiculars from $P$ are the heights of triangles $APB,$ $BPC,$ and $CPA,$ so their areas are $\dfrac{s}{2},$ $s,$ and $\dfrac{3s}{2}.$

Their sum equals the area of $\triangle ABC,$ which is also $\dfrac{\sqrt3}{4}s^2.$ Hence $3s=\dfrac{\sqrt3}{4}s^2.$

The positive solution is $s=4\sqrt3.$

Thus, the correct answer is D.

Solution:

Connect the centers of the four outer circles to form a square. Adjacent outer circles are tangent, so each side has length $2r.$

The diagonal of the square passes through the center circle, giving length $1+r+r+1=2+2r.$ Since a square with side $2r$ has diagonal $2r\sqrt2,$ we get $2(2r)^2=(2+2r)^2.$

Expanding gives $1+2r+r^2=2r^2,$ so $r^2-2r-1=0.$ The positive root is $r=1+\sqrt2.$

Thus, the correct answer is B.

Solution:

The shaded squares are those where the two remainders are both odd or both even. The first remainder is even (from the numbers $2$ and $6$) with probability $\dfrac13$ and odd with probability $\dfrac23.$

The second remainder is even with probability $\dfrac12$ and odd with probability $\dfrac12.$

The probability that they share parity is $\dfrac13\cdot\dfrac12+\dfrac23\cdot\dfrac12=\dfrac12.$

Thus, the correct answer is C.

Solution:

Choose $3$ of the $5$ rows in $\binom{5}{3}=10$ ways and $3$ of the $5$ columns in $\binom{5}{3}=10$ ways.

The three chosen blocks must occupy distinct rows and columns, so they form a matching between the three rows and three columns, which can be done in $3!=6$ ways.

The total is $10\cdot 10\cdot 6=600.$

Thus, the correct answer is C.

Solution:

Let $s$ be the side of the square and $h$ the altitude from $B$ to $AC.$ Then $h=\dfrac{AB\cdot BC}{AC}=\dfrac{3\cdot 4}{5}=\dfrac{12}{5}.$

The small triangle above the square is similar to $\triangle ABC$ with the square's top side as its base, giving $\dfrac{h-s}{h}=\dfrac{s}{AC},$ so $s=\dfrac{AC\cdot h}{AC+h}.$

Substituting, $s=\dfrac{5\cdot\frac{12}{5}}{5+\frac{12}{5}}=\dfrac{12}{37/5}=\dfrac{60}{37}.$

Thus, the correct answer is B.

Solution:

Each die shows the chosen number on the bottom with probability $\dfrac14.$ So the number appears $0,1,$ or $2$ times with probabilities $P(0)=\dfrac{9}{16},\ P(1)=\dfrac{6}{16},\ P(2)=\dfrac{1}{16}.$

The expected return is $(-1)\cdot\dfrac{9}{16}+(1)\cdot\dfrac{6}{16}+(2)\cdot\dfrac{1}{16}=\dfrac{-9+6+2}{16}=-\dfrac{1}{16}.$

Thus, the correct answer is B.

Solution:

Let $h$ be the altitude of the original pyramid; the smaller pyramid has altitude $h-2.$ The two pyramids are similar, so the ratio of their surface areas is the square of the ratio of their altitudes.

The smaller surface area is half the original, so $\left(\dfrac{h-2}{h}\right)^2=\dfrac12,$ giving $\dfrac{h}{h-2}=\sqrt2.$

Then $h=\sqrt2(h-2),$ so $h(\sqrt2-1)=2\sqrt2$ and $h=\dfrac{2\sqrt2}{\sqrt2-1}=4+2\sqrt2.$

Thus, the correct answer is E.

Solution:

Since $n$ is divisible by $9,$ its digit sum is a multiple of $9.$ With $k$ fours and $m$ nines, the digit sum is $4k+9m,$ so $9\mid 4k,$ forcing $9\mid k.$ Thus $k\ge 9,$ and with at least one $9,$ the number has at least ten digits.

For divisibility by $4,$ the last two digits must form a multiple of $4,$ and among $44,49,94, 99$ only $44$ works, so $n$ ends in $44.$

The smallest such ten-digit number places the single $9$ in the lowest available position, giving $4{,}444{,}444{,}944.$ Its last four digits are $4944.$

Thus, the correct answer is C.

Solution:

Combining, the expression is $\dfrac{9a^2+14b^2}{9ab}.$ For this to be an integer, $a$ must divide $9a^2+14b^2,$ hence $a\mid 14b^2.$ Since $\gcd(a,b)=1,$ we get $a\mid 14.$ Similarly $b\mid 9a^2$ gives $b\mid 9.$

So $a\in\{1,2,7,14\}$ and $b\in\{1,3,9\}.$ Checking these, only $b=3$ makes the expression an integer for each allowed $a.$

The valid pairs are $(1,3),(2,3),(7,3),$ and $(14,3),$ for a total of $4.$

Thus, the correct answer is A.