Answer: E

Solution:

The perimeter of each bedroom floor is $2(12+10)=44$ feet.

So the wall area in one bedroom is $44\cdot8-60=352-60=292$ square feet. Across all three bedrooms, Isabella paints $3\cdot292=876$ square feet.

Thus, the correct answer is E.

Answer: B

Solution:

The trip home uses $120/30=4$ gallons, and the return trip uses $120/20=6$ gallons.

The round trip covers $240$ miles on $10$ gallons, so the average is $$\dfrac{240}{10}=24 \text{ miles per gallon.}$$

Thus, the correct answer is B.

Answer: D

Solution:

The angles around $O$ sum to $360^\circ,$ so $$\angle AOC=360^\circ-140^\circ-120^\circ=100^\circ.$$

By the inscribed angle theorem, $\angle ABC$ subtends the same arc $AC$ as the central angle $\angle AOC,$ so $$\angle ABC=\tfrac12\angle AOC=50^\circ.$$

Thus, the correct answer is D.

Answer: B

Solution:

Since $3$ bananas cost as much as $2$ apples, $18$ bananas cost as much as $12$ apples.

Since $6$ apples cost as much as $4$ oranges, $12$ apples cost as much as $8$ oranges. Therefore $18$ bananas cost as much as $8$ oranges.

Thus, the correct answer is B.

Answer: D

Solution:

The three unanswered problems give $3\cdot1.5=4.5$ points.

So Sarah needs at least $100-4.5=95.5$ points from correct answers. Since $$15\lt\dfrac{95.5}{6}\lt16,$$ she must solve at least $16$ problems correctly, which would give her $16\cdot6+4.5=100.5$ points.

Thus, the correct answer is D.

Answer: D

Solution:

The perimeter is $5+6+7=18,$ so each bug crawls $9$ before they meet.

The bug going $A\to B\to C$ reaches $D$ on side $BC,$ having traveled $AB+BD=9.$ Since $AB=5,$ we get $BD=4.$

Thus, the correct answer is D.

Answer: E

Solution:

Because $AB=BC=EA$ and $\angle A=\angle B=90^\circ,$ quadrilateral $ABCE$ is a square, so $\angle AEC=90^\circ$ and $EC$ equals the common side length.

Then $CD=DE=EC,$ 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.

Answer: D

Solution:

Tom's age $N$ years ago was $T-N.$ His three children were each $N$ years younger, so their ages then totaled $T-3N.$

The condition gives $$T-N=2(T-3N),$$ so $5N=T$ and $T/N=5.$

Thus, the correct answer is D.

Answer: A

Solution:

Setting $3x-1=5$ gives $x=2.$

Then $$f(5)=2^2+2+1=7.$$

Thus, the correct answer is A.

Answer: C

Solution:

Since two girls leave while two boys arrive, the total group size is unchanged. The drop from $40\%$ to $30\%$ girls corresponds to the two girls who left.

So those two girls are $10\%$ of the group, meaning the group has $20$ people. The initial number of girls was $40\%$ of $20,$ or $8.$

Thus, the correct answer is C.

Answer: D

Solution:

Let $x=\angle A.$ Then $\angle B=\tfrac x2,$ $\angle C=\tfrac x3,$ and $\angle D=\tfrac x4.$

The angle sum gives $$x+\dfrac x2+\dfrac x3+\dfrac x4=\dfrac{25x}{12}=360,$$ so $x=\dfrac{12\cdot360}{25}=172.8\approx173.$

Thus, the correct answer is D.

Answer: C

Solution:

Take a class of $10$ students, so there is $1$ junior and $9$ seniors.

The total of all scores is $10\cdot84=840,$ and the seniors contribute $9\cdot83=747.$ So the junior scored $$840-747=93.$$

Thus, the correct answer is C.

Answer: D

Solution:

The cycle length is $30+3+30=63$ seconds, with three color changes per cycle.

Leah sees a change exactly when her three-second interval starts within the $3$ seconds before a switch. That gives $3\cdot3=9$ favorable seconds out of $63,$ a probability of $$\dfrac{9}{63}=\dfrac17.$$

Thus, the correct answer is D.

Answer: D

Solution:

Let $s=AB.$ Joining $P$ to the vertices splits the triangle into $\triangle PAB,$ $\triangle PBC,$ and $\triangle PCA,$ with areas $\tfrac{s}{2},$ $s,$ and $\tfrac{3s}{2}.$

Their total is $3s,$ which must equal the area $\tfrac{\sqrt3}{4}s^2$ of the equilateral triangle. So $$3s=\dfrac{\sqrt3}{4}s^2,$$ giving $s=\dfrac{12}{\sqrt3}=4\sqrt3.$

Thus, the correct answer is D.

Answer: E

Solution:

The odd-power terms are $ar+ar^3+\cdots=r(a+ar^2+\cdots),$ that is, $r$ times the even-power terms. The even-power terms sum to $7-3=4.$

So $3=4r,$ giving $r=\tfrac34.$ Then $a=7(1-r)=\tfrac74,$ and $$a+r=\dfrac74+\dfrac34=\dfrac52.$$

Thus, the correct answer is E.

Answer: A

Solution:

The rotation group of the tetrahedron has $12$ elements: the identity, $8$ rotations of order $3$ about a vertex-face axis, and $3$ rotations of order $2$ about an edge-midpoint axis.

The identity fixes all $3^4=81$ colorings. Each vertex rotation fixes one face and cycles the other three, so it fixes $3^2=9$ colorings; likewise each edge rotation swaps two pairs of faces and fixes $3^2=9.$

By Burnside's lemma the number of distinguishable colorings is $$\dfrac{81+8\cdot9+3\cdot9}{12}=\dfrac{180}{12}=15.$$

Thus, the correct answer is A.

Answer: D

Solution:

Because $b\lt10^b$ for all $b\gt0,$ it follows that $\log_{10}b\lt b.$ If $b\ge1,$ then $0\lt\dfrac{\log_{10}b}{b^2}\lt1,$ so $a$ could not be a nonzero integer.

Hence $0\lt b\lt1,$ so $\log_{10}b\lt0$ and $a=\dfrac{\log_{10}b}{b^2}\lt0.$ Thus $a\lt0\lt b\lt1\lt\dfrac1b,$ and the middle value of the sorted set is $b.$

Thus, the correct answer is D.

Answer: C

Solution:

Let the smaller square be $N^2,$ so the larger is $(N+1)^2$ and the gap is $2N+1.$ Then $$\overline{abc}=N^2+\dfrac{2N+1}{3},\qquad \overline{acb}=N^2+\dfrac{2(2N+1)}{3}.$$

Subtracting, $\overline{acb}-\overline{abc}=9(c-b)=\dfrac{2N+1}{3},$ so $27(c-b)=2N+1.$ If $c-b=0$ or $2,$ then $N$ is not an integer; if $c-b\ge3,$ then $N\ge40$ and the numbers are not three digits.

So $c-b=1,$ giving $N=13.$ The points one third and two thirds of the way from $13^2=169$ to $14^2=196$ are $178$ and $187,$ so $a+b+c=1+7+8=16.$

Thus, the correct answer is C.

Answer: A

Solution:

Let $\theta=\angle ABC.$ The base circle has circumference $6,$ so its radius is $\dfrac{6}{2\pi}=\dfrac3\pi.$ The height of the cylinder is the rhombus altitude $6\sin\theta.$

The volume is $$\pi\left(\dfrac3\pi\right)^2(6\sin\theta)=\dfrac{54}{\pi}\sin\theta=6,$$ so $\sin\theta=\dfrac{\pi}{9}.$

Thus, the correct answer is A.

Answer: D

Solution:

Two vertices of the first parallelogram lie at $(0,c)$ and $(0,d),$ and the other two have $x$-coordinates $\pm\dfrac{c-d}{b-a}.$ Its area works out to $\dfrac{(c-d)^2}{|b-a|}=18.$ The same computation for the second gives $\dfrac{(c+d)^2}{|b-a|}=72.$

So $(c-d)^2=18|b-a|$ and $(c+d)^2=72|b-a|.$ Subtracting, $4cd=54|b-a|,$ i.e. $2cd=27|b-a|.$

Thus $|b-a|$ is even, so $a+b$ is smallest with $\{a,b\}=\{1,3\};$ and $cd$ is a multiple of $27,$ so $c+d$ is smallest with $\{c,d\}=\{3,9\}.$ These satisfy all conditions, giving $a+b+c+d=1+3+3+9=16.$

Thus, the correct answer is D.

Answer: A

Solution:

A palindrome is fixed by its first half. Counting base-$3$ palindromes by length gives $2$ of length $1$ or $2,$ $6$ of length $3$ or $4,$ $18$ of length $5$ or $6,$ and $54$ of length $7.$

That totals $2+2+6+6+18+18+54=106$ palindromes with at most $7$ digits. Since $2007=2202100_3,$ the $7$-digit palindromes larger than it are $2210122,$ $2211122,$ $2212122,$ $2220222,$ $2221222,$ and $2222222,$ which is $6$ of them.

Therefore the count is $106-6=100.$

Thus, the correct answer is A.

Answer: A

Solution:

Track a third point always halfway between the two particles. Between the moments when the particles are at vertices/midpoints, both particles move linearly, so the midpoint moves linearly too, tracing straight segments that form a small triangle $XYZ.$

By symmetry $XYZ$ shares its center with $\triangle ABC.$ If $O$ is that center and $F$ is the midpoint of a side, then $$OZ=OC-ZC=\dfrac23 CF-\dfrac12 CF=\dfrac16 CF,$$ while $OC=\dfrac23 CF.$

So the ratio of circumradii is $\dfrac{OZ}{OC}=\dfrac14,$ and the area ratio is $\left(\dfrac14\right)^2=\dfrac{1}{16}.$

Thus, the correct answer is A.

Answer: A

Solution:

Let the legs be $a\le b.$ The condition is $\tfrac12 ab=3\left(a+b+\sqrt{a^2+b^2}\right),$ so $$ab-6a-6b=6\sqrt{a^2+b^2}.$$

Squaring and simplifying gives $ab(ab-12a-12b+72)=0,$ hence $(a-12)(b-12)=72.$ The positive integer solutions are $(a,b)=(3,4),(13,84),(14,48),(15,36),(16,30),(18,24),(20,21).$

The pair $(3,4)$ is extraneous (its area $6$ does not equal $3$ times its perimeter $12$), so exactly $6$ triangles work.

Thus, the correct answer is A.

Answer: A

Solution:

Multiplying by $b$ and subtracting $a$ gives $\dfrac{14b^2}{9a}=bk-a,$ an integer. Since $\gcd(a,b)=1,$ it follows that $a\mid14.$ Multiplying instead by $9a$ and subtracting $14b$ gives $\dfrac{9a^2}{b}=9ak-14b,$ so $b\mid9.$

Thus $a\in\{1,2,7,14\}$ and $b\in\{1,3,9\}.$ Checking the coprime candidates, the expression is an integer only for $(a,b)=(1,3),(2,3),(7,3),(14,3).$

So there are $4$ such pairs.

Thus, the correct answer is A.

Answer: C

Solution:

Set $D=(-1,0,0)$ and $E=(1,0,0),$ and let $\triangle ABC$ lie in the plane $z=k\gt0.$ Because $\angle CDE$ and $\angle DEA$ are right angles, $A$ and $C$ lie on radius-$2$ circles centered at $E$ and $D$ in the planes $x=1$ and $x=-1,$ so $A=(1,y_1,k),\ C=(-1,y_2,k)$ with $y_j=\pm\sqrt{4-k^2}.$

Since $\angle ABC=90^\circ,$ $AC=2\sqrt2,$ which forces $y_1=-y_2.$ Taking $y_1=1,$ $y_2=-1,$ gives $k=\sqrt3,$ so $A=(1,1,\sqrt3),$ $C=(-1,-1,\sqrt3),$ and $B$ is $(1,-1,\sqrt3)$ or $(-1,1,\sqrt3).$

In the first case $BE=2$ with $BE\perp DE;$ in the second $BD=2$ with $BD\perp DE.$ Either way $\triangle BDE$ has legs $2$ and $2,$ so its area is $\tfrac12(2)(2)=2.$

Thus, the correct answer is C.