Solution:

If $k$ of the baskets are worth $3$ points and the rest worth $2,$ the total is $2(5-k)+3k=10+k.$

As $k$ ranges over $0,1,\ldots,5,$ the total takes every integer value from $10$ to $15,$ giving $6$ possibilities.

Thus, the correct answer is E.

Solution:

After reversing the second row to $11,10,9,8$ and the fourth row to $25,24,23,22,$ the two diagonals are $1,10,17,22$ and $4,9,16,25.$

Their sums are $1+10+17+22=50$ and $4+9+16+25=54,$ so the positive difference is $54-50=4.$

Thus, the correct answer is B.

Solution:

Since $\sqrt{x}=x^{1/2},$ we have $x\sqrt{x}=x^{1}\cdot x^{1/2}=x^{3/2}.$

Taking the cube root multiplies the exponent by $\tfrac13,$ giving $\left(x^{3/2}\right)^{1/3}=x^{1/2}.$

Thus, the correct answer is D.

Solution:

One player's salary is largest when the other $20$ players each earn the minimum $\$15{,}000.$

That leaves $\$700{,}000-20\cdot\$15{,}000=\$700{,}000-\$300{,}000=\$400{,}000$ for the single player.

Thus, the correct answer is C.

Solution:

Since $(y-x)^2=(x-y)^2,$ the two inputs are identical.

Therefore $(x-y)^2\,\$\,(y-x)^2=\left((x-y)^2-(x-y)^2\right)^2=0^2=0.$

Thus, the correct answer is A.

Solution:

Since $AB=4\,BD$ and $AB+BD=AD,$ we get $5\,BD=AD,$ so $BD=\tfrac15 AD.$

Since $AC=9\,CD$ and $AC+CD=AD,$ we get $10\,CD=AD,$ so $CD=\tfrac{1}{10}AD.$

Then $BC=BD-CD=\tfrac15 AD-\tfrac{1}{10}AD=\tfrac{1}{10}AD.$

Thus, the correct answer is C.

Solution:

The large triangle has side length $10$ times that of a small triangle, so its area is $10^2=100$ times as large.

Since the small triangles tile it without overlap, exactly $100$ of them are required.

Thus, the correct answer is C.

Solution:

If $r$ roses and $c$ carnations are bought, then $3r+2c=50.$ Because $2c$ and $50$ are even, $3r$ must be even, so $r$ is even.

Also $3r\le 50,$ so $r\le 16.$ The even values $r=0,2,4,\ldots,16$ each give a valid $c,$ which is $9$ bouquets.

Thus, the correct answer is C.

Solution:

By Vieta's formulas, the sum of the roots of $ax^2-2ax+b=0$ is $\dfrac{-(-2a)}{a}=2.$

The average is half of this, namely $1.$

Thus, the correct answer is A.

Solution:

Let $O$ be the center and $D$ the midpoint of $AB.$ Then $OD\perp AB$ with $AD=3,$ so $OD=\sqrt{5^2-3^2}=4.$

Since $C$ is the midpoint of the minor arc, $O,$ $D,$ $C$ are collinear and $DC=OC-OD=5-4=1.$

Then $AC=\sqrt{AD^2+DC^2}=\sqrt{3^2+1^2}=\sqrt{10}.$

Thus, the correct answer is A.

Solution:

Using the recurrence, $u_5=2u_4+u_3=2u_4+9$ and $u_6=2u_5+u_4=2(2u_4+9)+u_4=5u_4+18.$

Setting $5u_4+18=128$ gives $u_4=22,$ so $u_5=2\cdot 22+9=53.$

Thus, the correct answer is B.

Solution:

Each flip counts $100000$ steps, so Pete took $44\cdot 100000+50000=4{,}450{,}000$ steps.

Dividing by $1800$ gives about $2472$ miles, which is closest to $2500.$

Thus, the correct answer is A.

Solution:

Since the mean of the first $n$ terms is $n,$ their sum is $n^2.$

The $n$th term is $n^2-(n-1)^2=2n-1,$ so the $2008$th term is $2\cdot 2008-1=4015.$

Thus, the correct answer is B.

Solution:

Because $\angle ABO=90^\circ,$ segment $AB$ is vertical, so $A=\left(5,\,5\tan 30^\circ\right)=\left(5,\,\tfrac{5\sqrt3}{3}\right).$

A $90^\circ$ counterclockwise rotation about the origin sends $(x,y)$ to $(-y,x),$ so the image of $A$ is $\left(-\tfrac{5\sqrt3}{3},\,5\right).$

Thus, the correct answer is B.

Solution:

From $a^2+b^2=(b+1)^2$ we get $a^2=2b+1,$ so $a$ is odd and $a^2$ is an odd perfect square.

Since $b\lt 100,$ we need $a^2=2b+1\lt 201,$ and $a^2\ge 9$ for $b\ge 4.$ The odd squares $9,25,49,81,121,169$ give $a=3,5,7,9,11,13,$ which is $6$ triangles.

Thus, the correct answer is A.

Solution:

Whenever at least one die is rolled, by symmetry the sum is odd with probability $\tfrac12.$

No die is rolled only when both coins are tails, with probability $\tfrac14;$ that sum $0$ is even. So the answer is $\left(1-\tfrac14\right)\cdot\tfrac12=\tfrac38.$

Thus, the correct answer is A.

Solution:

Exactly one approval among three occasions arises in $\binom{3}{1}=3$ ways, each with probability $(0.7)(0.3)(0.3)=0.063.$

The total is $3\cdot 0.063=0.189.$

Thus, the correct answer is B.

Solution:

Let $n$ be the number of bricks. Brenda lays $\tfrac{n}{9}$ per hour and Brandon $\tfrac{n}{10},$ so together they lay $\tfrac{n}{9}+\tfrac{n}{10}-10$ per hour.

Over $5$ hours this equals $n:$ $$5\left(\tfrac{n}{9}+\tfrac{n}{10}-10\right)=n.$$ Solving, $\tfrac{5n}{9}+\tfrac{n}{2}-50=n,$ which gives $n=900.$

Thus, the correct answer is B.

Solution:

The submerged cross-section is a circular segment. The chord is $4-2=2$ feet below the center, and $\cos\theta=\tfrac{2}{4}=\tfrac12,$ so the half-angle is $60^\circ$ and the central angle is $120^\circ.$

The sector area is $\tfrac{120}{360}\pi(4)^2=\tfrac{16\pi}{3},$ and the triangle formed by the two radii has area $\tfrac12(4)^2\sin 120^\circ=4\sqrt3.$ The segment area is $\tfrac{16\pi}{3}-4\sqrt3.$

Multiplying by the length $9$ gives $9\left(\tfrac{16\pi}{3}-4\sqrt3\right)=48\pi-36\sqrt3.$

Thus, the correct answer is E.

Solution:

Of the $36$ equally likely outcomes, the pairs giving sum $5$ are $(1,4),(2,3),(2,3),(4,1),$ which is $4$ outcomes.

Sum $7$ comes from $(1,6),(2,5),(2,5),(3,4),(3,4),(4,3),$ which is $6,$ and sum $9$ from $(1,8),(3,6),(3,6),(4,5),$ which is $4.$

The probability is $\tfrac{4+6+4}{36}=\tfrac{14}{36}=\tfrac{7}{18}.$

Thus, the correct answer is B.

Solution:

Seat the women first. The first woman may take any of the $10$ chairs, and since seats alternate, the remaining women fill their four seats in $4!$ ways, giving $10\cdot 4!=240$ arrangements.

Fix a woman in chair $1.$ Her spouse must sit in chair $4$ or chair $8;$ each choice then forces the placement of every other man consistently. So each seating of the women yields exactly $2$ valid seatings of the men.

The total is $2\cdot 240=480.$

Thus, the correct answer is C.

Solution:

There are $\tfrac{6!}{3!\,2!}=60$ distinguishable orderings. The three reds must occupy non-adjacent positions, and the possible red placements are $\{1,3,5\},\{2,4,6\},\{1,3,6\},$ and $\{1,4,6\}.$

For $\{1,3,5\}$ and $\{2,4,6\},$ the remaining seats are mutually non-adjacent, so the blue bead can go in any of the $3,$ giving $3+3=6.$ For $\{1,3,6\}$ and $\{1,4,6\},$ two remaining seats are adjacent, so the blue must separate the whites, giving $2+2=4.$

That is $10$ valid orderings, so the probability is $\tfrac{10}{60}=\tfrac16.$

Thus, the correct answer is C.

Solution:

The painted rectangle is $(a-2)\times(b-2),$ and it is half the floor, so $ab=2(a-2)(b-2).$

Expanding gives $0=ab-4a-4b+8,$ and adding $8$ yields $(a-4)(b-4)=8.$

With $b\gt a\gt 0,$ the factorizations $8=1\cdot 8=2\cdot 4$ give $(a,b)=(5,12)$ and $(6,8).$ So there are $2$ possibilities.

Thus, the correct answer is B.

Solution:

Let $M$ be the point with $\triangle BMC$ equilateral, on the same side of $BC$ as $A.$ Then $\angle ABM=70^\circ-60^\circ=10^\circ$ and $\angle MCD=170^\circ-60^\circ=110^\circ.$

Since $AB=BM$ and $MC=CD,$ triangles $ABM$ and $MCD$ are isosceles, giving $\angle AMB=85^\circ$ and $\angle CMD=35^\circ.$

Then $\angle AMD=360^\circ-85^\circ-60^\circ-35^\circ=180^\circ,$ so $M$ lies on $\overline{AD}$ and $\angle BAD=\angle BAM=85^\circ.$

Thus, the correct answer is C.

Solution:

Number the pails so Michael is at pail $0$ and the truck at pail $1.$ Michael reaches pail $n$ at $40n$ seconds. The truck leaves pail $n$ at $50(n-1)$ seconds and arrives there at $50(n-1)-30$ seconds.

Michael and the truck are together at pail $n$ when $50(n-1)-30\le 40n\le 50(n-1),$ which simplifies to $5\le n\le 8.$

At pail $5$ they meet as the truck departs, at pails $6$ and $7$ Michael passes it, and at pail $8$ they meet as the truck arrives. Between pails $6$ and $7$ the truck must overtake Michael once more, so in total they meet $5$ times.

Thus, the correct answer is B.