They ride toward each other, so their speeds add; find the time until they meet

The meeting time is $\frac{45}{18 + 12}$ hours; multiply Alice’s speed by that time

They ride toward each other, so their speeds add. That closes the $45$-mile gap at $18 + 12 = 30$ mph, and they meet after $\frac{45}{30} = 1.5$ hours. Alice starts at $A,$ so by then she’s gone $18 \cdot 1.5 = 27$ miles. Thus, E is the correct answer.

Let $p$ be the pizza’s weight and convert cups of orange slices to pounds using $\frac14$ pound per cup

Set $\frac13 p + \frac72 \cdot \frac14 = \frac34 p + \frac12 \cdot \frac14$ and solve for $p$

Let $p$ be the pizza’s weight. A cup of orange slices is $\frac14$ pound, so the two sides balance as $\frac13 p + \frac72 \cdot \frac14 = \frac34 p + \frac12 \cdot \frac14,$ that is $\frac13 p + \frac78 = \frac34 p + \frac18.$ Collect the pizza terms: $\frac68 = \left(\frac34 - \frac13\right)p = \frac{5}{12}p.$ So $p = \frac34 \cdot \frac{12}{5} = \frac95 = 1\frac45.$ Therefore, the answer is A.

A perfect square divisible by $5$ must be divisible by $25$

Write the square as $(5k)^2 = 25k^2 \lt 2023$ and count the valid $k$

If a perfect square is divisible by $5,$ it’s divisible by $25,$ so it looks like $(5k)^2 = 25k^2.$ We need $25k^2 \lt 2023,$ i.e. $k^2 \lt 80.9.$ That allows $k = 1, 2, \ldots, 8,$ which is $8$ squares. Thus, A is the correct answer.

In any quadrilateral, each side is less than the sum of the other three

If the longest side is $s,$ the other three sum to $26 - s,$ so $s \lt 26 - s$

In any quadrilateral each side is shorter than the sum of the other three. Call the longest side $s.$ The rest sum to $26 - s,$ so $s \lt 26 - s,$ which gives $s \lt 13$ and hence $s \le 12.$ Can we hit $12?$ The sides $4, 12, 9, 1$ work, since $12 \lt 4 + 9 + 1.$ So the greatest length is $12.$ Therefore, the answer is D.

Factor each base into primes: $8 = 2^3$ and $15 = 3 \cdot 5$

Combine into $10^{15} \cdot 3^5,$ which is $3^5$ followed by $15$ zeros

Factor everything into primes. $8^5 \cdot 5^{10} \cdot 15^5$ $= 2^{15} \cdot 5^{10} \cdot 3^5 \cdot 5^5$ $= 2^{15} \cdot 5^{15} \cdot 3^5$ $= 10^{15} \cdot 243.$ That’s $243$ followed by $15$ zeros, so it has $3 + 15 = 18$ digits. Thus, E is the correct answer.

Each edge belongs to $2$ faces, so the total face value is $2$ times the total edge value

Each vertex belongs to $3$ edges, so the total edge value is $3$ times the vertex total

Count by incidences. Each edge lies on $2$ faces, so the six face values together are $2$ times the total of all edge values. Each vertex lies on $3$ edges, so the total edge value is $3$ times the vertex sum. Chaining these, the cube’s value is $2 \cdot 3 \cdot 21 = 126.$ Therefore, the answer is D.

The running total reaches $3$ only through the first few rolls; list the disjoint ways to hit exactly $3$

The cases are first roll $3;$ rolls $1,2;$ rolls $2,1;$ rolls $1,1,1;$ add their probabilities

The total can only reach exactly $3$ through the opening rolls, and these ways are disjoint: $3$ alone (probability $\frac16$), then $1,2$ and $2,1$ (each $\frac1{36}$), and $1,1,1$ (probability $\frac1{216}$). Add them up: $\frac{36}{216} + \frac{6}{216} + \frac{6}{216} + \frac{1}{216} = \frac{49}{216}.$ Thus, B is the correct answer.

The Breadus reading is linear in Fahrenheit through $(110, 0)$ and $(350, 100)$

The slope is $\frac{100}{350 - 110};$ apply it to the $200 - 110$ degrees above the $0$ point

The Breadus reading is linear in Fahrenheit through $(110, 0)$ and $(350, 100),$ so $B = \frac{100}{350 - 110}(F - 110)$ $= \frac{5}{12}(F - 110).$ Plug in $F = 200:$ $B = \frac{5}{12} \cdot 90 = 37.5.$ Therefore, the answer is D.

The year $2023$ contributes two $2$s (even), one $0,$ and one $3$

The four month-and-day digits must supply one more $0$ and one more $3,$ with the remaining two digits equal to each other

The year $2023$ already gives two $2$s (even), one $0$, and one $3$. So to make every digit occur an even number of times, the four digits of $MM$ and $DD$ must supply one more $0$, one more $3$, and two equal digits, while keeping the number of $2$s even. Checking the legal dates gives $01/13,$ $01/31,$ $02/23,$ $03/11,$ $03/22,$ $10/13,$ $10/31,$ $11/03,$ and $11/30,$ exactly $9$ dates. Thus, E is the correct answer.

Let the current mean be $m$ over $n$ tests, so the current total is $mn$

Adding one $11$ gives mean $m + 1$ and adding three $11$s gives mean $m + 2;$ form two equations

Let $m$ be the current mean over $n$ quizzes. One more $11$ makes the mean $m + 1:$ $\frac{mn + 11}{n + 1} = m + 1,$ which tidies up to $m + n = 10.$ Three more $11$s make it $m + 2:$ $\frac{mn + 33}{n + 3} = m + 2,$ i.e. $3m + 2n = 27.$ Solve the pair and $m = 7.$ Therefore, the answer is D.

Each corner triangle has legs $a,b$ with $a+b=\sqrt3$ (a side of the big square) and $a^2+b^2=2$ (a side of the small square)

$(a+b)^2-(a^2+b^2)$ gives $2ab;$ then $a$ and $b$ are the roots of $t^2-\sqrt3\,t+\tfrac12=0$

Each corner right triangle has legs $a$ and $b.$ A side of the outer square gives $a+b=\sqrt3,$ and a side of the inscribed square gives $a^2+b^2=2.$ Subtract to find the product: $2ab=(a+b)^2-(a^2+b^2)$ $=3-2$ $=1,$ so $ab=\tfrac12.$ Then $a$ and $b$ are the roots of $t^2-\sqrt3\,t+\tfrac12=0,$ namely $\tfrac{\sqrt3\pm1}{2}.$ The ratio of the smaller leg to the larger is $\dfrac{\sqrt3-1}{\sqrt3+1}$ $=\dfrac{(\sqrt3-1)^2}{2}$ $=2-\sqrt3.$ Thus, C is the correct answer.

The reversed number ends in the first digit of $N,$ and for divisibility by $5$ that digit must be $0$ or $5$

Since $N$ is three digits it starts with $5;$ count the multiples of $7$ from $500$ to $599$

When we reverse $N,$ its last digit is the first digit of $N.$ For the reversal to be divisible by $5,$ that digit is $0$ or $5.$ A three-digit number can’t start with $0,$ so $N$ starts with $5,$ meaning $500 \le N \le 599$ (and the reversal ends in $5,$ always fine). Now just count multiples of $7$ here: from $7 \cdot 72 = 504$ to $7 \cdot 85 = 595,$ that’s $14$ numbers. Therefore, the answer is B.

Points that see segment $AC$ at a fixed $60^\circ$ lie on a circular arc through $A$ and $C$

The farthest such point from Abdul is a diameter endpoint; find the circle’s diameter with the law of sines

Let $A$ be Abdul, $C$ be Chiang with $AC = 48,$ and $B$ be Bharat with $\angle B = 60^\circ.$ Every point seeing $AC$ at $60^\circ$ lies on one circular arc, so all valid $B$ sit on a circle where chord $AC$ subtends $60^\circ.$ The law of sines gives its diameter, $\frac{AC}{\sin 60^\circ} = \frac{48}{\frac{\sqrt3}{2}} = 32\sqrt3.$ Now $AB$ is a chord, and a chord is longest when it’s a diameter. So $AB = 32\sqrt3$ and $AB^2 = 1024 \cdot 3 = 3072.$ Thus, C is the correct answer.