Let $t$ be the hours since 1:30. Andy has gone $8t$ miles north. Betsy starts an hour later, so she's gone $12(t-1)$ miles east. We want these equal: $8t = 12(t-1).$ That gives $4t = 12,$ so $t = 3.$ Three hours past 1:30 is 4:30. Thus, E is the correct answer.

The starting $10$-pound mix holds $5$ pounds of peanuts and $2$ pounds of cashews. Add $x$ pounds of the second mix, which is $20\%$ peanuts. We want the new peanut fraction to be $40\%,$ so $\frac{5 + 0.2x}{10 + x} = 0.4.$ This means $5 + 0.2x = 4 + 0.4x,$ giving $x = 5.$ Those $5$ pounds bring $0.4 \cdot 5 = 2$ more pounds of cashews, so the box now has $2 + 2 = 4.$ Therefore, the answer is B.

Split into two cases. Say two sides both equal $2025.$ Then the third side can be any integer from $1$ to $2025,$ which is $2025$ triangles. Now suppose $2025$ is the unique longest side. The two equal legs $s$ must satisfy $2s \gt 2025$ by the triangle inequality, and $s \le 2024.$ So $s$ runs from $1013$ to $2024,$ giving $1012$ triangles. Adding up, $2025 + 1012 = 3037.$ Thus, D is the correct answer.

Let $s$ be the number of students. If Ash joins them, his age is $14(s+1) - 12s = 2s + 14.$ If he joins the teachers instead (there are $15 - s$ of them), his age is $52(16 - s)$ $- 55(15 - s) = 3s + 7.$ Both describe the same Ash, so $2s + 14 = 3s + 7.$ That gives $s = 7,$ and Ash is $2 \cdot 7 + 14 = 28.$ Therefore, the answer is A.

Group the sequence into blocks. Block $k$ reads $k, k-1, \ldots, 2, 1, 2, \ldots, k-1, k,$ which is $2k - 1$ terms and ends on $k.$ So after block $k$ we've used $1 + 3 + \cdots + (2k-1) = k^2$ terms. Notice $2025 = 45^2.$ That's exactly the end of block $45,$ whose last term is $45.$ Thus, E is the correct answer.

Label the equilateral triangle $ABC.$ Each $60^\circ$ angle splits into three $20^\circ$ pieces. Take the outermost trisectors from $A$ and $B$: they meet at base angles $\tfrac23 \cdot 60^\circ = 40^\circ,$ so the hexagon vertex there has angle $180^\circ - 2\cdot 40^\circ = 100^\circ.$ The innermost trisectors from $A$ and $B$ meet at base angles $20^\circ,$ giving apex $180^\circ - 2\cdot 20^\circ = 140^\circ,$ and by vertical angles that's the opposite hexagon angle. So the six angles alternate $100^\circ$ and $140^\circ.$ The smallest is $100^\circ.$ Therefore, the answer is C.

By the Remainder Theorem, just plug in. We get $p(1) = 1 + 1 + a + b = 4,$ so $a + b = 2.$ And $p(2) = 8 + 4 + 2a + b = 6,$ so $2a + b = -6.$ Subtract the first from the second: $a = -8,$ hence $b = 10.$ Then $b - a = 10 - (-8) = 18.$ Thus, E is the correct answer.

Number them: (1) at least one true, (2) at least two true, (3) at least two false, (4) at least one false. Suppose (3) is true. Then at least two statements are false. But then (1), (2), and (4) all read as true, which leaves at most one false statement. That's a contradiction, so (3) must be false. Now (1), (2), and (4) are all true, and each matches reality with just one false statement. So exactly $1$ statement is false. Therefore, the answer is B.

The graph passes through $(1,25)$ exactly when $f(1 - a) = 25$. Let $t = 1 - a$, so we count solutions of $f(t) = 25$. Factor $f(x) = 100x(x-1)(x-2)$, with roots $0, 1, 2$. On $(0,1)$, the function is positive and $f(0.5) = 37.5 \gt 25$, so continuity gives one root on each side of $0.5$. On $(1,2)$ the function is negative, while for $x \gt 2$ it increases from $0$ to infinity, giving one more root. Thus $y = 25$ meets the cubic in $3$ points. Each gives one $a$, so there are $3$ values. Thus, C is the correct answer.

Let $O$ be the center on $AB$ and $P$ the midpoint of chord $CD.$ Set $r = OP$ for the small radius and $R = OD$ for the large one. Since $PD = 8,$ the Pythagorean theorem in triangle $OPD$ gives $R^2 - r^2 = 64.$ The shaded area is the big semicircle minus the small one: $\tfrac12\pi R^2 - \tfrac12\pi r^2$ $= \tfrac12\pi(R^2 - r^2)$ $= 32\pi.$ Therefore, the answer is C.

From the arithmetic sequence, $z = 1 + 3d,$ so $z \equiv 1 \pmod 3.$ From the geometric one, $z = p^3$ for some integer ratio $p \ge 2.$ We want the smallest such $z,$ so test $p = 2, 3, 4.$ Only $p = 4$ works, since $p^3 = 64 \equiv 1 \pmod 3.$ That forces $d = 21,$ and the sequences are $1, 22, 43, 64$ and $1, 4, 16, 64.$ So $x + y + z + p + q$ $= 22 + 43 + 64 + 4 + 16$ $= 149.$ Thus, E is the correct answer.

No digit is $0,$ so digits run from $1$ to $9.$ Put the single even digit in the first slot for now and multiply by $4$ at the end to place it. Split on that even digit. If it's the prime $2,$ then the three odd digits all have to be non-prime, so each is $1$ or $9,$ giving $2^3 = 8$ ways. Otherwise the even digit is $4, 6,$ or $8$ ($3$ choices), and exactly one of the odd digits is prime, worth $3, 5,$ or $7$ ($3$ choices) in one of the $3$ odd positions, while the other two odds come from $\{1, 9\}$ ($2^2$ ways). That's $3 \cdot 3 \cdot 3 \cdot 4 = 108.$ Altogether, $4(8 + 108) = 464.$ Therefore, the answer is D.

Let the outer side be $1.$ The squares have sides $1, k, k^2, \ldots,$ and the shaded rings alternate, so the shaded area is $1 - k^2 + k^4 - k^6 + \cdots = \frac{1}{1 + k^2}.$ We're told this equals $64\% = \frac{16}{25},$ so $1 + k^2 = \frac{25}{16}.$ Then $k^2 = \frac{9}{16},$ giving $k = \frac{3}{4}.$ Thus, D is the correct answer.

Seat the first student anywhere. The second student lands next to them with probability $\tfrac{2}{5},$ since two of the remaining five chairs are adjacent. Now the teachers fill two of the four leftover chairs. Of the $\binom{4}{2} = 6$ ways to do that, exactly $3$ are adjacent pairs. So the probability is $\tfrac{2}{5} \cdot \tfrac{3}{6} = \tfrac{1}{5}.$ Therefore, the answer is B.

Let $x = BC.$ Since $ABEF$ is a rectangle with $AB = 1$ and $AF = 7,$ and $AD = 5,$ we get $AC = \sqrt{1 + x^2},$ $CE = 7 - x,$ and $CD = 5 - \sqrt{1 + x^2}.$ The triangles $\triangle ABC$ and $\triangle EDC$ are similar, so $\frac{7 - x}{\sqrt{1 + x^2}} = \frac{5 - \sqrt{1 + x^2}}{x}.$ Clear denominators and square to get $24x^2 + 14x - 24 = 0,$ which factors as $(4x - 3)(3x + 4) = 0.$ The positive root is $x = \tfrac{3}{4}.$ So the area is $\tfrac12 \cdot \tfrac34 \cdot 1 = \tfrac{3}{8}.$ Thus, A is the correct answer.

There are $3^3 = 27$ equally likely placements. Of these, $3$ pile all coins into one jar (max $3$), and $6$ put one coin in each jar (max $1$). The other $18$ split $2$-$1$ (max $2$). So the expected maximum is $\frac{3 \cdot 3 + 2 \cdot 18 + 1 \cdot 6}{27} = \frac{51}{27} = \frac{17}{9}.$ Therefore, the answer is D.

Subtract off the remainders. Both $273420 = 273436 - 16$ and $272745 = 272760 - 15$ are multiples of $N,$ so their difference $675$ is too. Now $272745 = 404 \cdot 675 + 45,$ which makes $45$ a multiple of $N$ as well. The remainder $16$ means $N \gt 16,$ and the only divisor of $45$ bigger than $16$ is $45$ itself. So $N = 45,$ and its tens digit is $4.$ Thus, E is the correct answer.

Look at one factor $kx^2 - 4x - 3.$ Its discriminant $16 + 12k$ is positive, so it has two distinct real roots. By Vieta, their reciprocals sum to $\frac{x_1 + x_2}{x_1 x_2} = \frac{4/k}{-3/k} = -\tfrac{4}{3},$ and notice the $k$ cancels. Summing over all $2025$ factors, the reciprocals total $2025 \cdot \left(-\tfrac{4}{3}\right) = -2700.$ There are $4050$ roots in all, so the harmonic mean is $\frac{4050}{-2700} = -\tfrac{3}{2}.$ Therefore, the answer is B.

Each interior entry feeds two entries below, and the end values $-1$ and $1$ cancel in the sum. So every row's total doubles the one above. The top row sums to $3,$ and $12{,}288 = 3 \cdot 2^{12},$ so this is the $12$th row (counting the top as row $0$). Track the diagonals from the left. The second diagonal is $3, 2, 1, 0, -1, \ldots,$ dropping by $1$ each row. The third diagonal adds these up: for $n \ge 4$ it equals $7 - (0 + 1 + \cdots + (n-4))$ $= 7 - \frac{(n-4)(n-3)}{2}.$ Plug in $n = 12$: $7 - \frac{8 \cdot 9}{2} = -29.$ Thus, A is the correct answer.