1. FEROCIOUS. "Vulpine" evokes a fox’s stereotypical slyness; "lupine" evokes a wolf’s defining trait, its fierce, ferocious nature.
2. DOGMA. "Iconoclast" opposes tradition, whereas "heretic" opposes dogma.
3. LIGHT. Hemoglobin binds/transports oxygen while chlorophyll absorbs light for photosynthesis.
6. CIRCUMLOCUTION. An euphemism softens something offensive; a circumlocution avoids being direct. Both terms have Latin roots.
7. EQUIVALENCE CLASS. In math, an equivalence relation partitions a set into disjoint classes, and each element uniquely determines the class to which it belongs.
8. APHORISM. Just as "bittersweet" exemplifies an oxymoron, likewise, the given sentence is a well-known aphorism.
9. TRANSPARENT. "Cryptic" comes from the Greek word krýptein, meaning "to hide or conceal" (i.e., to obscure). "Diaphanous" derives from diaphanés, meaning "transparent."
10. WIRE. Serrated commonly describes an edge; barbed commonly describes wire.
11. WEAVER. An anvil is used by a blacksmith; a loom is used by a weaver.
12. BREVITY. Equivocal refers to ambiguity; laconic, to brevity.
13. Monday. Note that \(1000=7\times142+6\). Hence, \(142\) weeks from today will be a Tuesday, \(142\) weeks plus \(6\) days from today will be \(6\) days after today, namely Monday.
14. \(23\). Assume that the two holes that go all the way through the cube are the front and side holes, while the top hole only goes halfway through. By symmetry, there is no difference in their positions since the cube can be rotated to accommodate these holes. The solid has 6 faces derived from the original cube, though they are perforated. The front and side holes create \(4\) openings, each consisting of \(3\) faces. In addition, \(4\) faces surround the top hole opening, and a cross-shaped face that connects the bottom parts of the front and side hole openings. Therefore, the total number of faces in the solid is \(6 + 4 \times 3 + 4 + 1 = 23\). Note that a face is defined as a flat surface that forms part of the boundary of a solid.
15. \(1/2\) or 50 %. Let \(H\) represent the outcome "heads" and \(T\) represent the outcome "tails." We will denote the order of these letters to indicate the result of an observation, where the first letter represents the side observed, and the second letter represents the opposite side. For example, in the case of a fair coin, an outcome may be \(HT\), indicating that the observed side was heads and the opposite side was tails. Each of the following \(6\) outcomes is equally likely before we know anything: Coin \(1\)(fair): \(HT\) or \(TH\), Coin \(2\)(Fair): \(HT\) or \(TH\), Coin \(3\)(unfair): \(H_{1}H_{2}\) or \(H_{2}H_{1}\) (substrips indicate particular heads \(i\)). Because we know that the observed face was heads, we can eliminate two possibilities. However, the remaining four are still equally likely; In two of these scenarios, the other side is heads, and in the other two, the other side is tails. Thus, the desired probability is \(\frac{2}{4}=\frac{1}{2}\).
16. \(72\). Since all the books are distinct, there are \(5!\) possible arrangements, counting the order in which the objects appear in a sequence. Nonetheless, we want to subtract the arrangements in which two specific books, namely the red ones, are contiguous, to obtain the number of ways in which they are not contiguous or touching. There are \(4\) possible places in which they will be next to each other, and for each there are \(2!\) ways to arrange these books. Moreover, for each of these arrangements, there are \(3!\) ways to arrange the \(3\) blue books in the three remaining places. Thus, the total number of different ways the books can be arranged such that the red books are not contiguous with one another is simply \(5!-4\cdot2!\cdot3!=72\)
17. 40%. Imagine that the seven people are arranged in a line. Suppose the first person was born on Friday, and the other \(6\) were not. Since the probability of being born on Friday is \(\frac{1}{7}\) and the probability of not being born on that day is \(\frac{6}{7}\), the probability for this specific arrangement in the line is \(\left( \frac{1}{7} \right)\left( \frac{6}{7} \right)^{6}=\frac{6^{6}}{7^{7}}\). But the person who was born that day could be any one of the \(7\). It doesn't have to be the first person whose birthday falls on Friday; it could be the 2nd, or the 3rd, ..., or the 7th. There are \(7\) possible people to choose from. Hence, the desired probability is \(7\cdot\frac{6^{6}}{7^{7}} =\frac{6^{6}}{7^{6}}=\left( \frac{6}{7} \right)^{6}\approx 0.40\).
18. \(19\). It is left as an exercise for the reader to see why no more than \(19\) bounded areas are achievable. Hint: Look at the diagram below and ask yourself why this type of configuration? And how many intersection points are there? 
19. \(3.6\) km. The trains are exactly \(6\) km apart. Their combined speed is 20 kilometers per hour, which means it will take them \(18\) minutes to collide. During this time, the first train will travel \(2.4\) km. At the moment of collision, \(0.9\) km of the first train will be out of the tunnel in front, while \(0.6\) km will be out in the back, leaving \(1.5\) km still in the tunnel. The second train will have traveled \(3.6\) km by the time of the collision, so only \(0.9\) km of it will remain in the tunnel, with \(2.1\) km out of the tunnel.
20. \(30\) kg, \(50\) kg, and \(120\) kg. In a mobile in equilibrium, the torque on the left and right sides is equal to each other. For a system of only two hanging masses, we will have the equation \(m_{1}d_{1}=m_{2}d_{2}\). We can break down the system into simpler systems until we can solve for the missing weight values. Consider the system of the three masses below the \(200\)-kg mass on the left. Let \(m_{1}=x+y\), \(m_{2}=40\), \(d_{1}=3\), and \(d_{2}=6\). Then, \(m_{1}\cdot3=40\cdot6\Rightarrow m_{1}=80\). Now, we turn our attention to the two masses at the bottom on the left. Modeling as a system of equations involving torque results in \(x+y=80\) and \(5x=3y\), which implies that \(x=30\) and \(y=50\). Hence, the total mass on the left-hand side of the mobile is \(30+50+40+200=320\) kg. Consequently, we can solve for \(z\) by setting the torques equal, \(320\cdot3=z\cdot8\Rightarrow z=120\) kg.
21. \(21\). To verify this, you can start by drawing a regular dodecahedron or a two-dimensional projection of it, and numbering each of its \(30\) edges. Begin at one corner and on one edge, and then explore all the possible paths you can take along the edges of the dodecahedron. There’s no need to start from different corners or edges, as doing so would only result in repeating the same paths. While this process can be a bit more complex than it seems, it is useful for solving various spatial and logical puzzles.
22. \(12\). Let \(A\), \(B\), and \(C\) be the sets of all containers with alpha particles, beta particles, and gamma particles, respectively. By the inclusion-exclusion principle,
$$\left| A\cup B\cup C \right|=\left| A \right|+\left| B \right|+\left| C \right|-\left| A\cap B \right|-\left| A\cap C \right|-\left| B\cap C \right|+\left| A\cap B\cap C \right|=$$
But, we know that \(\left| A\cup B\cup C \right|+\left| A\cup B\cup C \right|^{c}=150\), where \(\left| A\cup B\cup C \right|^{c}=27\). Then, by substitution,
23. Yes! They can go back home, and either \(A\) or \(B\) may be the first to announce their color correctly. There are two possibilities: If humans \(B\) and \(C\) had their hats of the same color, then human \(A\) could immediately tell the color of their own hat (a different color from the observed one), and thus must have pronounced its color correctly. If, on the other hand, humans \(B\) and \(C\) had hats of different colors, then human \(A\) must have remained silent, which would be a clue for human \(B\), who could know by looking at human \(C\) what color the color of their own hat is (other than \(C\)'s hat color). In either case, all of them could be sent back home!
24. \(444\). Each term after is obtained by adding a successive perfect cube to the previous term. \(3+0^{3}=3\), \(3+1^{3}=4\), \(4+2^{3}=12\), \(12+3^{3}=39\), \(39+4^{3}=103\), \(103+5^{3}=228\), \(228+6^{3}=444\)\(,...\)
25. \(204\). Each term after \(720\) is obtained by alternating between adding and subtracting powers of three. \(720-3^{6}=21\), \(21+3^{5}=264\), \(264-3^{4}=183\), \(183+3^{3}=210\), \(210-3^{2}=201\),\(201+3^{1}=204\) \(,...\)
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