The first fully-correct set of answers was submitted by Chuck Takahashi. Congrats, Chuck.
1. 636, since every competitor (except the ultimate winner) must lose once. The two finalists may or may not have played the same number of games, but this was not stated as one of the conditions.
2. 6 x pi meters. The length of the inner comm line is 2 x pi x r meters, while the length of the outer line is 2 x pi x (r + 3) meters. The difference is 6 x pi meters. All the other numbers (distance from equator, diameter of sphere, etc.) are extraneous factoids.
3. 1 fignet is 4000 km. The formula for a sphere's volume is 4/3 x pi x r (cubed), where r is radius; that for a sphere's surface area is 4 pi x r (squared). If we express the planet's radius in fignets and recall that its surface area in square fignets is equal to its volume in cubic fignets, we can determine the radius by equating the two formulae and solving for r. Thus, r = 3 (fignets). We know the diameter (24,000 km), so 1 fignet must be 4000 km.
4. Picard only gained 5 Lwaxana-less minutes. Since the shuttlecraft arrived 10 minutes early, it must have shaved off 5 minutes from each leg (starbase to asteroid, asteroid to starbase). The Enterprise had been moving from the asteroid 55 minutes (one hour minus 5 minutes). Since Picard was going to get rid of Lwaxana an hour after the Enterprise arrived at the asteroid, he only parted with her five minutes earlier (but it was probably worth it). The speed is irrelevant.
5. There are three Vulcan and one human survivors. If there are n survivors, of which x are Vulcans, the probability that two random survivors are Vulcans is [x (x-1)] / [n (n - 1)]. We are told this ratio is 1/2. Next, we must find integral values of x and n that give this expression a value of 1/2. The only values that allow a total of no greater than 14 are n = 4 and x = 3.
6. 120.7 km. Solution one: For simplicity, equate 1 unit with the space between ships (50 km), and the time it takes them to travel this distance. Their speed (distance/time) is also 1. Let x be the total distance traveled by the query and reply bozons, and also their speed. The speed of the query bozons relative to the ships is x - 1. The speed of the reply bozons is x + 1. The reply and query bozons both travel 1 unit and make the round trip in unit time, so: 1/ (x - 1) + 1/ (x + 1) = 1, which can be expressed as x (squared) - 2x - 1 = 0, for which the positive value of x is 1 + square root of 2. Multiply this by 50 km to get the final answer of 120.7 km. The warp speed is irrelevant.
Solution two: Rob Carvalho said this about solution one: "Your method is way wierd. Now I'm no expert in hyperrelativity, but if we were to treat this problem in a nonrelativistic way, it just makes sense to say that in the time that the Enterprise traveled 7/10 of its distance, the bozons traveled 17/10 that distance, and then while the Enterprise traveled the remaining 3/10, the bozons traveled back the 7/10 to meet the Addison. This means that the bozons traveled at about 2.42 times the speed of the ships (which is irrelevant, they will be assimilated), and thus went about 2.42 times the distance that the ships went, which is 120km. Why bring in higher math when simple conjecture/trial and error will do? But your answer is more precise."
Kevin O'Connell noted: "Bozons: how far in what frame of reference? The starships? Do we assume that space isn't warped around the ships as they travel at Warp 9? The answer will be different if measured from a "motionless" frame of reference (I suppose, since I don't know how warp speed affects the Lorentz transformation!). If observed from a planet being passed, the answer will further depend on the velocity of the planet.
Kevin, is, of course, totally correct. The given answer is from the starships' (or bozons') frame of reference.
7. Yes. Assume there are a maximum of n forehead wrinkles on one (very wrinkly) Klingon. If there are n Klingons, it is possible (though unlikely) that all have different numbers of wrinkles. However, since there are at least n + 1 Klingons, the n + 1st Klingon must share the same number of forehead wrinkles as one of the other Klingons.
8. Jake is 13. Let B = Benjamin's age, J = Jake's age, and F = Ben's father's age. His father is five times Jake's age, so F = 5J. When Benjamin is as old as his father is now, Jake will be F - B years older than he is now; his age will then be J + F - B. Since this is 8 years more than Benjamin's present age, J + F - B = B + 8. The third equation is F + B = 100. Solving these equations gives J = 13, B = 35 and F = 65.
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