h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

[quick change of forum, kitty thinks it is a shame that we didn't manage to do it with a change of location, but never mind, answer cakewalker's first riddle here...]
[ to recap: the first riddle was set at http://www.h2g2.com/forumframe.cgi?forum=24306&thread=46208
and went along the lines of:
"First puzzle. When this house was still a stately home, though after it's vineyard had been sold, it had a butler known for dishonest dealings. He would water down his master's wine and profit by selling unused wine. Luckily, he was caught after only three thefts of such kind. However, to convict him, the police were faced, with the following puzzle. Each theft, he removed three pints of wine from the barrel, replacing them with water. Over the three thefts, he removed nine pints, reducing the wine in the barrel to just half it's strength. How much wine did the barrel originally hold? "]

*scratches head and starts to count fingers*

First he removes 3 pints of wine... if there were x to start with theres x-3 after the first theft
then he removes 3 pints of mixed wine and water...

[mathematical summary, and hint]

original no of pints of wine = x
concentration = 1

first theft of 3 pints
amount of wine removed = 3*1
number of pints of wine remaining = x-3
concentration of wine in mixture = (x-3)/x

second theft of 3 pints
amount of wine removed = 3*(x-3)/x
concentration of wine in mixture =

third theft of 3 pints
amount of wine removed =
concentration of wine in mixture = 1/2 =

Well, I fear the only answer I can manage for the concentration is (3((3((x-3)/x))/x))/x which I find rather cumbersome, not to mention probably incorrect. As you can see I am too in the dark over the answers to these puzzles - I was not party to their creation and have not had the opportunity as a portrait to solve them, or at least, not yet.

*starts counting toes too*

[mathematical summary, and improved hint]

original no of pints of wine = x
concentration = 1

first theft of 3 pints
amount of wine removed = 3*1
number of pints of wine remaining = x-3
concentration of wine in mixture = (x-3)/x

second theft of 3 pints
amount of WINE removed = 3 * (x-3) /x
concentration of WINE in mixture = initial amount of wine in barrel minus total amount of wine removed so far all over total amount of liquid in barrel = [x - 3 - 3(x-3)/x]/x

third theft of 3 pints
amount of WINE removed = ?
if this leaves half the original strength, then the TOTAL amount of WINE removed in the three thefts is also equal to x/2

Well, my good friends, while you are pondering that conundrum, here is another to bear in mind and possibly give inspiration, mathematically at least.

When I had been head of my household for, lets see, only around two years, a messager came to our manor and instructed me that it was the express wish of the King that an army be gathered. Thirty manors, including ours, were selected, with the choosing procedure to be followed as follows: from each manor there should be collected the same number of men collected as had been collected up until then. From the first manor was taken one man, from ours, the second, also one man. The army went to far away circuits to fight and won victoriously. How many people were in that army?

Correction:
the King's servant arrived at the first manor alone, to the second he went with one other...
[Cakewalker forgets that though the servant of the King only took one man from this manor, he had been duped into taking a girl dressed as a man (by the girl of course, Cakewalker would have sent 2 men if he had known, the girl had an unhealthy interest in soldiers and actually the manor settled down a bit after she left...)]

How about 2 raised to the power of 30?

a funny click sound comes from the top of Cakewalker's portrait, and it begins to hang slightly oddly, he looks worried

Localised Earthquake?

Forces beyond our collective comprehension at least. Either way, it would appear you are correct, my good sir, the answer is indeed 2 raised to the power of 30.

Time for the next puzzle. During the construction of the great hall that is now your prestigious establishment, there were numerous street urchins working in the less accessible parts of the building. In one such instance, a rope was supposed to be hung over a wheel fixed to the roof of the building. At one end of the rop a weight is fixed, which exactly counterbalances that of a young street urchin anging onto the other end of the rope. If the urchin begins to climb the rope, what will be the result?

Would the weight ascend the rope at the same rate as the urchin was climbing?

One would have thought that might be, though as they are both exactly the same weight they would be in equilibrium. I have an idea that would mean that the weight would stay in the same place.

Another strange click is heard and Cakewalker's portrait shifts more suddenly and has a pronounced tilt to it - he is aware that it only hangs off one pin now. A sheet of paper slips from behind it, it reads:

"Provided friction is neglected, the weight rises to match the urchin. Given friction in the pulley, the weight rises less or indeed not at all if the pulley is sufficiently stiff"

Does that mean my random guess was right?

The paper flips over in a sudden gust or breeze...

"Indeed you both were right... you have but one more riddle to solve to leave this room and you have even gained the right to take one object you may find useful for your journey. The objects from which you may choose will be listed when you gain egress"

I suppose taking more food is out of the question.

*Kitty thinks that they had better tuck in while the going is good...*

[A new tray of hot food has appeared at the table, and it smells good]