Which Alice?

Solutions to the Puzzles
the March Hare has twice as many tarts as the Dormouse (since the
Dormouse has half as many as the March Hare), so the March Hare
has two portions. The Hatter has three times as much as the March
Hare, so the Hatter has six portions. Since the Hatter has six
portions and the Dormouse has only one portion, then the Hatter
has five portions more than the Dormouse. Also, the Hatter has
twenty more tarts than the Dormouse, so five portions of tarts is the
same as twenty tarts. This means that there are four tarts in one
portion. So, the Dormouse has four tarts, the March Hare has eight,
and the Hatter has twenty-four, which is indeed twenty more than
the number the Dormouse has.
THE TABLES ARE TURNED! The March Hare took fivesixteenths
of the tarts, which left eleven-sixteenths. Then the
Dormouse took seven-elevenths of that—in other words, sevenelevenths
of eleven-sixteenths. Well, 7/11 x 11/16 = 7/16, so the
Dormouse took seven-sixteenths of all the tarts. Since the March
Hare took five-sixteenths of all the tarts, the two together took
seven-sixteenths and five-sixteenths, which is twelve-sixteenths.
This left four-sixteenths, which is one-quarter of the tarts, for
the Hatter. Also, eight tarts were left for the Hatter, so eight tarts
is one-quarter of all the tarts. Therefore there were thirty-two
tarts altogether. Now, one-sixteenth of thirty-two is two, so fivesixteenths
of thirty-two is ten. Therefore the March Hare ate ten
tarts. This left twenty-two tarts. Then the Dormouse ate sevenelevenths
of the twenty-two remaining tarts, which is fourteen
tarts (since one-eleventh of twenty-two is two, then sevenelevenths
must be fourteen). This left eight tarts for the Hatter,
so everything checks.
HOW MANY FAVORITES? This puzzle, usually solved by algebra,
is extremely simple if looked at in the following way: Let us first
give three tarts apiece to every one of the thirty guests. This leaves
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