16 cups tea (about half of pot A).
The amount of tea that can be kept within each pot is determined by the height of the spout opening. The tea level cannot rise above the spout opening since any extra tea would merely spill out from the spout.
Visual estimate would conclude that the spout of teapot B is approximately half the height of that of teapot A, therefore providing only half of the capacity, or 16 cups tea.

Sage A’s hat is white.
It cannot be blue.

1. The puzzle says that “Sage A sees that the other 2 sages are wearing blue hats.” So both sage B and sage C wear blue hats. That’s a given.

2. Neither B or C can deduce the color of their own hats, and therefore, they remain silent.

Here’s why:
If A’s hat were white, B would see C’s blue hat, and see A’s white hat. B’s hat could be either blue or white.

If A’s hat were blue, B would see C’s blue hat, and A’s blue hat. B’s hat could still be either blue or white.

If A’s hat were white, C now would see B’s blue hat, and see A’s white hat. C’s hat could be either blue or white.

If A’s hat were blue, C would see B’s blue hat, and see A’s blue hat. C’s hat could still be either blue or white.

So Sage B and C cannot speak up for sure.

3. After a while, when Sage A sees that neither sage B or sage C speak up, (and they speak up if they could solve the ridde), sage A reasons that neither sage B nor sage C, based on what they see, can deduce the color of their own hats.

Sage A reasons further that since one of the three sages should be able to speak to the king, sage A alone must be the one able to figure it out, since the other two cannot.

There must be something that makes him unique, something that he sees that the other two don’t see, something that he sees that makes the riddle solveable for him, while the other two cannot solve it. What does he see?
He sees two blue hats (B’s blue hat and C’s blue hat). So Sage A again reasons correctly that only he sees two blue hats. Since sage B’s hat is blue and sage C’s hat is blue, Sage A correctly reasons that his own hat cannot be blue. (Or else all three sages would see the other two sages wear blue hats).

So A’s hat must be white, and so A speaks up.

————————-

Note:
If A’s hat were blue, all three sages would be wearing blue, each sage would see the other two wearing blue hats, and there would be nothing to differentiate them, with all three silent. Nobody would speak up and they would all go home.

10.
Alternate solution-
Number the bottles, write that in binary, number prisoners from 1 to 10. Prisoner i drinks bottle n if bit i of n is 1. Each dead prisoner tells you that bit i of poisoned bottle is 1.

10 prisoners must sample the wine. Bonus points if you worked out a way to ensure than no more than 8 prisoners die.

Number all bottles using binary digits. Assign each prisoner to one of the binary flags. Prisoners must take a sip from each bottle where their binary flag is set.

Here is how you would find one poisoned bottle out of eight total bottles of wine.

          Bottle  | 1 |  2 | 3 | 4 |  5 |  6 | 7 |  8 |
Prisoner A | X |  X | X | X |  X |  X | X |  X |
Prisoner B | X |  X | X | X |  X |  X | X |  X |
Prisoner C | X |  X | X | X |  X |  X | X |  X |
In the above example, if all prisoners die, bottle 8 is bad. If none die, bottle 1 is bad. If A & B dies, bottle 4 is bad.

With ten people there are 1024 unique combinations so you could test up to 1024 bottles of wine.

Each of the ten prisoners will take a small sip from about 500 bottles. Each sip should take no longer than 30 seconds and should be a very small amount. Small sips not only leave more wine for guests. Small sips also avoid death by alcohol poisoning. As long as each prisoner is administered about a millilitre from each bottle, they will only consume the equivalent of about one bottle of wine each.

Each prisoner will have at least a fifty percent chance of living. There is only one binary combination where all prisoners must sip from the wine. If there are ten prisoners then there are ten more combinations where all but one prisoner must sip from the wine. By avoiding these two types of combinations you can ensure no more than 8 prisoners die.

Solution to the problem is as follows:

Move
Time
A & B Cross with torch
2
A Return with torch
1
C & D Cross with torch
10
B Return with torch
2
A & B Cross with torch
2
————————
————–
Total Time
17

Separate any 5 coins on side A and 5 coins on side B.

Lets say,
Side A – TTTHH
Side B – HHHTT

Revrse all coins of anyonen side, lets say side B
So
Side A – TTTHH
Side B – TTTHH

We got all similar facing coin, so simple

Johnny’s mother has a child i.e. Johnny

Other two childs are April and May.

So, third child is JOHNNY

“Incorrectly”

There are 110 squares without a rook.

There are 60 squares of size 1×1.
There are 35 squares of size 2×2.
There are 12 squares of size 3×3.
There are 3 squares of size 4×4.

A total of 60 + 35 + 12 + 3 = 110.

The German has the fish and drinks coffee in the green house, which is fourth on the block.