Riddle:

Rearrange all the letters in each of the sentences to form, in each case, a well known proverb.

1. I don't admit women are faint.

2. It rocks. The broad flag of the free.

3. Strong lion's share almost gone.

1. Time and tide wait for no man.

2. Birds of a feather flock together.

3. A rolling sone gathers no moss.

Riddle:

A mother has three sick children. She has a 24-ounce bottle of medicine and needs to give each child eight ounces of the medicine. She is unable to get to the store and has only three clean containers, which measure 5, 11 and 13 ounces. The electricity is out and she has no way of heating water to wash the containers and doesn't want to spread germs. How can she divide the medicine to give each child an equal portion without having any two children drink from the same container?

Fill the 5 oz. and 11 oz. Containers from the 24 oz. container. This leaves 8 oz. in the 24 oz. bottle. Next empty the 11 oz. bottle by pouring the contents into the 13 oz. bottle. Fill the 13 oz. bottle from the 5 oz. container (with 2 oz.) and put the remaining 3 oz. in the 11 oz. bottle. This leaves the 5 oz. container empty. Now pour 5 oz. from the 13 oz. bottle into the 5 oz. bottle leaving 8 oz. in the 13 oz. bottle. Finally pour the 5 oz. bottle contents into the 11 oz. bottle giving 8 oz. in this container.

Riddle:

I can come in a can,

I can come as a punch,

I can come as a win,

You can eat me for lunch.

What am I?

Beet/Beat

Riddle:

My first is often at the front door. My second is found in the cereal family. My third is what most people want. My whole is one of the United States.

Matrimoney. (mat + rye + money). Matrimony is certainly a "united state"!

Riddle:

0,1,2,3,4,5,6,7,8,9

Use the digits above once each only to compose two fractions which when added togeather equal 1.

35/70 + 148/296 = 1

Riddle:

There were five men going to church and it started to rain. The four that ran got wet and the one that stood still stayed dry. How did the one stay dry?

It was a body in a coffin with the bearers.

Riddle:

You are given a set of scales and 12 marbles. The scales are of the old balance variety. That is, a small dish hangs from each end of a rod that is balanced in the middle. The device enables you to conclude either that the contents of the dishes weight the same or that the dish that falls lower has heavier contents than the other. The 12 marbles appear to be identical. In fact, 11 of them are identical, and one is of a different weight. Your task is to identify the unusual marble and discard it. You are allowed to use the scales three times if you wish, but no more. Note that the unusual marble may be heavier or lighter than the others. You are asked to both identify it and determine whether it is heavy or light

Most people seem to think that the thing to do is weight six coins against six coins, but if you think about it, this would yield you no information concerning the whereabouts of the only different coin. As we already know that one side will be heavier than the other. So that the following plan can be followed, let us number the coins from 1 to 12. For the first weighing let us put on the left pan coins 1,2,3,4 and on the right pan coins 5,6,7,8. There are two possibilities. Either they balance, or they don't. If they balance, then the different coin is in the group 9,10,11,12. So for our second weighing we would put 1,2 in the left pan and 9,10 on the right. If these balance then the different coin is either 11 or 12. Weigh coin 1 against 11. If they balance, the different coin is number 12. If they do not balance, then 11 is the different coin. If 1,2 vs 9,10 do not balance, then the different coin is either 9 or 10. Again, weigh 1 against 9. If they balance, the different coin is number 10, otherwise it is number 9. That was the easy part. What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these coins could be the different coin. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings. Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the different coin is either 3 or 4. Weigh 4 against 9, a known good coin. If they balance then the different coin is 3, otherwise it is 4. Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy coin, or 1 is a different, light coin. For the third weighing, weigh 7 against 8. Whichever side is heavy is the different coin. If they balance, then 1 is the different coin. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy coin or 2 is a light different coin. Weigh 5 against 6. The heavier one is the different coin. If they balance, then 2 is a different light coin.

Riddle:

You can use me to stop,

You take me to smoke;

Not only do I stop, But I am a stop,

And the result of pool's first stroke.

What am I?

Brake/ Break

Riddle:

A great banquet was prepared for a Roman emperor and his courtiers. 22 Dormice, 40 Larks' Tongues, 30 Flamingos and 40 Roast Parrots were served.

How many portions of Boiled Ostrich were served?

42. Each vowel is worth 2 and each consonant 4, so Dormice gives 22, ect.