just going to post some random puzzles here, to keep people interested in maths!

Post your answers here, i might make it a quiz or something !
puzzle 1 )
4 astronauts must get from their pod to their ship in 12 mins

they have 1 oxygen tank which can support up to two people at a time

all astronauts travel at different speeds and if two astronauts share the oxygen take, they travel at the speed of the slowest astronaut
The speeds at which they can travel are as follows:

A:1
B:2
C:4
D:5

if A and D go and A returns, this is (Total 6 mins), then A and C go and A returns (Total 11 mins) then A and B go the total is 13 mins. How can it be done?

Puzzle 2)
[size=2]2 fathers and 2 sons went into a restaurant and ordered 3 whole pies, and they ate 1 whole pie each (no! 1 whole pie, not pi, because 1 pi radian is only half of a pie). How come?

Puzzle 3)
How can you get 4 liters of water EXACTLY from using an unlimited supply of water and a 5 liter and 3 liter bucket???

Astronaught A and B cross first (2 minutes). A returns (1 minute; total 3). C and D cross (5 minutes; total 8). B takes the oxygen tank back (2 minutes; total 10) and then A and B go back (2 minutes; Total 12)

There was an old man (Person A), a middle aged man (Person B), and a child (Person C). Person A was a father to Person B, and a grandfather to Person C. Person B was a father to Person C and a son to Person A. Person C was a son to Person B.

Fill the 5 litre bottle. The pour what you can of that into the 3 litre bottle. You’re now left with 2 litres of water in the 5 litre bottle. Empty the 3 litre bottle and pour those 2 litres into it. Fill the 5 litre bottle completely, and from that, fill up the 3 litre bottle to the top (1 litre). You’re now left with 4 litres in the 5 litre bottl.e

yep thats right for the second two,
you need to know some discrete mathmatics (thats like ordering things etc) for first one…
hmm, do i put some more up?

I was in the middle of figuring it out; I’ve seen the puzzle before (using a bridge instead) I just couldn’t quite remember the answer, and I wanted to post the other two before someone beat me.

hehe ok
here are the next couple
4)
I live in a village and I am a shopkeeper. I sell different things and all of them are sold in kg’s ( i.e 1 kg, 12 kg etc.). I only sell my goods in complete kg’s i.e only in whole numbers( 2.5 kg is out of scope). The problem is that I donot have
an electronic weighing machine and only have a traditional weighing machine. A friend of mine has told that I need only “4 different weights” to sell my goods if they are to be sold in the range of 1 to 40 kg. This is perfectly acceptable to me. Can you guess those 4 weights that I need to have so that I can weight anything between 1-40 kg? I am not allowed to use the weighing machine twice to weight a single request. i.e If someone wants to have 37 kg. of rice I will use those four wieghts (or less) to weigh 37 kg. of rice for him and not use the the weigh machine twise.
5)

Marbles

There are 8 marbles that weigh 1 ounce each, and 1 marble that weighs 1.1 ounces. The marbles are all uniform in size, appearance, and shape. You have a balance that contains 2 trays. You are only able to use the scale 2 times. How do you determine which marble is the heaviest using only the scale and marbles in 2 weighings

How do you six nines to equal exactlly 100(you can use addition, divistion, decimals… just no other digits… ( Oh and you cant say that 99+ (99/99), because thats lame lol… and you dont have to use all 9’s)

A smail is at the bottom of a 15 meter tree. He tries to reach the top. Each day, he manages to climb 4 meters, but during night, while he sleeps, he slips back 3 meters. How many days will he need to reach the top?

8)Mr and Mrs Smith have two children. You ring at their door, and one of their two children, a girl opens the door. What is the probability that the other child is also a girl? (Warning: medium difficulty )

I think it’s 8,9,11,12 kg scales… I gotta go to work though, so I only tested it to liek 15 (and 30 once). So let me know if I got it.

Choose six marbles and measure 3v3.

If one pile weighs more than the other:
Take two of those three marbles and measure them. If they’re equal, then the odd one out is the heavier one. If one weighs more than the other, than it’s the heavy one.

If the 3v3 weighs the same:
Weigh the last two and the heavier one will show.

You mean a balance scale, right? The key is to recognize that you can put weights on both sides of the scale to either add to or subtract from the item being weighed. A “balanced ternary” number system fits this problem perfectly, so the weights are 1, 3, 9, and 27 kg.

Day 1 : Start - 0 After Climb - 4 After Sleep - 1
Day 2 : Start - 0 After Climb - 5 After Sleep - 2
Day 3 : Start - 0 After Climb - 6 After Sleep - 3
Day 4 : Start - 0 After Climb - 7 After Sleep - 4
Day 5 : Start - 0 After Climb - 8 After Sleep - 5
Day 6 : Start - 0 After Climb - 9 After Sleep - 6
Day 7 : Start - 0 After Climb - 10 After Sleep - 7
Day 8 : Start - 0 After Climb - 11 After Sleep - 8
Day 9 : Start - 0 After Climb - 12 After Sleep - 9
Day 10 : Start - 0 After Climb - 13 After Sleep - 10
Day 11 : Start - 0 After Climb - 14 After Sleep - 11
Day 12 : Start - 0 After Climb - 15
(There must be some sort of equation, it's escaping me right now)

Well, if you take it as an independent case, it’s like a coin. It’s either a boy or a girl; 50/50. If you take genetics into acount, then if I remember right, there’s a slightly better chance of it being a girl.

Very fun puzzles…but here’s one that has stumped pretty much everyone I know (similar to marble puzzle but harder)…

You have 12 identical-looking coins. One of the coins is a fake, yet you do not know which one it is nor if it weighs more or less than the other 11. Using 3 turns with a regular balance scale, determine which coin is the fake and if it weighs more or less than the other 11. (props to anyone who gets this in under a day)

Another one that I heard on the radio awhile back…(easier but fun)

There is a tree with exemplary fruit in the middle of a courtyard that a farmer has decided must be guarded from people trying to take the fruit. There are 7 circular fences around the tree with a guard at each. Wanting to get at the fruit so badly, you go up to the first guard and tell him that if he lets you through, when you return you shall give him half of all the fruit you have taken, but then the guard must return one piece to you. You continue to bribe every guard you reach until you have reached the tree, whereupon you take the fruit and upon leaving fulfill your deal with all of the gaurds. How many pieces of fruit did you take?

I’ll try and come up with a few more or ask around for some…good luck!

Look at it like this: there are two children, one who opens the door, and one who doesn’t. Before she opens the door, the possibilities are:

Door Opener / Non Door opener

Boy / Girl
Boy / Boy
Girl / Girl
Girl / Boy

Once she opens the door, the first two possibilities are eliminated, leaving only the options that the shy child is either a Boy or a girl. In other words, the sex of the one child has nothing to do with the sex of the other.

Prove me wrong.

EDIT:

Even better, look at it like this:

FAMILIES WITH TWO CHILDREN:
25% have two boys - A girl will answer the door 0% of the time
50% have a boy and a girl - A girl will answer 50% of the time
25% have two girls - a girl will answer 100% of the time

If we take the product of the two percentages in each row, we get some more numbers to cloud the issue:

PROBABLILITY THAT A GIRL WILL ANSWER…
…and the family has two boys - 0%
…and there’s a boy and a girl - 25%
…and the family has two girls - 25%

So, a girl will answer (0% + 25% + 25%) = 50% of the time.
Exactly half of those responses will come from families with two girls, and exactly half will come from families with a boy and a girl.

As for the second, you forgot to disclaim that the person can’t carry more then reasonably possible, or something like that. For example, FizMan’s answer is the most efficient way to get 2, but if you want 3, all you have to do is start with 130. Or for 4, start with 258.

Another thing of interest is that if you were to start with 1, you’d cut it in half for the first guard, and then he’d have to somehow give you 1, and you’d have 1.5. After the next guard, you’d have 1.75. If there was an infinite number of guards, you would have 2, but it would probably be spoiled from having been cut open so long

I think you’re off Kris… I’m going to trust my highschool Finite math on this one…

We’ll do it your way:

Door Opener / Non Door opener

Boy / Girl
Boy / Boy

Girl / Girl
Girl / Boy

Okay, a girl opens the door… so you say that eliminates the first two possibilities…

Why? Remember, two combinations of boy/girl exist, and you’ve eliminated one of them for the wrong reason.

Whoever opens the door does not determine what children exist in the family (with the exception of the boy/boy possibility)

If you want to keep your method of reasoning, we have to add more possibilities. That is, the order of children. I’ll mention the first born child, the other is safely assumed to be the second born:

Door Opener / Non Door opener

Boy (1st born) / Girl
Boy (1st born)/ Boy

Boy / Girl (1st born)
Boy / Boy (1st born)

Girl (1st born)/ Girl
Girl (1st born)/ Boy

Girl / Girl (1st born)
Girl / Boy (1st born)

Eliminate the Boys answering the door, you’re left with the last four. We can eliminate one of the extraneous Girl/Girl possibilities as well.