Maths problems (interesting ones)

[Denman]

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)
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???

More to follow if there is any interest......

[FizMan]

1) 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)



2) 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.



3) 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

[Denman]

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?


ah you got the first one now

[FizMan]

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.

Yes yes more.

[Denman]

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

6) 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)


7)
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 )

[FizMan]

4) 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.

5) 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.


6) 9*9+9+9+9/9

That valid?


7) Twelve days?


8) 33%?

[Alan Anderson]

Originally Posted by Denman

...I...only have a traditional weighing machine.

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.

1 (given)
2 = 3 - 1
3 (given)
4 = 3 + 1
5 = 9 - 3 - 1
6 = 9 - 3
7 = 9 - 3 + 1
8 = 9 - 1
9 (given)
10 = 9 + 1
  :
38 = 27 + 9 + 3 - 1
39 = 27 + 9 + 3
40 = 27 + 9 + 3 + 1

[FizMan]

Right on each one? What aboot the scale one? Are both me and Alan right?

And Dave, not quite on the money there... consider all the possible combinations:


Boy -> Girl
Boy -> Boy

Girl -> Girl
Girl -> Boy

We are asked what's the probability that they have two girls. We KNOW that they have 1 girl already, so that eliminates the Boy->Boy possibility.

Which leaves us with 3 possible outcomes.

Boy -> Girl
Girl -> Girl
Girl -> Boy

So what are the chances they had Girl -> Girl? 1/3 (or 33%)

[Dave Scheck]

Originally Posted by FizMan

So what are the chances they had Girl -> Girl? 1/3 (or 33%)

Good catch. Never was very good with stats

[Astronouth7303]

Originally Posted by Dave Scheck

Good catch. Never was very good with stats

me 2. Those sort of problems always throw me. I say someone should go out and take a study!

[Steve Howland]

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!

[FizMan]

I'll start thinking aboot the coin one, but right now with the fruit, I'm pretty sure you take just two pieces of fruit.

Each guard you give him half (1/2 of two fruit is 1 piece) then he gives you one piece back (back up to two). You never lose any fruit.

[Alan Anderson]

Originally Posted by diobsidian

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.

The details are simple but hard to explain without graphics. Label the twelve coins A through I. Weigh ABC against DEF, then ABC against GHI. The results will tell you which group of three is different and whether it is heavy or light. Then you weigh any two of the different group's coins against each other to find out which is the odd one; if they match, the third one is the different one.

if ABC < DEF and ABC < GHI, one of ABC is light
if ABC < DEF and ABC = GHI, one of DEF is heavy
if ABC = DEF and ABC < GHI, one of GHI is heavy
if ABC = DEF and ABC > GHI, one of GHI is light
if ABC > DEF and ABC = GHI, one of DEF is light
if ABC > DEF and ABC > GHI, one of ABC is heavy

The other three combinations are not possible if exactly one coin is different.

Once it's narrowed down to three coins, just compare two of them. If they don't balance, you already know whether the fake is heavy or light; just choose that one. If they do balance, the other must be the fake.

Originally Posted by diobsidian

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?

If I take two, then I can give one (half of my two) to each guard and according to the agreement he'll give it back to me, leaving me with the two I took.

[DaBruteForceGuy]

Originally Posted by Steve Howland

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)

AHHHHH my brain hurts ... i need the solution!

[Daniel Brim]

Originally Posted by Steve Howland

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)

You don't need to use the scale at all.

First, if you need a small graduated cylinder. 11 coins should be exactly the same, so they all have the same density. The fake, although not necessarily weighing the same, will have a different density than the others. Simply put in X mL of water in the cylinder and calculate the displacement of the water for each coin dropped, and the fake coin will be different.