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Qtec
06-06-2003, 09:49 PM
There are three light switches outside a room. Two are false.You may only look in room once. When you look,only one of the switches may be in the ''ON'' position.
Which switch is the real one ??

Q

highsea
06-10-2003, 04:26 AM
The one in the room

Ken
06-11-2003, 03:06 PM
Can't tell unless you have infrared vision. Then, if a switch was just turned off to get to the one-on state the light would still be hot. Then you can tell. If, besides looking in, I am permitted to feel the temperature of the light then I can tell which is the real switch.

If the light is not hot then it is the switch that was always off.
KenCT

Qtec
06-11-2003, 06:09 PM
Well done Ken. If you liked this one ,try this.


You have 12 coins,
one of the coins is slightly heavier or lighter than the others. You have only a simple scale to help you. No ozs or grams -you can only weigh one against the other.
After 3 weighings you should be able to tell which coin is different and if it is heavier /lighter than the rest.
Each weighing must be done seperately.
this is tricky ,

Q

Deniel
06-11-2003, 11:09 PM
Separate the 12 coins into 4 groups of 3 coins (A,B,C,D) Compares A with B (that's 1), if it weights even than the fake coin is in either C or D, if it doesn't weight even then it's either in A or B. Say it weights even, which means the culprit is in either C or D, grab either C or D (Take C) and then weight it against A or B (Take A, that's 2) If C and A weights even, then the fake is in D. If it doesn't weight even then it is in C. Let's say it's in D There are 3 coins in the D group, grab any two of them and compare them. If it's even then the fake is the one not being compared, if it's not even, then we're screwed /ccboard/images/graemlins/grin.gif

Qtec
06-12-2003, 05:57 AM
Almost ,you are going in the right direction. /ccboard/images/graemlins/cool.gif
Not bad though.


[ Kunt u Nederlands praten ]

Q

Deniel
06-13-2003, 02:55 AM
Yeah, I thought I got it covered until I get to the last part and was really puzzled ^__^
Can't wait to find out the solution

highsea
06-13-2003, 04:06 AM
<blockquote><font class="small">Quote Qtec:</font><hr>
You have 12 coins,
one of the coins is slightly heavier or lighter than the others. You have only a simple scale to help you. No ozs or grams -you can only weigh one against the other.
After 3 weighings you should be able to tell which coin is different and if it is heavier /lighter than the rest.
Each weighing must be done seperately.
this is tricky ,
Q <hr /></blockquote>


First Solution:

Divide the coins into three 4 coin main groups, and then each of these into sub-groups of 3 and 1 coins. Place two of the main groups on the pans of the balance. Observe the condition of the balance. This is the first weighing.

Rotate the 3 coin groups, and observe the condition of the balance. This is the second weighing.

If there is no change, all 3 coin groups are good coins. Clear them from the balance, and rotate the single coins. This is the third weighing, and will identify the odd coin and determine its relative weight. Problem solved!

If there is a change, it will identify the group that contains the odd coin, and determine its relative weight. Clear the balance of all other coins, and put one coin from the odd 3 coin group on each pan, and the third one on the table. This is the third weighing, and will identify the odd coin. As you already know its relative weight - Problem solved!

Second Solution:

Number the coins. Weigh them in this sequence to identify the odd coin and determine it's relative weight.

(left pan)/ (right pan)
first weighing (5,6,8,10)/ (7,9,11,12)
second weighing (2,3,4,7)/ (5,6,11,12)
third weighing (1,4,10,11)/ (2,5,7,8)

-CM

Qtec
06-13-2003, 05:22 AM
Goddam your one smart m.......r.Excellent. I do think that your second ex. is a bit the same as your first.
I also found two ways . The other is to split the coins into 2 groups of 6.[ A and B] You leave out 1 coin from each and weigh 5 against 5 .If they are not equal you take 2 coins off of B . You take 2 coins from A and add them to B The 2 good coins [ the ones left out ] add them to A . This tells you in which group of three the odd one out is and whether it is lighter or heavier.

Q &gt; mumble ,mumble , this took me a week ....

Q.

highsea
06-13-2003, 05:30 AM
<blockquote><font class="small">Quote Qtec:</font><hr> Goddam your one smart m.......r.Excellent. I do think that your second ex. is a bit the same as your first. <hr /></blockquote>

The first method reduces the third weighing to a pair of coins. The second method has 3 weighings of 4 coins each, so they are mathematically different.

-CM

highsea
06-13-2003, 05:46 AM
Ok, here's another one.

You have ten boxes; each contains nine balls. The balls in one box weigh 0.9 kg; the rest weigh 1.0 kg. You have one weighing on an accurate scale to find the box containing the light balls. How do you do it?

-CM

highsea
06-13-2003, 06:43 AM
<blockquote><font class="small">Quote Qtec:</font><hr> I also found two ways . The other is to split the coins into 2 groups of 6.[ A and B] You leave out 1 coin from each and weigh 5 against 5 .If they are not equal you take 2 coins off of B . You take 2 coins from A and add them to B The 2 good coins [ the ones left out ] add them to A . This tells you in which group of three the odd one out is and whether it is lighter or heavier.

Q &gt; mumble ,mumble , this took me a week ....

Q. <hr /></blockquote>

Say after your first weighing, A is heavier. You replace 2 coins from B with 2 coins from A and add the 2 coins that were left out to A. A is still heavier. How do you know if the A group has a heavy coin, or the B group has a light coin?
/ccboard/images/graemlins/confused.gif
-CM

Qtec
06-13-2003, 07:11 AM
I told you you were smart . My mistake .you take 3 off side B . You take 2 coins from A + 1 of the good coins [ from the 2 left out ]. N ow it works

Q

Qtec
06-13-2003, 08:03 AM
If i do this again , you might start thinking I,m clever !
The trick is that you empty one box . You take one ball from box number 1. 2 balls from box number 2 .etc you empty box 10 , throw all the balls in . If the diff. is 0.1 you know its boxnumber 1 etc. If there is no diff. then its the box you emptied out.


Q

UWPoolGod
06-13-2003, 08:50 AM
The easier of the coin problem that I had seen says: You ahve 9 rings, one is heavier that the rest. You can use the scale twice. Find which ring it is the heavy one. That one is way easier.

UWPoolGod
06-13-2003, 08:52 AM
There were a bunch of these problem solvers that Microsoft gives for one if its interviews for employment. Just to see how your thinking process works, and if you can think your way through problems. I have a few more if I can think of them.

UWPoolGod
06-13-2003, 09:00 AM
I think this is right...

You have three baskets at a store on a high shelf so you can't look inside. One filled with only apples, one filled with only oranges, and one filled with both apples and oranges. All of the three baskets are mislabeled with "Apples", "Oranges", or "Apples+Oranges". What is the least amount of grabs of one fruit into any of the baskets that would allow you to correctly switch the lables to the right positions?

UWPoolGod
06-13-2003, 09:07 AM
Here's another one-

You are 1/3 of the way through a train tunnel when you hear the whistle blow behind you. If you turn and run back to the opening towards the train it will hit you right at the opening. If you turn and run to the other end away from the train it will hit you precisely at the other ends opening. Assuming that you run 20MPH (Hauling @$$), what is the the speed of the train?

Qtec
06-13-2003, 09:32 AM
1. 3 groups of 3 rings. Weigh 3 against 3. This shows the heavy group of 3. Put one to the side ,weigh 1 against the other. If they are the same the heavy one is the one you left out.

2. I would just feel what is in the baskets ! So , 1 go .

Q

highsea
06-13-2003, 10:04 AM
<blockquote><font class="small">Quote Qtec:</font><hr> My mistake .you take 3 off side B . You take 2 coins from A + 1 of the good coins [ from the 2 left out ]. N ow it works

Q <hr /></blockquote>

Nope. Now you are don't have enough coins to balance A unless you use 2 coins from B. Even if you remove 3 coins from each side, it will take an additional weighing, and you still haven't determined the relative weight! /ccboard/images/graemlins/mad.gif

highsea
06-13-2003, 10:11 AM
<blockquote><font class="small">Quote Qtec:</font><hr> If i do this again , you might start thinking I,m clever !
The trick is that you empty one box . You take one ball from box number 1. 2 balls from box number 2 .etc you empty box 10 , throw all the balls in . If the diff. is 0.1 you know its boxnumber 1 etc. If there is no diff. then its the box you emptied out.


Q <hr /></blockquote>

Essentially correct, but you take 0 balls from the first box (box 0), 1 from 1, 2 from 2, 3 from 3, etc. up to 9 from 9. The weight of the total then tells you which box was light. i.e. 45 KG means box 0 was the light one.44.9 indicates box 1, 44.8 indicates box 2, etc.

-CM

highsea
06-13-2003, 10:26 AM
<blockquote><font class="small">Quote UWPoolGod:</font><hr> I think this is right...

You have three baskets at a store on a high shelf so you can't look inside. One filled with only apples, one filled with only oranges, and one filled with both apples and oranges. All of the three baskets are mislabeled with "Apples", "Oranges", or "Apples+Oranges". What is the least amount of grabs of one fruit into any of the baskets that would allow you to correctly switch the lables to the right positions?

<hr /></blockquote>

It takes 2 grabs. One grab of the apples or oranges basket, and one grab in the mixed basket.

highsea
06-13-2003, 11:03 AM
<blockquote><font class="small">Quote UWPoolGod:</font><hr> Here's another one-

You are 1/3 of the way through a train tunnel when you hear the whistle blow behind you. If you turn and run back to the opening towards the train it will hit you right at the opening. If you turn and run to the other end away from the train it will hit you precisely at the other ends opening. Assuming that you run 20MPH (Hauling @$$), what is the the speed of the train? <hr /></blockquote>

The train is going 60 mph. It's gonna hurt!

Qtec
06-13-2003, 12:13 PM
Your right . i need to think about this one .I,m almost sure there was another solution.

Q

highsea
06-13-2003, 01:53 PM
<blockquote><font class="small">Quote Qtec:</font><hr> Your right . i need to think about this one .I,m almost sure there was another solution.

Q <hr /></blockquote>

Try this way.

Label the balls 1 through 12. Weigh 1, 2, 3, 4 against 5, 6, 7, 8 This is the first weighing.

1. They balance, so 9, 10, 11, 12 contain the odd ball. Weigh 6, 7, 8 against 9, 10, 11. This is the second weighing.

The final weighing is determined by the results of the second weighing:

a. They balance, therefore 12 is the odd ball and so weigh 12 against any other to discover whether it is heavy or light.

b. 9, 10, 11 are heavy and so they contain an odd heavy ball. Weigh 9 against 10. If they balance, 11 is the odd heavy ball, otherwise the heavier of 9 and 10 is the odd ball.

c. If 9, 10, 11 are light, we use the same procedure to reach the same conclusion for the odd light ball.

2. 5, 6, 7, 8 are heavy and so either they contain an odd heavy ball or 1, 2, 3, 4 contain an odd light ball. Weigh 1, 2, 5 against 3, 6, 10.

a. They balance, so the odd ball is 4 (light) or 7 or 8 (heavy). Thus weigh 7 against 8. If they balance 4 is light, otherwise the heavier of 7 and 8 is the odd heavy ball.

b. 3, 6, 10 are heavy, so the odd ball can be 6 (heavy) or 1 or 2 (light). Thus weigh 1 against 2. If they balance 6 is heavy, otherwise the lighter of 1 and 2 is the odd light ball.

c. 3, 6, 10 are light, so the odd ball is 3 and light or 5 and heavy. We thus weigh 3 against 10. If they balance 5 is heavy, otherwise 3 is light.

3. If 5, 6, 7, 8 are light we use a similar procedure to that in 2.

There are several ways to do this, but the rule is: If n weighings are allowed, the odd ball can be determined in (3^n - 3)/2 apparently identical balls. So for 3 weighings,(you have (27-3)/2=12 balls.

-CM &lt;~~~Now can any one tell me why the side pockets attract the cue ball?