You have 9 balls...

Deiwos said:
You fill the 3 up and pour it in the 5. You fill the 3 again and pour it in to the 5 to leave you with 1 gallon in the 3. Empty the 5, fill it with the 1 from the 3 and then fill the 3 again. Put this into the 5 and you have 4 gallons.

I didn't google this :)
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That makes a lot more sense... :)
 
simulatorman said:
1 - Split the balls into three groups of three balls each.

2 - First use of the scale: Weigh any two groups against each other.

3 - If the groups weigh the same then the heavier ball is in the third group, otherwise it is in the group that weighs more.

4 - We now know that the heavy ball is one of three.

5 - Second use of the scale: Using the group of balls that we know contains the heavy ball, weight any two balls against each other.

6 - If one of the balls is heavier then we have our answer. If the balls weigh the same then the third ball is the heavy one and we again have our answer.

:p:D
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There is no point to these threads because people just google it and copy/paste the answers.:confused:

From Wiki

1 - Split the balls into three groups of three balls each.
2 - First use of the scale: Weigh any two groups against each other.

3 - If the groups weigh the same then the heavier ball is in the third group, otherwise it is in the group that weighs more.

4 - We now know that the heavy ball is one of three.

5 - Second use of the scale: Using the group of balls that we know contains the heavy ball, weight any two balls against each other.

6 - If one of the balls is heavier then we have our answer. If the balls weigh the same then the third ball is the heavy one and we again have our answer.
 
i-bert said:
There is no point to these threads because people just google it and copy/paste the answers.:confused:

From Wiki

1 - Split the balls into three groups of three balls each.
2 - First use of the scale: Weigh any two groups against each other.

3 - If the groups weigh the same then the heavier ball is in the third group, otherwise it is in the group that weighs more.

4 - We now know that the heavy ball is one of three.

5 - Second use of the scale: Using the group of balls that we know contains the heavy ball, weight any two balls against each other.

6 - If one of the balls is heavier then we have our answer. If the balls weigh the same then the third ball is the heavy one and we again have our answer.
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LOL BUSTED!!!
 
Q3

A man is trapped in a room. The room has only two possible exits: two doors. Through the first door there is a room constructed from magnifying glass. The blazing hot sun instantly fries anything or anyone that enters. Through the second door there is a fire-breathing dragon. How does the man escape?
 
Piece of cake. Took me around a minute. :p

Split them into three groups of 3
balance the first two groups on the balance and if they are of equal weight then the third group contains the heavier ball, different weights then the heavier group contains it...

Split that into three balls and balance two balls on the balance, if equal the heaviest ball is the one not on the balance if not then the heavier ball is the one tipping the balance.

Ta da! :p

EDIT: Seems we're on question three...

so.. get the dragon to chase you into the other room, slip round it and escape as the dragon goes up in smoke! :p
 
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I was thinking along the lines of an electronic scale where you could only weigh one group at a time hehe, took me a little while to figure out i misread balance for scale, so I failed at the first one.
 
Q4


500 men are arranged in an array of 10 rows and 50 columns according to their heights. Tallest among each row of all are asked to come out. And the shortest among them is A. Similarly after resuming them to their original positions, the shortest among each column are asked to come out. And the tallest among them is B. Now who is taller A or B ?
 
sedm1000 said:
Q3

A man is trapped in a room. The room has only two possible exits: two doors. Through the first door there is a room constructed from magnifying glass. The blazing hot sun instantly fries anything or anyone that enters. Through the second door there is a fire-breathing dragon. How does the man escape?
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He waits until night time then just leaves through magnified room?
 
Q5
You are in a game of Russian Roulette with a revolver that has 3 bullets placed in three consecutive chambers. The cylinder of the gun will be spun once at the beginning of the game. Then, the gun will be passed between two players until it fires. Would you prefer to go first or second?


[feel free to contribute your 'own' problems]
 
If its a 6 shooter.. surely first gives you the best odds and a chance to survive. Second gives you worst odds and certain death..
 
Second, if the "spun" means moved round one slot, as that means the chamber fired will be empty, second empty but third (first person to shoot) will be full.

I take it A was correct?
 
sedm1000 said:
Q5
You are in a game of Russian Roulette with a revolver that has 3 bullets placed in three consecutive chambers. The cylinder of the gun will be spun once at the beginning of the game. Then, the gun will be passed between two players until it fires. Would you prefer to go first or second?


[feel free to contribute your 'own' problems]
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First. Assuming 6 chambered gun, you have 50/50 chance of dying. If you go second, assuming the guy didn't die on the first go, you have 3/5 chance of dying.
 
gord said:
If its a 6 shooter.. surely first gives you the best odds and a chance to survive. Second gives you worst odds and certain death..
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Why though?

The other option is the bullets will be at the bottom of the revolver chambers (heavier), therefore first?

If the chamber is due to the hammer being in the middle of the three empty ones, the last empty one will be when it fires as the first pulls the trigger (revolver goes round once to a new chamber) and the second will get shot as that is the first bullet filled chamber?
 
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