These questions are not too difficult, only the 6th is for non-native english speaking people a little puzzle.

1. 9 PM 2. 10 3. 1600 4. 17 5. E 6. BANALITIES

Another question in return:

You have four snooker-balls, all looking equal. You know that at most one ball has a different weight, but you are not sure if there is such a ball, and you also don't know if the weigth is less or more, and you don't know which one it is. You own a accurate digital balance which can be used to exactly measure the weigth of one or some of the balls. You have to find out if one of the balls has a different weigth, if that is true, which ball it is and the exact weigth of each of the balls. You may use the balance three times and all the balls have to stay in one piece....

Each number system has exactly 10 different digits.

Helen Thomas
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I took those questions from the Mensa workout.

Answer 30 questions in half an hour. I got 21 out of 30 and they said I "could" qualify for Mensa. These posted here were some of the questions I couldn't fathom. Well done!

Helen Thomas
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Originally posted by Peter van de Riet:

Another question in return:

You have four snooker-balls, all looking equal. You know that at most one ball has a different weight, but you are not sure if there is such a ball, and you also don't know if the weigth is less or more, and you don't know which one it is. You own a accurate digital balance which can be used to exactly measure the weigth of one or some of the balls. You have to find out if one of the balls has a different weigth, if that is true, which ball it is and the exact weigth of each of the balls. You may use the balance three times and all the balls have to stay in one piece....

First weigh two balls

then the other two balls

If the weights are equal , then given only one ball could have a different weight, all balls weigh the same weight.

If not equal, add the two weighings.

Remove one ball's weight from the equation by weighing just one ball from the heavier side.(Third weighing)

Multiply that weight by 2. If it's equal to the weight of the other side then the other ball is the odd ball, if it's not equal the removed ball is the odd ball. [ January 31, 2005: Message edited by: Helen Thomas ]

Helen, This works only if one of the balls is heavier, if one ball weights less then your solution won't work.

Helen Thomas
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Oh well, kiss goodbye to Mensa

I think I know where it's wrong

"If it's equal to the weight of the other side then the other ball is the odd ball, if it's not equal the removed ball is the odd ball."

Change that to Remove one ball's weight from the equation by weighing just one ball from weighing 1 or weighing 2 picked randomly (Third weighing)

Multiply that weight by 2. If the weight x 2 is equal to the weight of the other side or it's weight is half the weight of it's side ( calculated from weighing 1 and weighing 2) , you can determine which three balls have the same weight.

( That's removed the problem with too many if's)

Still NOT right!!

Weigh three balls. If they have the same weight ...... Monday morning [ January 31, 2005: Message edited by: Helen Thomas ]

ball 3 + ball 2 = 11 (weight of ball 3 = 6 and ball 4 = 5)

or weight of ball 3 = 5 and ball 4 = 6 and you still don't know

Helen Thomas
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Originally posted by Peter van de Riet:

or weight of ball 3 = 5 and ball 4 = 6 and you still don't know

I do so know because ball 3 and ball 2 were weighed together.

ball 2 + ball 3 = 11 ; ball 2 = 5

so ball 3 must be 6 and by default ball 4 is 5 as at most one ball is different.

But I do accept if the first three balls have the same weight you aren't sure what's the weight of the fourth ball.

You cannot believe how many tasks I've done in between . Drove 20 miles too. This is my final answer. Someone else have a go. [ February 01, 2005: Message edited by: Helen Thomas ]

weighing 1 : ball a weighing 2 : ball a + ball b + ball c

situation 1: (weighing 1) equals (weighing 2 divided by 3) situation 2: (weighing 1) not equals (weighing 2 divided by 3)

weighing 3: situation 1 -> ball d situation 2 -> ball b

Results: ball a: result of weighing 1 ball b: situation 1 -> result of weighing 1 situation 2 -> result of weighing 3 ball c: weighing 2 minus (ball a) minus (ball b) ball d: situation 1 -> weighing 3 situation 2 -> if (ball a) equals (ball b or ball c) -> (ball a) else (ball b)

Helen Thomas
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That looks correct. That was a good problem. Thanks for posting it Peter.