puzzle for all..

R

roran

Registered Users (C)
No good postings..
No arguments...
Long time waiters are slowly moving out..
To cheer up, I thought I will post a puzzle, which was sent by my manager..

I have four cards:

+---+ +---+ +---+ +---+
| | | | | | | |
| A | | B | | 2 | | 3 |
| | | | | | | |
+---+ +---+ +---+ +---+

On each of these cards, there is a letter on one side, and a number on the other side.

I have a theory about these cards. My theory is: "If there is a vowel on one side of a card, there is an even number on the other side."

I would like to test my theory. Which card or cards must I turn over, and look at what's on the other side, to completely test my theory?
 
Turn over card A and card 3.

You need to test original statement - if vowel --> even number
and contrapositive statement - if not even number -- > not vowel.

your hypothesis which is a sufficient condition, is proved if and only if:

(1) Card A has an even number on the other side AND
(2) Card 3 has a non-vowel letter on the other side.

ric


Originally posted by roran
No good postings..
No arguments...
Long time waiters are slowly moving out..
To cheer up, I thought I will post a puzzle, which was sent by my manager..

I have four cards:

+---+ +---+ +---+ +---+
| | | | | | | |
| A | | B | | 2 | | 3 |
| | | | | | | |
+---+ +---+ +---+ +---+

On each of these cards, there is a letter on one side, and a number on the other side.

I have a theory about these cards. My theory is: "If there is a vowel on one side of a card, there is an even number on the other side."

I would like to test my theory. Which card or cards must I turn over, and look at what's on the other side, to completely test my theory?
Click to expand...
 
Last edited by a moderator:
ONLY CARD "A" , BECAUSE

"vowel- even" means only that.

If there's a picture of Sahib_Q on the other side of card "2" it will not contradict the condition.

If there's even number on the other side of "B" it's OK, too...
 
Shahibq

Your argument has one loop hole... if there a vowel on the back of 3, the theory will be wrong. The same arguement for handsome.

I cannot quite understand Chicken65's answer. Maybe I am missing something. Kozhi... can you explain why you said that?

I agree with ric2.

(I confused ric2 and handsome the first time i posted)
 
Last edited by a moderator:
I agree with ric2

--
A - to make sure other side is even number
3 - to make sure other side is not a vowel


--
B - the other side can be odd or even, so there is no use in looking at other side
2 - the other side can be a vowel or not a vowel, so there is no use in looking at other side.
 
Haaa.. It is a funny one. It took me 2 seconds to figure out. Welcome much harder puzzles.
 
Just A

Well ric2 according to the theory "If there is a vowel on one side of the card there is an even number on the other side of the card".
For that hypothesis you need to just turn over A. According to rules of exclusion all you have to check is vowel=even not constanant=odd. Constanant can also be even but all vowel should be even.
so if "all vowel" = even then hypothesis is true.
If Constants are even then hypothesis can yet be true or false.
 
It should be A and 3 since all of them have to be checked to verify our condition
A to see if it is an even number.If not FALSE
3 to be sure it is not an vowel.IF not FALSE
It does not matter what B has
It does not matter what 2 has
 
Good one!

A and 3 is definitely right.

But Roran, your boss might be checking out how much time you have to spend on these kinds of puzzle.
 
Re: Good one!

LOL! Or, your boss could be using this as a way to check whether you spend your time working on "extra-curricular" activities like this ;)

Originally posted by gc791188
A and 3 is definitely right.

But Roran, your boss might be checking out how much time you have to spend on these kinds of puzzle.
Click to expand...
 
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