Simple logic test

[stan394]

Originally posted by: zimu
A and 1 without a doubt, only those two can disprove the theory, which is all thats required.

wrong

 

[NanoStuff]

A and 1 is the correct answer. Straight forward really, no trick question here.

 

[swtethan]

all

you never know whats on the back of those cards, they could be random

 

[stan394]

Originally posted by: NanoStuff
A and 1 is the correct answer. Straight forward really, no trick question here.

wrong

 

[Dacalo]

Originally posted by: GrantMeThePower

Originally posted by: Qosis

Originally posted by: Slew Foot
A and 2.

The B and 1 are irrelevant, you flip the A (vowel) to make sure there's an odd on the other side. You flip the 2 to make sure there is NOT a vowel on the other side.

It doesn't matter if there is a vowel on the other side of the 2, the statement can still remain true. It doesn't say that the others can NOT have a vowel on the other side, for example. You have to flip the 1 so you can check if the vowel is on the other side to prove it true. And you have to flip the A so you can see if the odd number is on the other side.

Wrong. You need to check number 2 becuase it says "If there is a vowel on one side, then there must be an odd number on the other side." which means "if there is a vowel on one side, then there must not be an even number because an even number is not an odd number" which means "if there is an even number it must NOT have a vowel on the other side" so you need to check number 2. You do NOT need to check number 1 because it doesn't say whether an odd can have a vowel or not, just that a vowel needs an odd number.

By this logic, the two number cards are useless.

The statement says "If there is a vowel on one side, then there must be an odd number on the other side."

When you are stating that 2 needs to be checked, you are assuming that "If there is an odd number on one side, then there must be a vowel on the other side." You are trying to disapprove this by turning the 2 right? So how is this different from turning the 1 to affirm it?

 

[xtknight]

Originally posted by: Mark R
You are presented with 4 cards. The cards have a letter on one side, and a number on the other side. They are laid in front of you as follows:

[*]A[*]B[*]1[*]2

Which cards must be flipped over in order to test the following statement:

If there is a vowel on one side, then there must be an odd number on the other side.

A and 1. B is not a vowel and 2 is not an odd number, obviously. I don't need to test them because they're just wrong. The other two on the other hand must be tested because I don't know what's on the other side but the first side is correct. Hopefully that'll clear it up for the confused people. Mmm...A1 steak sauce.

Blah. Okay, I confess. Only A needs to be tested. I'm wrong.

Remember, the converse, inverse, and contrapositive of a statement are NOT ALWAYS TRUE! That's key in solving this problem.

TRUE/MUST (original statement) - If there is a vowel on one side, then there must be an odd number on the other side.

UNKNOWN (converse) - If there is an odd number on one side, then there must be a vowel on the other side.

UNKNOWN (inverse) - If there is not a vowel on one side, then there must not be an odd number on the other side.

UNKNOWN (contrapositive) - If there is not an odd number on one side, then there must not be a vowel on the other side.

There is a vowel on one side in:
A

Thus, only A must be flipped to reveal the original statement as being true. Jeez, you gotta be careful with this stuff man, this causes religious wars in ATOT.

 

[stan394]

Originally posted by: Dacalo

Originally posted by: GrantMeThePower

Originally posted by: Qosis

Originally posted by: Slew Foot
A and 2.

The B and 1 are irrelevant, you flip the A (vowel) to make sure there's an odd on the other side. You flip the 2 to make sure there is NOT a vowel on the other side.

It doesn't matter if there is a vowel on the other side of the 2, the statement can still remain true. It doesn't say that the others can NOT have a vowel on the other side, for example. You have to flip the 1 so you can check if the vowel is on the other side to prove it true. And you have to flip the A so you can see if the odd number is on the other side.

Wrong. You need to check number 2 becuase it says "If there is a vowel on one side, then there must be an odd number on the other side." which means "if there is a vowel on one side, then there must not be an even number because an even number is not an odd number" which means "if there is an even number it must NOT have a vowel on the other side" so you need to check number 2. You do NOT need to check number 1 because it doesn't say whether an odd can have a vowel or not, just that a vowel needs an odd number.

By this logic, the two number cards are useless.

The statement says "If there is a vowel on one side, then there must be an odd number on the other side."

When you are stating that 2 needs to be checked, you are assuming that "If there is an odd number on one side, then there must be a vowel on the other side." You are trying to disapprove this by turning the 2 right? So how is this different from turning the 1 to affirm it?

let's break it down this way:

for the 2:
- if vowel other side, statement is false
- if no vowel other side, statement is true

so you have to check that

for the 1:
- if vowel other side, statement is true
- if no vowel other side, statement is true

so you don't have to check that.

 

[NanoStuff]

Originally posted by: swtethan
all

you never know whats on the back of those cards, they could be random

It doesn't matter what's behind B and 2. 2 could have a vowel behind it or not, they're not a part of the 'test'.

Unless I'm wrong

 

[jagec]

Originally posted by: tjaisv
But what if there was a Y on the back of the 2?

haha, awesome :thumbsup:

Originally posted by: Dacalo

When you are stating that 2 needs to be checked, you are assuming that "If there is an odd number on one side, then there must be a vowel on the other side." You are trying to disapprove this by turning the 2 right? So how is this different from turning the 1 to affirm it?

Turning the 1 doesn't give you any additional information. If you turn the 1 and it's a vowel, the rule stands. If you turn the 1 and it's a consonant, the rule stands. No reason to turn the one.

 

[thesurge]

a and 2

 

[NanoStuff]

Ok I change my argument. A and 2. I'm leaving now

 

[Schfifty Five]

Definitely A and 2 ONLY.

Card A - Flip it over to make sure there is an ODD number behind it.

Card B - Do not need to flip because constanants can have ODD or EVEN. We are only concerned of Vowels.

Card 1 - Do not need to flip because the statement never says an ODD card must be a Vowel...it says Vowels must have ODD behind it.

Card 2 - Flip it over to make sure there is NO Vowel behind it, if there is, statement is false.

 

[Schfifty Five]

I can't believe there are more votes for A and 1 than there are for A and 2!

 

[swtethan]

there is nothing saying that there is a rule attached to what the card MUST have on them, it could all be random so you would have to turn them all to test if the statement was true

 

[JujuFish]

Possibilities:
A:
Other side is odd: Statement is true
Other side is even: Statement is untrue

B:
Other side is odd: Statement is true
Other side is even: Statement is true

1:
Other side is vowel: Statement is true
Other side is consonant: Statement is true

2:
Other side is a vowel: Statement is false
Other side is a consonant: Statement is true

So, the only two cards that affect the outcome are A and 2.
The key with B and 1 is that while it requires vowels to have odd numbers, no such stipulation is made regarding consonants. So B can have anything behind it. 1 can have anything behind it, because the converse of the logic statement is not required.

 

[tjaisv]

Test is invalid or paradoxical, because it's undefined for Y (which could be both a vowel and a non-vowel) and this could therefore invalidate any testing of the number 2 card, as it could be Y and therefore a vowel (which would necessarily contradict the inferred parameter that any even numbered card not have a vowel on the other end).

 

[TuxDave]

Originally posted by: xtknight

Originally posted by: Mark R
You are presented with 4 cards. The cards have a letter on one side, and a number on the other side. They are laid in front of you as follows:

[*]A[*]B[*]1[*]2

Which cards must be flipped over in order to test the following statement:

If there is a vowel on one side, then there must be an odd number on the other side.

A and 1. B is not a vowel and 2 is not an odd number, obviously. I don't need to test them because they're just wrong. The other two on the other hand must be tested because I don't know what's on the other side but the first side is correct. Hopefully that'll clear it up for the confused people. Mmm...A1 steak sauce.

Blah. Okay, I confess. Only A needs to be tested. I'm wrong.

Remember, the converse, inverse, and contrapositive of a statement are NOT ALWAYS TRUE! That's key in solving this problem.

TRUE/MUST (original statement) - If there is a vowel on one side, then there must be an odd number on the other side.

UNKNOWN (converse) - If there is an odd number on one side, then there must be a vowel on the other side.

UNKNOWN (inverse) - If there is not a vowel on one side, then there must not be an odd number on the other side.

UNKNOWN (contrapositive) - If there is not an odd number on one side, then there must not be a vowel on the other side.

There is a vowel on one side in:
A

Thus, only A must be flipped to reveal the original statement as being true. Jeez, you gotta be careful with this stuff man, this causes religious wars in ATOT.

You need to flip 2 also. If it has an A on the back the statement will be proven wrong. If it doesn't have an A on the back, then the statement is still true because this case doesn't provide a counter example.

 

[cpacini]

Originally posted by: NanoStuff


Unless I'm wrong

You are

 

[Schfifty Five]

Originally posted by: swtethan
there is nothing saying that there is a rule attached to what the card MUST have on them, it could all be random so you would have to turn them all to test if the statement was true

wrong.

Yes it could be random what the cards on the back have....but even so, you do not need to test Card B or Card 1 to determine whether or not the Logic statement is True or False.

 

[Schfifty Five]

Originally posted by: JujuFish
Possibilities:
A:
Other side is odd: Statement is true
Other side is even: Statement is untrue

B:
Other side is odd: Statement is true
Other side is even: Statement is true

1:
Other side is vowel: Statement is true
Other side is consonant: Statement is true

2:
Other side is a vowel: Statement is false
Other side is a consonant: Statement is true

So, the only two cards that affect the outcome are A and 2.
The key with B and 1 is that while it requires vowels to have odd numbers, no such stipulation is made regarding consonants. So B can have anything behind it. 1 can have anything behind it, because the converse of the logic statement is not required.

Clearest most logical answer yet.

 

[mugs]

A, and 2.

(damnit!)

 

[zanieladie]

I pick...look the answer up in the back of the book!

 

[AnonymouseUser]

Originally posted by: JujuFish
Possibilities:
A:
Other side is odd: Statement is true
Other side is even: Statement is untrue

B:
Other side is odd: Statement is true
Other side is even: Statement is true

1:
Other side is vowel: Statement is true
Other side is consonant: Statement is true

2:
Other side is a vowel: Statement is false
Other side is a consonant: Statement is true

So, the only two cards that affect the outcome are A and 2.
The key with B and 1 is that while it requires vowels to have odd numbers, no such stipulation is made regarding consonants. So B can have anything behind it. 1 can have anything behind it, because the converse of the logic statement is not required.

Nowhere does the statement state that their must not be a vowel behind an even number, just that there must be an odd number behind a vowel.

The only cards that need to be turned are A and 1.

Edit >> I stand corrected below. A and 2 it is.

 

[Number1]

It doesn?t mater what?s behind B and 2.
A and 1 must be tested to confirm statement: ?If there is a vowel on one side, then there must be an odd number on the other side?

Another way of saying it:. If there is an odd number on one side, then there must be a vowel on one side.

 

[swtethan]

back of the cards
1
1
a
a


what are you gonna do now? right....