Simple Logic Quiz
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e and 4
i know my answer was right so this thread is OVER -
now if you change one word or letter from the OP, then you changed it -
and the whole thread is invalid -
now if you change one word or letter from the OP, then you changed it -
and the whole thread is invalid -
Ask Mr. Religion
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They are each a box by the precision of your statement. Hence, each box comprises six sides. Any of the boxes could meet the condition as it is stated and hence all of them must be inspected to verify the truthfulness of the statement as a universal for all four boxes. But there is nothing in the way the problem is stated to imply a universal truth for the four box population. Hence, the truthfulness can apply to "a box", one that actually shows a vowel.
Answer: Box E is a must. Choosing any other box may or may not validate the assumption so stated, but if the statement is true, Box E must show and even number on the side opposite in the six-sided box in question.
AMR
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You're overthinking it, they're only two-sided (originally posed as cards). Plus, you're half right.....
Huckleberry
New member
You're overthinking it, they're only two-sided (originally posed as cards). Plus, you're half right.....
In that case it must be either E or 4. We can't know which without seeing the other sides. :idunno:
Edit: Never mind. Reread the OP and now I have no idea. E and...maybe 4. :idunno:
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What makes you think so?
E and 4?
exminister
Well-known member
The first two
The first box had a vowel on one side so the hidden other side must have an even number. The second an even number, the hidden side a vowel.
The other two show neither a vowel or an even number so the hidden would be the same.
annabenedetti
like marbles on glass
I'm terrible at logic puzzles (or puzzles of any kind besides jigsaw puzzles) so I'm going to take a guess expecting I'm wrong:
E and 7.
7 would disprove the rule if there's a vowel on the other side.
and that's how i concluded - i was right - EOT
Ask Mr. Religion
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You're overthinking it, they're only two-sided (originally posed as cards). Plus, you're half right.....
Then be more precise.
"If a box..." speaks volumes. :AMR:
AMR
at the point that you said box i already assumed a box doesn't matter. the point was vowels and even numbers. a box, circle, triangle or octogon doesn't matter. it's logic
If a box shows an even number on one side, then it can still show anything on the other side and the rule would hold. In other words IF A, then B <> IF B then A.
So this eliminates box 4.
Obviously Box E is required to test the rule.
Clearly, box K will not test anything about boxes with vowels on them.
This leaves box 7 by elimination.
But if we wanted a stronger argument rather than the elimination argument, it is this: Box 7 tests the rule because if the other side were a vowel, then the rule would be proved false. Therefore Box 7 does provide a valid test. If the other side were a consonant, it does not disprove the rule. And in this case (considering the OP asked for simple logic) a consonant with an odd number proves the rule merely by not disproving it.
Answer E and 7.
This also proves that Annabenedetti and I are the most intelligent people in the world.
Also, it doesn't really matter that the OP was edited and PJ answered the original OP. The answer would be the same in both cases. You would still need box 7 to prove the rule was true, assuming it was true.
PJ, you are wrong because the rule to be tested is a rule about boxes with vowels on them, not about boxes with even numbers on them.
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exminister
Well-known member
DR good job, but since you had to work through it :scripto: and Annabenedetti just knew it she is the most intelligent :BRAVO:
- from the peanut gallery
E is obviously the first box.
If there is a vowel on four it proves the rule. If there is a consonant then we don't know anything necessarily, since consonants may also produce for even numbers, there not being a rule established to control them.
If 7 has a vowel on the other side then using the rule we have our answer and the rule is invalidated. If, however, it has a consonant we don't necessarily know the rule to be true since we have no rule for consonants and can't necessarily infer from it.
K doesn't work for the same reason as 4.
I'd take E and 7. At least the seven, with a vowel, would invalidate the rule and the other options don't really tell you anything.
true, good answer
conversely
Ask Mr. Religion
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He has borrowed from a classic selection task and botched it by adding the notion of "box" to the problem to make it seem, er, original. :AMR: Had it been cards, planes not cubics, the answer I gave would have been different.
View attachment 20325
[Click to enlarge]
Put an "E" on one of the sides of the box in the pix above. For the statement to be true its opposing side must have an even number.
You are left with 4, 7, K that exist on one of the six sides of three other boxes just like the one shown above. One of those three boxes must have 4 on one of its sides, so that box could be chosen as well to prove the statement...for that box.
The statement claims "a box" not "all boxes". The other two remaining boxes may in fact meet the condition, as with only 7 on one box and K on the last box, the rule could still be true given there are two opposing sides on each of these last two boxes that could have the proper conditions to make the statement true.
AMR
View attachment 20325
[Click to enlarge]
Put an "E" on one of the sides of the box in the pix above. For the statement to be true its opposing side must have an even number.
You are left with 4, 7, K that exist on one of the six sides of three other boxes just like the one shown above. One of those three boxes must have 4 on one of its sides, so that box could be chosen as well to prove the statement...for that box.
The statement claims "a box" not "all boxes". The other two remaining boxes may in fact meet the condition, as with only 7 on one box and K on the last box, the rule could still be true given there are two opposing sides on each of these last two boxes that could have the proper conditions to make the statement true.
AMR
e4
ec
e9
ea
42
4a
4c
49
72
7a
7c
79
kc
ka
k9
k2
from this we see that
the e-card can give you one valid and 3 rule breaker results
the 4-card can give you one valid and 3 meaningless results
the 7-card can return one rule breaker and 3 meaningless results
the k-card can return one rule breaker and 3 meaningless results
If we add the caveat that the back side of a lettered card will always be a number and the backside of a numbered card will always be a letter then
we see that
the e-card can return one valid and one invalid result (e4;e9)
the 4- card can return one valid and one meaningless result (4a;4c)
the 7 card can return one invalid and one meaningless result (7a;7c)
the k card can return only 2 meaningless results (k4;k9)
therefore the only two cards which can invalidate the rule are the e-card and the 7-card. The 4-card and the k-card can not disprove the rule which is what a test of logic requires. This is, of course, a long winded way of saying anna is correct.
ec
e9
ea
42
4a
4c
49
72
7a
7c
79
kc
ka
k9
k2
from this we see that
the e-card can give you one valid and 3 rule breaker results
the 4-card can give you one valid and 3 meaningless results
the 7-card can return one rule breaker and 3 meaningless results
the k-card can return one rule breaker and 3 meaningless results
If we add the caveat that the back side of a lettered card will always be a number and the backside of a numbered card will always be a letter then
we see that
the e-card can return one valid and one invalid result (e4;e9)
the 4- card can return one valid and one meaningless result (4a;4c)
the 7 card can return one invalid and one meaningless result (7a;7c)
the k card can return only 2 meaningless results (k4;k9)
therefore the only two cards which can invalidate the rule are the e-card and the 7-card. The 4-card and the k-card can not disprove the rule which is what a test of logic requires. This is, of course, a long winded way of saying anna is correct.