Simple Logic Quiz

Desert Reign

LIFETIME MEMBER
LIFETIME MEMBER
the e-card can return one valid and one invalid result (e4;e9)

The issue is whether the E card can test the hypothesis that the rule of vowel-even is true. To test means to allow for possibilities which are consistent with the rule or which are inconsistent with it. The E card can be in either of 2 types: an even number or an odd number (and not just 4 or 9). The even number would be consistent with the rule but the odd number would disprove it. It is this possibility which makes the E card is a valid test. Note that if the E card had an even number on it, this does not prove the rule. The rule must be shown to work in all 4 cases in order to be proven. This is what I think you are missing in your logic. The problem is about testing the rule, not proving it. If the test is to be valid then it must be possible to disprove the rule. Validity of the results is not the issue. Validity of the test is the issue.

the 4- card can return one valid and one meaningless result (4a;4c)
The result "4c" is not meaningless. It is consistent with the rule. The result 4a is also consistent with the rule. It is the fact that both possibilities are consistent with the rule that makes this card invalid as a test. Because neither of the outcomes disproves the rule. The 4 card can also return a lot of other results such as 4j, 4o, 4z and so on, which you seem to have ignored.

the 7 card can return one invalid and one meaningless result (7a;7c)
Again, the result 7c is not meaningless. It is consistent with the rule. The result 7a is a possibility and this possibility would disprove the rule. Because one of the outcomes would disprove the rule, this card is a valid test of the rule. And again, there are other possible results which you have ignored.

the k card can return only 2 meaningless results (k4;k9)
Again, these results are not meaningless. They are consistent with the rule. There are also other possible results but because all of them would be consistent with the rule (as with the 4 card) then it cannot test the rule.

You seem to have confused provability with testability. Proving a rule is a great deal harder than disproving it and is why almost no scientific theories are claimed to be proven.

There is another issue of logic here which is bothersome. In what I said above, and in your list of permutations, we are assuming that there are only 4 cards in total and that the rule only applies to those 4 cards. You are further assuming that only certain numbers and certain letters are allowed. However, we are only presented with one side of 4 cards, not with their reverse sides. The only assumption we need to make is (as you said) that a card must have a letter on one side and a number on the other side.
Then, if the the number of cards was infinite, we could never prove the rule, though we could easily disprove it. If the number of cards was just the four posed in the problem, or some other finite number, then we could prove the rule by examining the backs of all cards. We don't necessarily need to examine the backs of all cards to disprove the rule. We just need to find the first instance of a card that is inconsistent with the rule. That is why we look at the E and the 7 to test this. We are not asked to prove the rule.

Finally, if you stated the problem slightly differently and said that the test should only consist of cards with consonants on the letter side, then the rule is self-proving and cannot be tested. All the cards would by definition be consistent with the rule so the rule would be proved. But the rule cannot be tested in this way. This is what (supposedly) distinguishes a scientific theory from a non-scientific one.
 

quip

BANNED
Banned
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. ;)

AMR

Not exactly...you were given the direction that: "If a box shows a vowel on one side, then it has an even number on the other side.". Which, of course, infers two-sides (at least as far as the puzzle is concerned) Your error in logic; my graphic limitations here on ToL. :eek:
 

Desert Reign

LIFETIME MEMBER
LIFETIME MEMBER
Oh wow, thank you. I was afraid I'd embarrassed myself. :chuckle:

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

Well, I just knew it as well. I just like splainin things. And here's another splanation: if you intuit something, you could be wrong. Intuitions are frequently wrong. But they are also very quick and often it is better to make a wrong decision than to make no decision at all. But if you double check your intuitions afterwards through logic and research, then you gain more confidence in them and that confidence makes your next intuitions more powerful until in the end you become a mental powerhouse capable of playing chess to a mediocre level. (JK.)
 

Town Heretic

Out of Order
Hall of Fame
So if you got the right answer without reading anyone else's answer but after they gave it you might as well have read all the answers and let the thread die the moment you saw the right one?

:mmph: What is this, Congress? Okay, I'm only guessing you get in trouble there for doing your own work. :eek:
 

quip

BANNED
Banned
So if you got the right answer without reading anyone else's answer but after they gave it you might as well have read all the answers and let the thread die the moment you saw the right one?

:mmph: What is this, Congress? Okay, I'm only guessing you get in trouble there for doing your own work. :eek:

No refundin' of tickets bub! :cigar:

Otherwise, a fine job! :third:
 

musterion

Well-known member
Two trains at a station on the same track leave at the same time for the same destination 400 miles away. Which one will arrive last?
 

Town Heretic

Out of Order
Hall of Fame
Bob always takes the train home from work and arrives at the station at five o'clock. His wife meets him at the station and drives him home. On Friday, he left work early and took an earlier train home, arriving at the station at four o'clock. He began walking home. On his way home he met his wife, driving to pick him up. They drove back together and saved ten minutes on her regular drive.

Assuming a constant rate of speed for the wife, how long was Bob walking before he met his wife on the road?

:D
 

Desert Reign

LIFETIME MEMBER
LIFETIME MEMBER
Bob always takes the train home from work and arrives at the station at five o'clock. His wife meets him at the station and drives him home. On Friday, he left work early and took an earlier train home, arriving at the station at four o'clock. He began walking home. On his way home he met his wife, driving to pick him up. They drove back together and saved ten minutes on her regular drive.

Assuming a constant rate of speed for the wife, how long was Bob walking before he met his wife on the road?

:D

I assume that when you say she saved ten minutes on her regular drive that you mean they arrived back home 10 minutes before they would usually.

In that case the answer would be 55 minutes. No explanations this time. I just knew it!
 

Dan Emanuel

Active member
Bob always takes the train home from work and arrives at the station at five o'clock. His wife meets him at the station and drives him home. On Friday, he left work early and took an earlier train home, arriving at the station at four o'clock. He began walking home. On his way home he met his wife, driving to pick him up. They drove back together and saved ten minutes on her regular drive.

Assuming a constant rate of speed for the wife, how long was Bob walking before he met his wife on the road?

:D
Five minute's?


DJ ("Dumb jock")
1.0
 

patrick jane

BANNED
Banned
Bob always takes the train home from work and arrives at the station at five o'clock. His wife meets him at the station and drives him home. On Friday, he left work early and took an earlier train home, arriving at the station at four o'clock. He began walking home. On his way home he met his wife, driving to pick him up. They drove back together and saved ten minutes on her regular drive.

Assuming a constant rate of speed for the wife, how long was Bob walking before he met his wife on the road?

:D

what's the answer TH ?
 

Town Heretic

Out of Order
Hall of Fame
Oh, sorry. I saw the right answer and forgot to ring the bell. Bob's wife needed to drive ten minutes less than usual, which is five minutes less in the direction of the station and five minutes less back home. So Bob met his wife at five minutes to five. He had walked already since four o'clock, so he walked 55 minutes in total.
 

Ask Mr. Religion

☞☞☞☞Presbyterian (PCA) &#9
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Hall of Fame
Not exactly...you were given the direction that: "If a box shows a vowel on one side, then it has an even number on the other side.". Which, of course, infers two-sides (at least as far as the puzzle is concerned) Your error in logic; my graphic limitations here on ToL. :eek:

Er, no.

Plain rules of grammar dictate that "If a box shows a vowel on one side, then it has an even number on the other side" means "If a box shows a vowel on one side, then it {box} has an even number on the other side {of the box}".

Your complaint to me implies you intended:
"If a box shows a vowel on one side, then it {box} has an even number on the other side {of that aforemenioned one side of the box}"

The error here lies in imprecise grammar on your part, not my logic.

AMR
 
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