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Re: Logic Game Challenge #34: Party Problem (Hard)

by jbert Fri Jan 06, 2012 12:31 pm

Q7: The answer key lists B as the correct answer. Is not answer A the contrapostive of the original rule? As such it by definition has the same logical implications. Furthermore, answer B is too specific: it omits the information that if Sara or Pam is the only woman that comes, then Nick will not come.
 
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Re: Logic Game Challenge #34: Party Problem (Hard)

by timmydoeslsat Fri Jan 06, 2012 3:35 pm

jbert wrote:Q7: The answer key lists B as the correct answer. Is not answer A the contrapostive of the original rule? As such it by definition has the same logical implications. Furthermore, answer B is too specific: it omits the information that if Sara or Pam is the only woman that comes, then Nick will not come.


I just did this game. For #7, answer A would not be acceptable because of its sufficient condition. If two or more women do not come to the party...

Well, could it not be the case that T and S both go to the party with Nick? That would be two women going to the party and Nick can validly go according to the original rule. This new rule in answer choice (A) would tell us that Nick cannot go.


On choice B, I am a little concerned with the wording although it is listed as a correct answer.

If R is the only woman at the party ---> ~ N

N ---> R is not the only woman at the party.

This would seem to allow for the opportunity of P being the only woman at the party after placing N at the party, while the original rule specified that at least 2 women must be present.

I would like clarification on that.
Last edited by timmydoeslsat on Fri Jan 06, 2012 6:11 pm, edited 1 time in total.
 
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Re: Logic Game Challenge #34: Party Problem (Hard)

by jbert Fri Jan 06, 2012 6:01 pm

Ah, I think there is ambiguity in the wording of answer A.
Once could legitimately parse the phrase "if two or more women do not come to the party" to mean either:
1) if there are two or more women in the set of people who do not come to the party
or
2) if there are not two or more women in the set of people who do come to the party

TimmyDoesLSAT interpreted it as the first version, and I interpreted it as the second. Semantically, both interpretations are valid. Thus, the correctness of the answer is inherently ambiguous.

Or am I missing something?
 
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Re: Logic Game Challenge #34: Party Problem (Hard)

by timmydoeslsat Fri Jan 06, 2012 6:17 pm

The original rule was:

N ---> At least two women go

Contrapositive of that rule is:

Less than two women go ---> ~N


(A) states if two or more women do not go ---> ~N

That is not an adequate substitution because it does not have the same effect.

Lets juxtapose the two rules:


Original: Less than two women go ---> ~N
Choice A: two or more women do not go ---> ~N


(A) allows for two women to go while the original specifies less than two.
 
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Re: Logic Game Challenge #34: Party Problem (Hard)

by jbert Sat Jan 07, 2012 10:50 pm

Again, your argument depends on a particular semantic interpretation of "if two or more women do not go." There are two legitimate semantic interpretations. One being
~2 or more women go (less than 2 women go)
the other being
2 or more women stay home (2 or fewer women go)

Both interpretations are valid. Hence the answer is ambiguous.

timmydoeslsat wrote:The original rule was:

N ---> At least two women go

Contrapositive of that rule is:

Less than two women go ---> ~N


(A) states if two or more women do not go ---> ~N

That is not an adequate substitution because it does not have the same effect.

Lets juxtapose the two rules:


Original: Less than two women go ---> ~N
Choice A: two or more women do not go ---> ~N


(A) allows for two women to go while the original specifies less than two.
 
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Re: Logic Game Challenge #34: Party Problem (Hard)

by timmydoeslsat Mon Jan 09, 2012 2:59 pm

jbert wrote:Again, your argument depends on a particular semantic interpretation of "if two or more women do not go." There are two legitimate semantic interpretations. One being
~2 or more women go (less than 2 women go)
the other being
2 or more women stay home (2 or fewer women go)

Both interpretations are valid. Hence the answer is ambiguous.

The original condition we want to substitute for is this:

Fewer than 2 women go to the party ---> ~N



You can reword that sufficient condition in any way you like. I chose to form the condition in terms of going to the party, you can reword it in terms of not going to the party.

This is the same exact rule, no difference.

At least 3 women do not go to the party ---> ~N

That is the same rule as our original condition, those two forms are equally valid interpretations of the contrapositive.


The problem is the rule that answer (A) would give us:

If 2 or more women do not come to the party --->~N


Notice how "at least 3" and "2 or more" are different things.

It is not a valid substitution to incur the same effects as did our original rule.
 
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Re: Logic Game Challenge #34: Party Problem (Hard)

by efcaley Sun Jan 29, 2012 8:12 am

For question 6, doesn't the rule that Tracy comes unless both Lance and Reena come mean that Lance cannot come? That would make B,C, and D correct answer choices.

If all 4 women come, that means that both Tracy and Reena come. If Lance comes, then Tracy can't come.

Am I missing something?
 
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Re: Logic Game Challenge #34: Party Problem (Hard)

by timmydoeslsat Sun Jan 29, 2012 5:41 pm

efcaley wrote:For question 6, doesn't the rule that Tracy comes unless both Lance and Reena come mean that Lance cannot come? That would make B,C, and D correct answer choices.

If all 4 women come, that means that both Tracy and Reena come. If Lance comes, then Tracy can't come.

Am I missing something?

The rule you are talking about is:

~T ---> L and R


We do not have any information regarding what happens when T goes.

We have information about what happens if T does not go, we know that L and R both go.

This question stem tells us that all of the women go to the party and it wants to know what cannot be a complete and accurate list of the men that go. So we are going to be looking for an answer choice that either has to have someone going but does not show it, or puts in people that cannot both be present at the party.

We will start by what we know for sure: 4 women are in

P R S T

This triggers that K will be in. Any answer choice that does not have K would be our answer. However, they all do! So we must go a step beyond this.

We know that M is a random/floater/free agent/etc. It has no rules about it. In this circumstance it can be validly in or out. The list could be complete without M, so it does not have to show up in an answer choice. So M does not matter right now.

Let us look at the rest of our rules.

Look at this rule:

O and P ---> ~L

We know that P is in because it is a woman. O and L are both men.

What this rule is basically telling us is not both O and L.

O ---> ~L

So we know that O and L cannot both be present at the party.

(C) includes both and it is our answer because it could never happen.
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Re: Logic Game Challenge #34: Party Problem (Hard)

by noah Tue Feb 07, 2012 7:05 pm

noah wrote:
timmydoeslsat wrote:Your answers: CEADACA
Correct Answers: BAEDCC

I believe those are off. Timmy, are you setting them on a wild goose chase?!

Ah! Those are for the easy version. I think this thread is just for the harder one.
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Re: Logic Game Challenge #34: Party Problem (Hard)

by noah Tue Feb 07, 2012 7:12 pm

timmydoeslsat wrote:I just did this game. For #7...

On choice B, I am a little concerned with the wording although it is listed as a correct answer.

If R is the only woman at the party ---> ~ N

N ---> R is not the only woman at the party.

This would seem to allow for the opportunity of P being the only woman at the party after placing N at the party, while the original rule specified that at least 2 women must be present.

I would like clarification on that.

This is a tough question! I helped on writing this game and it took me a while to re-figure it out! I suggest you write out all the options for limiting the women to just 1 person. That should clear up the issue. Tell me how it goes.
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Re: Logic Game Challenge #34: Party Problem (Hard)

by noah Tue Feb 07, 2012 7:15 pm

Great conversation everyone!

I know this is a tough game - it was intended to stretch you!

With #7, the semantics is tough, and perhaps the LSAT wouldn't play with that fire, but I believe the numbers work out so that either interpretation works out.

By the way, nobody has yet written up a full explanation, so for the intrepid prepper, there's a challenge for you! :ugeek:

(Don't make Timmydoeslsat do it!)
 
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Re: Logic Game Challenge #34: Party Problem (Hard)

by timmydoeslsat Wed Feb 08, 2012 12:21 pm

I went back to this game again and did #7 over again.

The original rule:

N ---> 2 or more women

Answer choice B:

N ---> ~R only woman at party

So what I am going to do here is ask myself if this new found rule would force me to have two women go to the party and if it is consistent with how the original rule operated.

I know from the third rule that at least one of T and R must be present.

Both of these are women. So I know that if T is chosen, S will be chosen, thus taking care of the condition concerning the requirement for Nick to go.

I could of course have both T and R present, which would tag along S, giving us 3 women. So that is consistent with this new rule, as it states R is not the only woman.

This new rule is really simply precluding the situation of one R being alone, which would be valid without the two rules in question.

I will presume that the rule of "R is not the only woman at the party" does not necessitate R's presence. It only precludes R being the only woman present. So we can have R be out, in which case we would have P, S, T try to be the one woman rule breakers.

To try and break this new rule, I would try to have only one of those 3 be present so as to show it being inconsistent with the original rule of at least 2 women being present if Nick goes.

I know I have to have at least one of T and R, so that means I must have T. In which case I must have S. I could never have just P or just S or just T, thus this rule is consistent.

I would like clarification on the bold above however.
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Re: Logic Game Challenge #34: Party Problem (Hard)

by noah Wed Feb 08, 2012 12:27 pm

Looks like you got deep with #7 last night! Nice work.

timmydoeslsat wrote:I will presume that the rule of "R is not the only woman at the party" does not necessitate R's presence. It only precludes R being the only woman present. So we can have R be out, in which case we would have P, S, T try to be the one woman rule breakers.

I would like clarification on the bold above however.


You're right - it doesn't require R to be in. It's like saying "I can't be the only person at the party in a costume!" That doesn't mean I have to go in a costume, but that if I do go in a costume, we know there are others.
 
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Re: Logic Game Challenge #34: Party Problem (Hard)

by jeffr Tue Jun 05, 2012 8:06 pm

The right answer B is right. If R is out there at least two women from the contrapositive of the rule "Tracy unless Lance and Reena come." This yields the contrapositive that if Lance or Reena don't come, Tracy will. Thus, none of the woman other than Reena can be alone. Sarah can't, or the contra is triggered and Tracy would have to come. Pam can't because it would trigger the contra and Tracy and, based on the first rule, Sarah would have to come. Tracy obviously can't be the only woman, based on the first rule. We know that either Tracy or Reena must be in (from the Tracy and Reena/Lance rule). So the necc condition 2 or more women must come to the party would obtain unless R were alone, in which case Nick would not be able to come.. Which is exactly what "Nick will come to the party only if two or more women come." would imply. If there are fewer than 2 women, which would only be if R were not alone), he can not come. If there are more R is not the lone woman, there are two or more and thus Nick would not be precluded from coming.

Still, while this is complex and elegant as a choice, and is correct, A is correct too and much more direct. As the first poster noted it is the contrapositive of "Nick will come to the party only if two or more women come." Nc---> +2 W( two or more than two women). This generates answer choice: not +2W (two or more than 2 women) then not Nc. This is the implication of the rule, and is an identical statement. Yes, not +2 is the sufficient, not the necessary condition,to Jbert's point, but it is properly negated, so the identity of the logical statement is the same,
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Re: Logic Game Challenge #34: Party Problem (Hard)

by ohthatpatrick Wed Oct 24, 2012 4:09 pm

I was thwarted by the obnoxious wording of (A) as well, all the while thinking "wait-a-sec ... this is too good to be true ... they're just giving me the contrapositive?"

My take on the potential ambiguity is this:

What all of us who were thinking (A) was the contrapositive wanted it to mean was
If it is not the case that two or more women are coming to the party ...

That would be logically equivalent to "less than 2" attend the party.

The way the answer choice is written, though, makes it really about the OUT column, not about the IN column.

It's talking about how many women DON'T go to the party, not about how many women do.

And, naturally, 2 IN, 2 OUT would force N to be OUT according to this new rule, while 2 IN, 2 OUT was happily compatible with N being IN before.

I agree that LSAT would not take a chance with this ambiguity.

A rule that would say,
"If Halloween does not occur on a Wednesday ... "
means the same as
"If it is not the case that Halloween occurs on a Wednesday ..."

While a rule that says,
"If two or more holidays do not occur on a Wednesday"
could be interpreted to mean
"Check Wednesday and see if two or more holidays occur"
or
"Check all the holidays, and see if you can find at least two of them that don't occur on a Wednesday"

Those are distinct conditions, but both seem like legitimate interpretations.
 
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Re: Logic Game Challenge #34: Party Problem (Hard)

by christrfx Fri Jul 10, 2015 4:30 pm

timmydoeslsat, or anybody else, please clarify the rule in question for #6. The rule states "Tracy is coming to the party unless both Lance and Reena come." In my mind, this means that Tracy will always come to the party unless the condition of LR is satisfied. Meaning T ≠ LR. The rule means that Tracy IS coming if Lance and Reena are both absent, which would be L/R = T and -LR = T. Please tell me how my logic is flawed.

timmydoeslsat wrote:
efcaley wrote:For question 6, doesn't the rule that Tracy comes unless both Lance and Reena come mean that Lance cannot come? That would make B,C, and D correct answer choices.

If all 4 women come, that means that both Tracy and Reena come. If Lance comes, then Tracy can't come.

Am I missing something?

The rule you are talking about is:

~T ---> L and R


We do not have any information regarding what happens when T goes.

We have information about what happens if T does not go, we know that L and R both go.

This question stem tells us that all of the women go to the party and it wants to know what cannot be a complete and accurate list of the men that go. So we are going to be looking for an answer choice that either has to have someone going but does not show it, or puts in people that cannot both be present at the party.

We will start by what we know for sure: 4 women are in

P R S T

This triggers that K will be in. Any answer choice that does not have K would be our answer. However, they all do! So we must go a step beyond this.

We know that M is a random/floater/free agent/etc. It has no rules about it. In this circumstance it can be validly in or out. The list could be complete without M, so it does not have to show up in an answer choice. So M does not matter right now.

Let us look at the rest of our rules.

Look at this rule:

O and P ---> ~L

We know that P is in because it is a woman. O and L are both men.

What this rule is basically telling us is not both O and L.

O ---> ~L

So we know that O and L cannot both be present at the party.

(C) includes both and it is our answer because it could never happen.
 
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Re: Logic Game Challenge #34: Party Problem (Hard)

by alicia.granse Tue Oct 20, 2015 12:03 pm

I second this. If both Lance and Rena come, Tracy doesn't. I chose C because it was the most incorrect, but I'm still not satisfied.


christrfx wrote:timmydoeslsat, or anybody else, please clarify the rule in question for #6. The rule states "Tracy is coming to the party unless both Lance and Reena come." In my mind, this means that Tracy will always come to the party unless the condition of LR is satisfied. Meaning T ≠ LR. The rule means that Tracy IS coming if Lance and Reena are both absent, which would be L/R = T and -LR = T. Please tell me how my logic is flawed.

timmydoeslsat wrote:
efcaley wrote:For question 6, doesn't the rule that Tracy comes unless both Lance and Reena come mean that Lance cannot come? That would make B,C, and D correct answer choices.

If all 4 women come, that means that both Tracy and Reena come. If Lance comes, then Tracy can't come.

Am I missing something?

The rule you are talking about is:

~T ---> L and R


We do not have any information regarding what happens when T goes.

We have information about what happens if T does not go, we know that L and R both go.

This question stem tells us that all of the women go to the party and it wants to know what cannot be a complete and accurate list of the men that go. So we are going to be looking for an answer choice that either has to have someone going but does not show it, or puts in people that cannot both be present at the party.

We will start by what we know for sure: 4 women are in

P R S T

This triggers that K will be in. Any answer choice that does not have K would be our answer. However, they all do! So we must go a step beyond this.

We know that M is a random/floater/free agent/etc. It has no rules about it. In this circumstance it can be validly in or out. The list could be complete without M, so it does not have to show up in an answer choice. So M does not matter right now.

Let us look at the rest of our rules.

Look at this rule:

O and P ---> ~L

We know that P is in because it is a woman. O and L are both men.

What this rule is basically telling us is not both O and L.

O ---> ~L

So we know that O and L cannot both be present at the party.

(C) includes both and it is our answer because it could never happen.
 
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Re: Logic Game Challenge #34: Party Problem (Hard)

by Didius Falco Mon Sep 05, 2016 7:58 pm

The rule does not preclude Tracey from appearing; whether or not R and L appear. It states, in other terms, that [Any time T is absent (~T)---> Both L and R are present].

We can think about this simplistically as two worlds;



World 1: T In (this rule is irrelevant)

World 2: T Out (R and L are In)


Thats all; not that hard to parse in this form. In world 1, we don't have to worry about this rule at all. T can be in or out, and it has no effect on our placement decisions. Thus, It is perfectly acceptable for all 4 women to attend the party.

(For the record, I wouldn't recommend framing this rule in the actual game situation----just used the device to help answer the previous poster's question)
 
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Re: Logic Game Challenge #34: Party Problem (Hard)

by AustinB970 Fri Jun 22, 2018 1:21 pm

I agree entirely. The rule is not ~T --> L ^R ... The rule states that she will go unless L^R both go. So that has to mean that if L^R both go, the "unless" condition is satisfied. That means the first clause is now reversed. "T will go unless L^R both go", means that the first clause "T will go" is going to be satisfied, if the second clause is not the case. So it is more so T --> ~(L^R). With the premise that T^R both go, that means that L cannot or else you have a flawed argument. L^R being true, the not cannot be satisfied to allow T to go. I chose C because it also had the Oscar rule violated, but I believe 2-4 are all invalid combinations

christrfx wrote:timmydoeslsat, or anybody else, please clarify the rule in question for #6. The rule states "Tracy is coming to the party unless both Lance and Reena come." In my mind, this means that Tracy will always come to the party unless the condition of LR is satisfied. Meaning T ≠ LR. The rule means that Tracy IS coming if Lance and Reena are both absent, which would be L/R = T and -LR = T. Please tell me how my logic is flawed.

timmydoeslsat wrote:
efcaley wrote:For question 6, doesn't the rule that Tracy comes unless both Lance and Reena come mean that Lance cannot come? That would make B,C, and D correct answer choices.

If all 4 women come, that means that both Tracy and Reena come. If Lance comes, then Tracy can't come.

Am I missing something?

The rule you are talking about is:

~T ---> L and R


We do not have any information regarding what happens when T goes.

We have information about what happens if T does not go, we know that L and R both go.

This question stem tells us that all of the women go to the party and it wants to know what cannot be a complete and accurate list of the men that go. So we are going to be looking for an answer choice that either has to have someone going but does not show it, or puts in people that cannot both be present at the party.

We will start by what we know for sure: 4 women are in

P R S T

This triggers that K will be in. Any answer choice that does not have K would be our answer. However, they all do! So we must go a step beyond this.

We know that M is a random/floater/free agent/etc. It has no rules about it. In this circumstance it can be validly in or out. The list could be complete without M, so it does not have to show up in an answer choice. So M does not matter right now.

Let us look at the rest of our rules.

Look at this rule:

O and P ---> ~L

We know that P is in because it is a woman. O and L are both men.

What this rule is basically telling us is not both O and L.

O ---> ~L

So we know that O and L cannot both be present at the party.

(C) includes both and it is our answer because it could never happen.
[/quote]
 
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Re: Logic Game Challenge #34: Party Problem (Hard)

by MichaelC134 Sun Nov 04, 2018 11:13 pm

I think that Rule 3 is (~L or ~R) -> T or the contrapositive ~T -> (L and R) which would mean that if R is not there, T is there and the contrapositive if T is not there R is there. So, either R or T has to be there.

So for B to be correct, we cannot have a situation where only P, S or T is at the party with no other woman. Rule 1 tells us that T - > S so T cannot be alone at the party without another woman. P or S cannot be alone at the party because either R or T has to be at the party. Thus, "if Reena is alone at the party, Nick will not come." is equivalent to saying "if only one woman is at the party, Nick will not come" (because P, S or T cannot be alone at the party w/o another woman) which is the contrapositive of "Nick will come to the party only if two or more women come". Thus, B is the answer.