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- ohthatpatrick
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Atticus Finch
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Re: Logic Game Challenge #40: Volleyball Game
Hi! I'm not able to submit my answer and check to see if they're right, because the page says "submit answer to #1" even though I did select an answer.
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- jgutella
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Vinny Gambini
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Re: Logic Game Challenge #40: Volleyball Game
Question # 7 needs revision.
And +1 to the above poster. I got the same error. Can't submit answers.
And +1 to the above poster. I got the same error. Can't submit answers.
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Atticus Finch
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Re: Logic Game Challenge #40: Volleyball Game
We'll look into it. Apologies!
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Vinny Gambini
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Re: Logic Game Challenge #40: Volleyball Game
Note: ~ means not, v means or, ^ means and
The rules:
At least two people must attend the game: IN >/= 2
Ron and Tim cannot attend the game together: [R/T]in
But even better: R "”> ~T and T "”> ~R.
If Paul attend the game, so does Tim: P "”> T and ~T "”> ~P.
If Ron does not attend the game, neither does Vanessa: ~R "”> ~V and V "”> R.
If Wanda or Tim attend the game, they attend together: W "”> (W ^ T) and T "”> (W ^ T).
Or (W v T) "”> (W ^ T)
~(W ^ T) "”> ~(W v T)
(~W v ~T) "”> (~W ^ ~T)
You could probably figure this out without all the logical steps and save a lot of time, but we know that T and W are either in together or out together.
So T <"”> W or T o"”o W.
Sammy attends the game if, and only if, Quita does not attend: S <"”> ~Q or S o"”o ~Q.
Or S "”> ~Q and Q "”> ~S
and ~S "”> ~Q and ~Q "”> S.
Connecting inferences:
The best way to do this is with a diagram, but in case my picture doesn’t upload, I’ll connect them a bit in text.
R "”> ~T and T "”> ~R
~R "”> ~V and V "”> R
P "”> T and ~T "”> ~P
S o"”o ~Q
T o"”o W
P "”> T "”> [W and (~R "”> ~V)]
V "”> R "”> ~T "”> (~P and ~W)
1. Which of the following would be a complete and accurate list of students who attend the game?
(A) Quita, Ron, Sammy, Vanessa
We know right off the bat that Quita and Sammy can’t be together because of the rule S o"”o ~Q.
(B) Paul, Sammy, Tim
If Paul is in, Tim is in because P "”> T. That’s good. But if Tim is in, Wanda is in because T o"”o W. Nope.
(C) Ron, Tim, Vanessa
If Ron is in, Tim is out because R "”> ~T. Nope.
(D) Quita, Tim, Wanda
If Quita is in, Sammy is out because Q o"”o ~S. If Tim is in, Ron is out and Vanessa is out because T "”> ~R "”> ~V. And Wanda is in because T o"”o W. Looks good. We don’t have any inferences that begin with ~P, so Paul can be out.
(E) Wanda, Sammy, Paul
If Wanda is in, Tim is in because T o"”o W. Nope.
2. If Wanda does not attend the game, which of the following must be true?
Before looking at the answers, we know for sure ~W "”> ~T "”> ~P.
So we have OUT: W, T, P, S/Q. We have to have at least two IN. So we could have IN: R, S/Q or IN: V, R, S/Q but we can’t have IN: V, S/Q because V "”> R.
(A) Ron does not attend the game.
This is definitely impossible.
(B) Vanessa does not attend the game.
This could be true, but we’re looking for MUST BE TRUE.
(C) Ron and Vanessa do not both attend the game.
This would leave us with only one IN, and we can’t have that.
(D) Either Ron or Vanessa attend the game.
This wording tricky because of the V "”> R rule. But know that Ron or both Ron and Vanessa have to be in, so this is the right answer.
(E) Either Ron or Vanessa, but not both, attend the game.
We know Ron has to be in, so this isn’t true.
3. Which of the following could be true?
(A) Both Vanessa and Paul attend the game.
This can’t be true because P "”> T "”> ~R "”> ~V or V "”> R "”> ~T "”> ~P. A picture makes this problem a lot easier!
(B) Both Vanessa and Wanda attend the game.
This can’t be true because V "”> R "”> ~T "”> ~W or W "”> T "”> ~R "”> ~V.
(C) Neither Ron nor Tim attends the game.
This answer is super tempting! I almost immediately went for it because it seems to be testing whether you recognize that R "”> ~T and T "”> ~R allow for ~R and ~T. But ~R "”> ~V and ~T "”> ~P and ~W. This leaves us with OUT: R, T, V, P, W. So IN: S, Q, which breaks the rule S o"”o ~Q.
(D) Neither Sammy nor Quita attends the game.
This can’t be true because S o"”o ~Q.
(E) Neither Paul nor Ron attends the game.
~P doesn’t give us anything. ~R "”> ~V. So we have OUT: P, R, V, S/Q and to keep the two IN rule, IN: S/Q, T, W. Yes.
4. What is the minimum number of students who do not attend the game?
We can take this one straight from the rules. We know at least one has to be out because S o"”o ~Q. How about two? If T or W is OUT, then they both are. So it makes sense to try leaving them in. But that gives us T "”> ~R "”> ~V. That leaves three out with S/Q. And we’d have IN: W, P, T, S/Q. But if T or W were out, we’d have ~W and ~S/Q and ~T "”> ~P. That’s four out. Three is our answer.
(B) 3
5. If Tim attends the game, which of the following could be true.
First, T"”> W and T "”> ~R "”> ~V.
(A) Quita does not attend the game but Ron does.
We don’t immediately know about Quita, but Ron definitely doesn’t. Nope.
(B) Sammy does not attend the game but Vanessa does.
Again, we don’t know about Sammy, but Vanessa definitely doesn’t. No.
(C) Neither Quita nor Sammy attends the game.
We know at least one of them does because S o"”o ~Q. No.
(D) Neither Quita nor Paul attends the game.
So we’d have IN: T, W, S and OUT: R, V, Q, P. Looks good.
(E) Exactly five students attend the game.
We know from the last problem that at least three have to be out, which leaves a maximum of four in. No.
6. If the condition that requires that at least two students attend the game is replaced by a condition that requires that exactly three students attend the game, how many combinations of students attending the game are possible?
Don’t forget the "exactly!"
We know for sure that IN: S/Q and OUT: S/Q. That leaves us with at least two options.
Then I tried moving around the [T W] block. We could have IN: S/Q, T, W and OUT: S/Q, P, R, V. Or if [T W] are out, the only option is OUT: S/Q, T, W, P, leaving IN: S/Q, R, V. So that leaves us with four combinations.
(C) 4
7. Which of the following, if substituted for the condition that Paul attends the game, so does Tim, would have the same effect in determining who can attend the game together?
First, the rule we’re trying to replace is P "”> T.
(A) Either Paul and Wanda attend the game together or neither attends.
This gives us (P ^ W) or (~P ^ ~W). With our rule P "”> T "”> W, we could have T ^ W and ~P, so this doesn’t work.
(B) Wanda does not attend the game only if Paul does not attend the game.
This gives us ~W "”> ~P. Since P "”> T "”> W, we know ~W "”> ~T "”> ~P, so ~W "”> ~P. This one looks good.
(C) Tim does not attend the game if Paul does not attend the game.
This gives ~P "”> ~T, and we can’t infer that from P "”> T.
(D) Paul cannot attend the game unless Vanessa does not attend.
So ~P "”> ~V. But with our original rule, we could definitely have ~P and V. No.
(E) If Paul attends the game, then at most four students attend the game.
This is true, because P "”> T "”> ~R "”> ~V. So IN: S/Q, P, T, W and OUT: S/Q, R, V. But without P "”> T, we couldn’t be sure which were in and which were out, so this isn’t the answer.
The rules:
At least two people must attend the game: IN >/= 2
Ron and Tim cannot attend the game together: [R/T]in
But even better: R "”> ~T and T "”> ~R.
If Paul attend the game, so does Tim: P "”> T and ~T "”> ~P.
If Ron does not attend the game, neither does Vanessa: ~R "”> ~V and V "”> R.
If Wanda or Tim attend the game, they attend together: W "”> (W ^ T) and T "”> (W ^ T).
Or (W v T) "”> (W ^ T)
~(W ^ T) "”> ~(W v T)
(~W v ~T) "”> (~W ^ ~T)
You could probably figure this out without all the logical steps and save a lot of time, but we know that T and W are either in together or out together.
So T <"”> W or T o"”o W.
Sammy attends the game if, and only if, Quita does not attend: S <"”> ~Q or S o"”o ~Q.
Or S "”> ~Q and Q "”> ~S
and ~S "”> ~Q and ~Q "”> S.
Connecting inferences:
The best way to do this is with a diagram, but in case my picture doesn’t upload, I’ll connect them a bit in text.
R "”> ~T and T "”> ~R
~R "”> ~V and V "”> R
P "”> T and ~T "”> ~P
S o"”o ~Q
T o"”o W
P "”> T "”> [W and (~R "”> ~V)]
V "”> R "”> ~T "”> (~P and ~W)
1. Which of the following would be a complete and accurate list of students who attend the game?
(A) Quita, Ron, Sammy, Vanessa
We know right off the bat that Quita and Sammy can’t be together because of the rule S o"”o ~Q.
(B) Paul, Sammy, Tim
If Paul is in, Tim is in because P "”> T. That’s good. But if Tim is in, Wanda is in because T o"”o W. Nope.
(C) Ron, Tim, Vanessa
If Ron is in, Tim is out because R "”> ~T. Nope.
(D) Quita, Tim, Wanda
If Quita is in, Sammy is out because Q o"”o ~S. If Tim is in, Ron is out and Vanessa is out because T "”> ~R "”> ~V. And Wanda is in because T o"”o W. Looks good. We don’t have any inferences that begin with ~P, so Paul can be out.
(E) Wanda, Sammy, Paul
If Wanda is in, Tim is in because T o"”o W. Nope.
2. If Wanda does not attend the game, which of the following must be true?
Before looking at the answers, we know for sure ~W "”> ~T "”> ~P.
So we have OUT: W, T, P, S/Q. We have to have at least two IN. So we could have IN: R, S/Q or IN: V, R, S/Q but we can’t have IN: V, S/Q because V "”> R.
(A) Ron does not attend the game.
This is definitely impossible.
(B) Vanessa does not attend the game.
This could be true, but we’re looking for MUST BE TRUE.
(C) Ron and Vanessa do not both attend the game.
This would leave us with only one IN, and we can’t have that.
(D) Either Ron or Vanessa attend the game.
This wording tricky because of the V "”> R rule. But know that Ron or both Ron and Vanessa have to be in, so this is the right answer.
(E) Either Ron or Vanessa, but not both, attend the game.
We know Ron has to be in, so this isn’t true.
3. Which of the following could be true?
(A) Both Vanessa and Paul attend the game.
This can’t be true because P "”> T "”> ~R "”> ~V or V "”> R "”> ~T "”> ~P. A picture makes this problem a lot easier!
(B) Both Vanessa and Wanda attend the game.
This can’t be true because V "”> R "”> ~T "”> ~W or W "”> T "”> ~R "”> ~V.
(C) Neither Ron nor Tim attends the game.
This answer is super tempting! I almost immediately went for it because it seems to be testing whether you recognize that R "”> ~T and T "”> ~R allow for ~R and ~T. But ~R "”> ~V and ~T "”> ~P and ~W. This leaves us with OUT: R, T, V, P, W. So IN: S, Q, which breaks the rule S o"”o ~Q.
(D) Neither Sammy nor Quita attends the game.
This can’t be true because S o"”o ~Q.
(E) Neither Paul nor Ron attends the game.
~P doesn’t give us anything. ~R "”> ~V. So we have OUT: P, R, V, S/Q and to keep the two IN rule, IN: S/Q, T, W. Yes.
4. What is the minimum number of students who do not attend the game?
We can take this one straight from the rules. We know at least one has to be out because S o"”o ~Q. How about two? If T or W is OUT, then they both are. So it makes sense to try leaving them in. But that gives us T "”> ~R "”> ~V. That leaves three out with S/Q. And we’d have IN: W, P, T, S/Q. But if T or W were out, we’d have ~W and ~S/Q and ~T "”> ~P. That’s four out. Three is our answer.
(B) 3
5. If Tim attends the game, which of the following could be true.
First, T"”> W and T "”> ~R "”> ~V.
(A) Quita does not attend the game but Ron does.
We don’t immediately know about Quita, but Ron definitely doesn’t. Nope.
(B) Sammy does not attend the game but Vanessa does.
Again, we don’t know about Sammy, but Vanessa definitely doesn’t. No.
(C) Neither Quita nor Sammy attends the game.
We know at least one of them does because S o"”o ~Q. No.
(D) Neither Quita nor Paul attends the game.
So we’d have IN: T, W, S and OUT: R, V, Q, P. Looks good.
(E) Exactly five students attend the game.
We know from the last problem that at least three have to be out, which leaves a maximum of four in. No.
6. If the condition that requires that at least two students attend the game is replaced by a condition that requires that exactly three students attend the game, how many combinations of students attending the game are possible?
Don’t forget the "exactly!"
We know for sure that IN: S/Q and OUT: S/Q. That leaves us with at least two options.
Then I tried moving around the [T W] block. We could have IN: S/Q, T, W and OUT: S/Q, P, R, V. Or if [T W] are out, the only option is OUT: S/Q, T, W, P, leaving IN: S/Q, R, V. So that leaves us with four combinations.
(C) 4
7. Which of the following, if substituted for the condition that Paul attends the game, so does Tim, would have the same effect in determining who can attend the game together?
First, the rule we’re trying to replace is P "”> T.
(A) Either Paul and Wanda attend the game together or neither attends.
This gives us (P ^ W) or (~P ^ ~W). With our rule P "”> T "”> W, we could have T ^ W and ~P, so this doesn’t work.
(B) Wanda does not attend the game only if Paul does not attend the game.
This gives us ~W "”> ~P. Since P "”> T "”> W, we know ~W "”> ~T "”> ~P, so ~W "”> ~P. This one looks good.
(C) Tim does not attend the game if Paul does not attend the game.
This gives ~P "”> ~T, and we can’t infer that from P "”> T.
(D) Paul cannot attend the game unless Vanessa does not attend.
So ~P "”> ~V. But with our original rule, we could definitely have ~P and V. No.
(E) If Paul attends the game, then at most four students attend the game.
This is true, because P "”> T "”> ~R "”> ~V. So IN: S/Q, P, T, W and OUT: S/Q, R, V. But without P "”> T, we couldn’t be sure which were in and which were out, so this isn’t the answer.
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- naddant2
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Vinny Gambini
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Re: Logic Game Challenge #40: Volleyball Game
40. Volleyball Game
Given (7) Seniors: P, Q, R, S, T, V, W may or maynot attend volleyball game.
Rule 0: { } attend >=2
Rule 1: R-->~T or T-->~R
Rule 2: P--> T or ~T-->~P
Rule 3: ~R-->~V or V-->R
Rule 4: W<-->T or ~W<-->~T
Rule 5: ~Q<-->S or Q<-->~S
Q1: Which of the following could be a complete, accurate list of all attending?
(A)Q, R, S, V (No _ Violates Rule 5: Q<-->~S)
(B)P, S, T (No _ Violates Rule 4 W<-->T)
(C)R,T,V (No _ Violates Rule 1: R-->~T)
(D)Q,T,W (Yes _ No Rule Violations)
(E)W,S,P (No _ Violates Rule 2: P-->T)
Q2: If ~W, which of the following must be true?
(A)~R (No _ Violates all possible solutions)
(B)~V (No _ Violates {Q/S, R, V})
(C)~(RV) (No _ Violates {Q/S. R. V})
(D)R/V or RV (Yes _ Satisfies all possible solutions)
(E) R/V & ~(RV) (No _ Violates {Q/S. R, V})
Rule 4: ~W<-->~T {Q/S, P, R, V}
Rule 2: ~T-->~P {Q/S, R, V}
Apply Rule 0: {}>=2 & Rule 3: ~R-->~V to yield possible solutions.
Possible Solutions: {Q/S, V, R} (Consider Rule 3 & Rule 0)
{Q/S, R} (Consider Rule 3 & Rule 0)
Q3: Which of the following could be true?
(A)VP (No _ Violates Rule 3-->1-->2:V-->R-->~T-->~P)
(B)VW (No _ Violates Rule 3-->1-->4: V-->R-->~T<-->~W)
(C)~(RT) (No _ Violates Rule 0: {}>=2)
(D)~(SQ) (No _ Violates Rule 5: ~Q<-->S)
(E)~(PR) (Yes _ doesn't violate any rules)
Q4: What is the min {} who do not attend?
(A)4 (No)
(B)3 (Yes)
(C)2 (No)
(D)1 (No)
(E)0 (No)
Maximize {} attending
Rule 1: T-->~R is optimal, as T allows P & W, R only allows V: {T,~R)
Rule 2: P-->T is optimal, as it allows P: {T, P, ~R)
Rule 3: ~R-->~V is forced: {T, P, ~R, ~V)
Rule 4: W<-->T is forced: {T, W, P, ~R,~V)
Rule 5: ~S<-->Q either or: {T, W, P, Q, ~S, ~R, ~V)
Q5: If T which of the following could be true?
(A)~QR (No _ Violates Rule 1: T-->~R)
(B)~SV (No _ Violates Rule 1-->3: T-->~R-->~V)
(C)~(QS) (No _ Violates Rule 5: ~S<-->Q)
(D)~(QP) (Yes _ No Rule Violations)
(E) {} = 5 (No _ Violates {T, W, P, Q/S} or {T, W,Q/S})
Q6: Permutations of {}=3?
(A)2 (No)
(B)3 (No)
(C)4 (Yes)
(D)5 (No)
(E)6 (No)
Possible Solutions: {T, W, Q/S}
{V, R, Q/S}
Q7: Replacement For Rule 2?
(A) P<-->W (No -- Violates {T, W, Q/S solution})
(B) ~P-->~W or W-->P (Yes -- supports Rules 2 through Rule 4)
(C) ~P-->~T or T-->P (No -- Violates {T, W, Q/S} solution)
(D) V-->~P, P-->~V (No -- doesn't address Rule 2)
(E) P--> {} <=4 (No -- doesn't address Rule 2)
Given (7) Seniors: P, Q, R, S, T, V, W may or maynot attend volleyball game.
Rule 0: { } attend >=2
Rule 1: R-->~T or T-->~R
Rule 2: P--> T or ~T-->~P
Rule 3: ~R-->~V or V-->R
Rule 4: W<-->T or ~W<-->~T
Rule 5: ~Q<-->S or Q<-->~S
Q1: Which of the following could be a complete, accurate list of all attending?
(A)Q, R, S, V (No _ Violates Rule 5: Q<-->~S)
(B)P, S, T (No _ Violates Rule 4 W<-->T)
(C)R,T,V (No _ Violates Rule 1: R-->~T)
(D)Q,T,W (Yes _ No Rule Violations)
(E)W,S,P (No _ Violates Rule 2: P-->T)
Q2: If ~W, which of the following must be true?
(A)~R (No _ Violates all possible solutions)
(B)~V (No _ Violates {Q/S, R, V})
(C)~(RV) (No _ Violates {Q/S. R. V})
(D)R/V or RV (Yes _ Satisfies all possible solutions)
(E) R/V & ~(RV) (No _ Violates {Q/S. R, V})
Rule 4: ~W<-->~T {Q/S, P, R, V}
Rule 2: ~T-->~P {Q/S, R, V}
Apply Rule 0: {}>=2 & Rule 3: ~R-->~V to yield possible solutions.
Possible Solutions: {Q/S, V, R} (Consider Rule 3 & Rule 0)
{Q/S, R} (Consider Rule 3 & Rule 0)
Q3: Which of the following could be true?
(A)VP (No _ Violates Rule 3-->1-->2:V-->R-->~T-->~P)
(B)VW (No _ Violates Rule 3-->1-->4: V-->R-->~T<-->~W)
(C)~(RT) (No _ Violates Rule 0: {}>=2)
(D)~(SQ) (No _ Violates Rule 5: ~Q<-->S)
(E)~(PR) (Yes _ doesn't violate any rules)
Q4: What is the min {} who do not attend?
(A)4 (No)
(B)3 (Yes)
(C)2 (No)
(D)1 (No)
(E)0 (No)
Maximize {} attending
Rule 1: T-->~R is optimal, as T allows P & W, R only allows V: {T,~R)
Rule 2: P-->T is optimal, as it allows P: {T, P, ~R)
Rule 3: ~R-->~V is forced: {T, P, ~R, ~V)
Rule 4: W<-->T is forced: {T, W, P, ~R,~V)
Rule 5: ~S<-->Q either or: {T, W, P, Q, ~S, ~R, ~V)
Q5: If T which of the following could be true?
(A)~QR (No _ Violates Rule 1: T-->~R)
(B)~SV (No _ Violates Rule 1-->3: T-->~R-->~V)
(C)~(QS) (No _ Violates Rule 5: ~S<-->Q)
(D)~(QP) (Yes _ No Rule Violations)
(E) {} = 5 (No _ Violates {T, W, P, Q/S} or {T, W,Q/S})
Q6: Permutations of {}=3?
(A)2 (No)
(B)3 (No)
(C)4 (Yes)
(D)5 (No)
(E)6 (No)
Possible Solutions: {T, W, Q/S}
{V, R, Q/S}
Q7: Replacement For Rule 2?
(A) P<-->W (No -- Violates {T, W, Q/S solution})
(B) ~P-->~W or W-->P (Yes -- supports Rules 2 through Rule 4)
(C) ~P-->~T or T-->P (No -- Violates {T, W, Q/S} solution)
(D) V-->~P, P-->~V (No -- doesn't address Rule 2)
(E) P--> {} <=4 (No -- doesn't address Rule 2)
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- tsstein
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Vinny Gambini
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Re: Logic Game Challenge #40: Volleyball Game
Sketch: PQRSTVW
1. Never {RT} 2. P-> T = ~T->~P 3. ~R->~V = V->R 4. W or T-> {WT} = ~{WT}-> ~W and~T
5. S or Q, but not both
There are obviously deductions to be made. However, in selection games it is better to remember there are causal chains and use them as the questions demand.
Questions-
1. An acceptability question. Use the rules to see if they are being violated by any options:
Rule 1 is being violated by C, it includes both R and T. Rule 2 is being violated by option E, it includes P without T. Rule 3 is not being violated, move on. Rule 4 is being violated by option B, it includes T without W. Rule 4 is being violated by option A, it has both S and Q. Hence, our correct answer is D.
2. If W does not attend the game, we know that T cannot attend the game (by rule 4). If not T, than not P (Rule 2). Here it would be helpful to draw a list of entities that are in and those that are out. Some of you could probably do it in your head. If we know that W,T and P are out then our only remaining options are Q R S V. We know that either Q or S will attend but not both. We still need at least one more to satisfy the minimum requirement of having 2 entities. Hence, either V or R will attend. Which is choice D.
3. An open ended "could be true" question should be skipped until you have finished all your "If" questions, as they will help you answer this question. Read this after you have gone through question 6. When doing open ended "could be true" questions, it is best to wait and then use the sketches you did for previous questions. This always works and saves you a lot of time!. Lets go through the answers and see if they appeared in any of the work we did for previous sketches. (Note that just because an answer didn't appear in your previous work it does not mean it is false.)
answer choice A, doesn't appear in any of our previous work so lets move on (it is also demonstrably false but we should't care to show that in this stage). Answer choice B doesn't appear in any of our previous work so lets move on (it is also demonstrably false but we should't care to show that in this stage). Answer choice C doesn't appear in any of our previous work so lets move on. Answer choice D doesn't appear in any of our previous work so lets move on. Answer choice E, does appear! remember the work we did for question 6? two of our outcomes included only W, T and either Q or S but not both. So E must be our answer. I could have shown why A-D is false, but there is no reason to do the work twice! This is not luck, it works in a very high percentage of LSAT questions.
4. When asked about a minimum, look for causal relationship by which the existence of one entity excludes the other's. We know that either Q or S will attend but not both (Rule 5), so we know that at least 1 will not attend. We know that If R or T attend, the other does not (Rule 1), so we we know that least 2 will not attend. We also know that if R does not attend, V does not attend (Rule 3). That by itself is not enough. However, we also know that if T does not attend W does not attend (Rule 4). Hence, since R and T cannot both attend and they affect V and W, at least 3 will not attend. which is choice B.
5. If Tim attends, we know that W attends (rule 4). If Tim attends we know That R doesn't attend (Rule 1). If R doesn't attend, than V does not attend (Rule 3). We also know that either Q or S will attend but not both. Lets make a list of in's and out's.
In- T, W, Q/S Out- Q/S, T, V Left to be determined- P. Now let's look at the answers. A. wrong. We already deduced that Ron doesn't attend the game. B. Wrong. We already deduced that V doesn't attend the game. C. Wrong. We know that one of Q or S must attend the game. D. Correct. This fits perfectly into our sketch. P and Q are out, so we are left with T, W and S. Circle and move on.
6. The question asks for the number of possible combinations if we have exactly 3. lets make a list of in's and outs.
In- Q/S _ _ Out- Q/S _ _ _
We know either Q or S but not both will attend because of Rule 5.Now ask "Is there an entity that forces us to go over 3? Yes, it is P. If P than T (Rule 2) and if T than W (Rule 4). P makes us have 4 entities so we know P must be out.
In- Q/S _ _ Out- Q/S P _ _
Who is Left? R,T,V,W remember that by rule 4 W and T always go together. In addition, by rule 3 If we have V we have R. So the other two slots will be filled with either T and W or V and T. thats two options, but lets not forget that each of those options can have either Q or S. Hence, we have 4 options. Which is answer C.
7. Now I would go back to question 3. I would always leave a rule substitution question to last. The rule we want to replace by an equivalent rule is Rule 2. P->T ~T->~P. lets examine each answer. A. Should jump at you as saying too much. Perviously we had answers that included W but not P so this can't be the answer. B. this rule says ~W-> ~P or P->W. By rule 4 we know that W and T go together. so we now know that P->W and T. Is that the same as before when we had p->T and W? Yes! circle and move on.
Thanks for reading,
Tomer Stein.
1. Never {RT} 2. P-> T = ~T->~P 3. ~R->~V = V->R 4. W or T-> {WT} = ~{WT}-> ~W and~T
5. S or Q, but not both
There are obviously deductions to be made. However, in selection games it is better to remember there are causal chains and use them as the questions demand.
Questions-
1. An acceptability question. Use the rules to see if they are being violated by any options:
Rule 1 is being violated by C, it includes both R and T. Rule 2 is being violated by option E, it includes P without T. Rule 3 is not being violated, move on. Rule 4 is being violated by option B, it includes T without W. Rule 4 is being violated by option A, it has both S and Q. Hence, our correct answer is D.
2. If W does not attend the game, we know that T cannot attend the game (by rule 4). If not T, than not P (Rule 2). Here it would be helpful to draw a list of entities that are in and those that are out. Some of you could probably do it in your head. If we know that W,T and P are out then our only remaining options are Q R S V. We know that either Q or S will attend but not both. We still need at least one more to satisfy the minimum requirement of having 2 entities. Hence, either V or R will attend. Which is choice D.
3. An open ended "could be true" question should be skipped until you have finished all your "If" questions, as they will help you answer this question. Read this after you have gone through question 6. When doing open ended "could be true" questions, it is best to wait and then use the sketches you did for previous questions. This always works and saves you a lot of time!. Lets go through the answers and see if they appeared in any of the work we did for previous sketches. (Note that just because an answer didn't appear in your previous work it does not mean it is false.)
answer choice A, doesn't appear in any of our previous work so lets move on (it is also demonstrably false but we should't care to show that in this stage). Answer choice B doesn't appear in any of our previous work so lets move on (it is also demonstrably false but we should't care to show that in this stage). Answer choice C doesn't appear in any of our previous work so lets move on. Answer choice D doesn't appear in any of our previous work so lets move on. Answer choice E, does appear! remember the work we did for question 6? two of our outcomes included only W, T and either Q or S but not both. So E must be our answer. I could have shown why A-D is false, but there is no reason to do the work twice! This is not luck, it works in a very high percentage of LSAT questions.
4. When asked about a minimum, look for causal relationship by which the existence of one entity excludes the other's. We know that either Q or S will attend but not both (Rule 5), so we know that at least 1 will not attend. We know that If R or T attend, the other does not (Rule 1), so we we know that least 2 will not attend. We also know that if R does not attend, V does not attend (Rule 3). That by itself is not enough. However, we also know that if T does not attend W does not attend (Rule 4). Hence, since R and T cannot both attend and they affect V and W, at least 3 will not attend. which is choice B.
5. If Tim attends, we know that W attends (rule 4). If Tim attends we know That R doesn't attend (Rule 1). If R doesn't attend, than V does not attend (Rule 3). We also know that either Q or S will attend but not both. Lets make a list of in's and out's.
In- T, W, Q/S Out- Q/S, T, V Left to be determined- P. Now let's look at the answers. A. wrong. We already deduced that Ron doesn't attend the game. B. Wrong. We already deduced that V doesn't attend the game. C. Wrong. We know that one of Q or S must attend the game. D. Correct. This fits perfectly into our sketch. P and Q are out, so we are left with T, W and S. Circle and move on.
6. The question asks for the number of possible combinations if we have exactly 3. lets make a list of in's and outs.
In- Q/S _ _ Out- Q/S _ _ _
We know either Q or S but not both will attend because of Rule 5.Now ask "Is there an entity that forces us to go over 3? Yes, it is P. If P than T (Rule 2) and if T than W (Rule 4). P makes us have 4 entities so we know P must be out.
In- Q/S _ _ Out- Q/S P _ _
Who is Left? R,T,V,W remember that by rule 4 W and T always go together. In addition, by rule 3 If we have V we have R. So the other two slots will be filled with either T and W or V and T. thats two options, but lets not forget that each of those options can have either Q or S. Hence, we have 4 options. Which is answer C.
7. Now I would go back to question 3. I would always leave a rule substitution question to last. The rule we want to replace by an equivalent rule is Rule 2. P->T ~T->~P. lets examine each answer. A. Should jump at you as saying too much. Perviously we had answers that included W but not P so this can't be the answer. B. this rule says ~W-> ~P or P->W. By rule 4 we know that W and T go together. so we now know that P->W and T. Is that the same as before when we had p->T and W? Yes! circle and move on.
Thanks for reading,
Tomer Stein.
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- yvetterc23
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Vinny Gambini
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Re: Logic Game Challenge #40: Volleyball Game
6/7 correct. What a bummer. 7. Which of the following, if substituted for the condition that if Paul attends the game, so does Tim, would have the same effect in determining who can attend the game together?
I felt it was between A or B, but B seems to be the wrong answer in my eyes. If Tim goes than so does Wanda and vis-versa. But it doesn't mean the same for "if Paul attends the game, so does Tim. I felt that is Paul didn't attend the game Tim still MIGHT attend the game. So if tim still might attend the game so will Wanda, so I felt Wanda might be at the game even if Paul didn't go.
So I couldn't prove A to be wrong so I choose A.
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- mornincounselor
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Elle Woods
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Re: Logic Game Challenge #40: Volleyball Game
6/7 correct, missed number six.
We know either Q/S will be at the dance occupying one of our 3 slots, we know P and the other of S/Q occupy two of the four out slots.
I came up with six possibilities:
TWQ
TWS
RVQ
RVS
RWQ
RWS
Of course returning to our rules we see that if W -> T and therefore we can eliminate the third set of possibilities, leaving us with 4.
We know either Q/S will be at the dance occupying one of our 3 slots, we know P and the other of S/Q occupy two of the four out slots.
I came up with six possibilities:
TWQ
TWS
RVQ
RVS
RWQ
RWS
Of course returning to our rules we see that if W -> T and therefore we can eliminate the third set of possibilities, leaving us with 4.
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- evantheo
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Vinny Gambini
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Re: Logic Game Challenge #40: Volleyball Game
yvetterc23 Wrote::(
7. Which of the following, if substituted for the condition that if Paul attends the game, so does Tim, would have the same effect in determining who can attend the game together?
I felt it was between A or B, but B seems to be the wrong answer in my eyes. If Tim goes than so does Wanda and vis-versa. But it doesn't mean the same for "if Paul attends the game, so does Tim. I felt that is Paul didn't attend the game Tim still MIGHT attend the game. So if tim still might attend the game so will Wanda, so I felt Wanda might be at the game even if Paul didn't go.
So I couldn't prove A to be wrong so I choose A.
I thought the same thing. I'm still beginning my LSAT prep, but can anyone explain why:
If Paul attends -> Tim attends = If Paul doesn't attend -> Tim doesn't attend
I took a logic course this year, and this looks like p->q = ~p->~q, which I believe is invalid. If Paul attends, then Tim must attend. But it doesn't say that Tim attends "only if" Paul attends. Therefore, shouldn't Tim still be able to attend even if Paul does not attend?
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- david.GOCANADIENSGO
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Vinny Gambini
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Re: Logic Game Challenge #40: Volleyball Game
6/7, I got question 2 wrong. Could anyone explain to me what I did wrong?
Thanks.-david.GOCANADIENSGO
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- tommywallach
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Atticus Finch
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Re: Logic Game Challenge #40: Volleyball Game
You'd have to explain your process. You can see the correct process in many of the answers above, so if you got something different, we'd have to see your process before we could explain where you went wrong. Thanks!
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- equatick
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Vinny Gambini
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Re: Logic Game Challenge #40: Volleyball Game
I'm having trouble with the condition "Sammy attends the game if, and only if, Quita does not attend."
Why can't neither Quita nor Sammy attend? If Sammy were to attend, it can only be when Quita does not attend. It seems that the condition should be "Sammy MUST attend the game if, and only if, Quita does not attend" in order for this to be absolutely true.
Why can't neither Quita nor Sammy attend? If Sammy were to attend, it can only be when Quita does not attend. It seems that the condition should be "Sammy MUST attend the game if, and only if, Quita does not attend" in order for this to be absolutely true.
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- chikiblues
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Vinny Gambini
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Re: Logic Game Challenge #40: Volleyball Game
7. Which of the following, if substituted for the condition that if Paul attends the game, so does Tim, would have the same effect in determining who can attend the game together?
(A) Either Paul and Wanda attend the game together or neither attends.
(B) Wanda does not attend the game only if Paul does not attend the game.
(C) Tim does not attend the game if Paul does not attend the game.
(D) Paul cannot attend the game unless Vanessa does not attend.
(E) If Paul attends the game, then at most four students attend the game.
I don't understand why the answer is B and not D.
In the original scenario the solutions are:
TWQ, TWS, PTWQ, PTWS, RQ, RS, RVQ, RVS
If Answer B is true that eliminates the solutions TWQ and TWS
However if D is true the solutions stay the same, since P and V cannot attend together, and D states that V has to be absent for P to attend.
Can someone explain why this is wrong?
(A) Either Paul and Wanda attend the game together or neither attends.
(B) Wanda does not attend the game only if Paul does not attend the game.
(C) Tim does not attend the game if Paul does not attend the game.
(D) Paul cannot attend the game unless Vanessa does not attend.
(E) If Paul attends the game, then at most four students attend the game.
I don't understand why the answer is B and not D.
In the original scenario the solutions are:
TWQ, TWS, PTWQ, PTWS, RQ, RS, RVQ, RVS
If Answer B is true that eliminates the solutions TWQ and TWS
However if D is true the solutions stay the same, since P and V cannot attend together, and D states that V has to be absent for P to attend.
Can someone explain why this is wrong?
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- SoniaF29
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Vinny Gambini
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Re: Logic Game Challenge #40: Volleyball Game
Too convuluted to explain via text, needs a video of person explaining
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- N.H816
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Vinny Gambini
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Re: Logic Game Challenge #40: Volleyball Game
This was my work that helped me divide and conquer the statements.
Explanation - if first statement in string (if R attends = R+), then all the other statements.
(---) = "can go, but not must go"
Max and Min determined by + and - in string, but obviously, 2 must attend.
Q/S means one of these, but not both.
+R,-T,(+V),-P,-W,Q/S max +3 min +2
-R,(+T),-V,(+P),(+W),Q/S max +4 min +1
-T,(+R),(+V),-P,-W,Q/S max +3 min +1
+T,-R,-V,(+P),+W,Q/S max +4 min +3
+P,+T,-V,-R,+W,Q/S max +4 min +4
-P
+V,-T,+R,-P,-W,Q/S max +3 min +3
-V
+W,+T,-V,(+P),-R,Q/S max +4 min +3
-W,-T,(+V),-P,(+R),Q/S max +3 min +1
Explanation - if first statement in string (if R attends = R+), then all the other statements.
(---) = "can go, but not must go"
Max and Min determined by + and - in string, but obviously, 2 must attend.
Q/S means one of these, but not both.
+R,-T,(+V),-P,-W,Q/S max +3 min +2
-R,(+T),-V,(+P),(+W),Q/S max +4 min +1
-T,(+R),(+V),-P,-W,Q/S max +3 min +1
+T,-R,-V,(+P),+W,Q/S max +4 min +3
+P,+T,-V,-R,+W,Q/S max +4 min +4
-P
+V,-T,+R,-P,-W,Q/S max +3 min +3
-V
+W,+T,-V,(+P),-R,Q/S max +4 min +3
-W,-T,(+V),-P,(+R),Q/S max +3 min +1
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- N.H816
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Vinny Gambini
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Re: Logic Game Challenge #40: Volleyball Game
equatick Wrote:I'm having trouble with the condition "Sammy attends the game if, and only if, Quita does not attend."
Why can't neither Quita nor Sammy attend? If Sammy were to attend, it can only be when Quita does not attend. It seems that the condition should be "Sammy MUST attend the game if, and only if, Quita does not attend" in order for this to be absolutely true.
if, and only if = means "either both statements are true, or both statements are false"
were it to say "Sammy attends the game if Quita does not attend" , Sammy could still attend the game if Quita attends. They are not mutually exclusive. But when you bring in "if, and only if" that takes away that other option. If Quita attends then Sammy does not attend, because Sammy attends only if Quita does not. They become equally weighted values, both changing the other with a change of self. Mutually Exclusive.
If it were to say "Sammy attends the game if, and only if, Quita attends the game" this would be Mutually Inclusive.
Another thought, were it to say "Sammy attends the game only if Quita does not attend", this is still lacking because it does not mandate that Sammy attend, but only that Sammy is free to attend only with Quita stays.
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