LSAT Logic Game setup help

blipblop

Senior member
Jun 23, 2004
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For the school paper, five students - jiang, kramer, lopez, megregian, and o'neill- each review one or more of exactly three plays: Sunset, Tamerlane, and Undulation, but do not review any other plays. The following conditions apply:

Kramer and Lopez each review fewer of the plays than Megregrian.

Neither Lopez nor Megregian reviews any play Jiang reviews.

Kramer and O'Neill both review Tamerlan.

Exactly two of the students review exactly the same play or plays as each other.

How do I set this up? this is the last games on the December 2003 LSAT I'm completely lost...

Edited to add questions...

19. which one of the following could be an accurate and complete list of the students who review only Sunset?

A)Lopez
B)O'Neill
C)Jiang, Lopez
D)Kramer, O'Neill
E)Lopez, Megregian

20. which one of the following must be true?

a. Jiang reviews more of the plays than Lopez does
b. Megregrian reviews more of the plays than Jiang does
c. Megregrian reviews more of the plays than O'neill does
d. O'neill reviews more of the plays than Jiang does
e. O'neill reviews more of the plays than Kramer does

Only two questions out of 6, but I need to find a way to properly setup this games because come saturday if something like this appears, i'm dead... thanks guys.
 

Legendary

Diamond Member
Jan 22, 2002
7,019
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What do they want? What plays are reviewed by each person, or how many?

Edit: Also, your 2nd condition has a typo "and" -> "any"
 

Cookie

Golden Member
Jul 3, 2001
1,759
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81
19. which one of the following could be an accurate and complete list of the students who review only Sunset?

C)Jiang, Lopez



20. which one of the following must be true?

b. Megregrian reviews more of the plays than Jiang does


Just write down what you know and everything falls into place.
eg)Megregrian has to review at least 2 plays, Jiang can only review 1, therefore b is necessary
 

dullard

Elite Member
May 21, 2001
25,543
4,038
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Start with #20, as it is needed for the rest of the problem. Here are the conditions (I labeled them for easier referencing):
A) Each review one or more of exactly three plays.
B) K and L each review fewer of the plays than M.
C) Neither L nor M reviews any play J reviews.
D) K and O both review T.
E) Exactly two of the students review exactly the same play or plays as each other.

From condition B, we know that M reviewed at least 2 of the three plays. From condition C, we know that M didn't review at least one of the three plays. Combined, conditions B and C say that M reviewed 2 plays. Also, we then know that J, K, and L each reviewed only 1 play.

With that in mind, #20 is easy to solve. B is the correct answer for #20.

For the other problems, draw out the possible charts. Put a column of J, K, L, M, and O. Put a row of S, T, and U. Fill in what you know. For example, K only reviewed one play and condition D said that one and only play was T. So write Yes in the cell for K and T and No for the other cells for K.

In this problem, you aren't given enough information to solve it all. So you have to guess. A lot of the conditions rely on J. So make three grids, in each you give J a different play that you guess. From there, you should easilly be able to write a lot of yesses and a lot of nos and a few maybes.

That'll answer question #19 quite quickly. A is the only possible answer.
 

Cookie

Golden Member
Jul 3, 2001
1,759
2
81
Originally posted by: dullard
Start with #20, as it is needed for the rest of the problem. Here are the conditions (I labeled them for easier referencing):
A) Each review one or more of exactly three plays.
B) K and L each review fewer of the plays than M.
C) Neither L nor M reviews any play J reviews.
D) K and O both review T.
E) Exactly two of the students review exactly the same play or plays as each other.

From condition B, we know that M reviewed at least 2 of the three plays. From condition C, we know that M didn't review at least one of the three plays. Combined, conditions B and C say that M reviewed 2 plays. Also, we then know that J, K, and L each reviewed only 1 play.

With that in mind, #20 is easy to solve. B is the correct answer for #20.

For the other problems, draw out the possible charts. Put a column of J, K, L, M, and O. Put a row of S, T, and U. Fill in what you know. For example, K only reviewed one play and condition D said that one and only play was T. So write Yes in the cell for K and T and No for the other cells for K.

In this problem, you aren't given enough information to solve it all. So you have to guess. A lot of the conditions rely on J. So make three grids, in each you give J a different play that you guess. From there, you should easilly be able to write a lot of yesses and a lot of nos and a few maybes.

That'll answer question #19 quite quickly. E is the only possible answer.

E can't be the answer because Megregian reviewed 2 plays.
 

dullard

Elite Member
May 21, 2001
25,543
4,038
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Originally posted by: Cookie
E can't be the answer because Megregian reviewed 2 plays.
Doh, I missed the ONLY part of the question. I rushed and I shouldn't have. Change that to A.

 

brikis98

Diamond Member
Jul 5, 2005
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Originally posted by: Cookie
19. which one of the following could be an accurate and complete list of the students who review only Sunset?

C)Jiang, Lopez



20. which one of the following must be true?

b. Megregrian reviews more of the plays than Jiang does


Just write down what you know and everything falls into place.
eg)Megregrian has to review at least 2 plays, Jiang can only review 1, therefore b is necessary

yup, i agree with these answers.
 

Cookie

Golden Member
Jul 3, 2001
1,759
2
81
Originally posted by: brikis98
Originally posted by: Cookie
19. which one of the following could be an accurate and complete list of the students who review only Sunset?

C)Jiang, Lopez



20. which one of the following must be true?

b. Megregrian reviews more of the plays than Jiang does


Just write down what you know and everything falls into place.
eg)Megregrian has to review at least 2 plays, Jiang can only review 1, therefore b is necessary

yup, i agree with these answers.

Actually, after reviewing dullards points, I think the answer might be 19) A
because I forgot to take into account one of the criteria
"Exactly two of the students review exactly the same play or plays as each other. "
and also
"Neither Lopez nor Megregian reviews any play Jiang reviews"

19) A
FTW
 

kumanchu

Golden Member
Feb 15, 2000
1,471
4
81
i think blipblop is looking for more than the answer, he's asking for a method to find the answer to similar problems consistently.

i always made charts for the complicated logic games and drew out the information the problem gave in the margin of the testbook. if you haven't learned to chart problems like this, a start is to make simple drawings of the relations, and list the rules next to them.

hope it helps.
 

bahbahbooey

Member
Nov 18, 2006
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19) can NOT be A. If it is A, then J is the only one that does S, and that means M must be at T and U(because M must be at LEAST 2 review cuz of rule 1)...HOWEVER, J must be at a T or U, and J cannot be with M

the Answer is E <---- DISREGARD THIS. --if u want me to go thru why all the others are wrong, PM me


This is hwo i Set it up...i did not draw out hte last rule, so u gotta remember the last rule in ur head ----> http://img161.imageshack.us/my.php?image=lsat19ny5.jpg

 

Cookie

Golden Member
Jul 3, 2001
1,759
2
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Originally posted by: bahbahbooey
19) can NOT be A. If it is A, then J is the only one that does S, and that means M must be at T and U(because M must be at LEAST 2 review cuz of rule 1)...HOWEVER, J must be at a T or U, and J cannot be with M

the Answer is E. --if u want me to go thru why all the others are wrong, PM me

It can not be E for the reason stated above. (M needs 2 reviews)
 

DaShen

Lifer
Dec 1, 2000
10,710
1
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That is kind of easy knowing that there are only 3 plays. :)

Megregian observed the 2 other plays that Jiang didn't see.

Jiang only saw one play, since Megregian saw two plays, and they mutally exclude each other. Kramer and Lopez only saw one play sas well.

Kramer only saw Tamerlian.

O'Neill saw Tamerlian, and maybe something else....

**EDIT**
BTW

19) A
20) B

19 by process of elimination. It is the only viable choice, becaue it states who only reviewed Sunset. We know everyone else in that list is eliminated because O'Neill already reviewed one other play, and the other ones are eliminated by one person or the other.

20 is easy based on the first things given.

Anyone who says E for 19 didn't read the question right.
 

dullard

Elite Member
May 21, 2001
25,543
4,038
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Originally posted by: bahbahbooey
19) can NOT be A. If it is A, then J is the only one that does S, and that means M must be at T and U(because M must be at LEAST 2 review cuz of rule 1)...HOWEVER, J must be at a T or U, and J cannot be with M

the Answer is E. --if u want me to go thru why all the others are wrong, PM me


This is hwo i Set it up...i did not draw out hte last rule, so u gotta remember the last rule in ur head ----> http://img161.imageshack.us/my.php?image=lsat19ny5.jpg
You too misread the question. Option A says that L reviews S only, not J.

 

Cookie

Golden Member
Jul 3, 2001
1,759
2
81
19. which one of the following could be an accurate and complete list of the students who review only Sunset?

A)Lopez
B)O'Neill
C)Jiang, Lopez
D)Kramer, O'Neill
E)Lopez, Megregian

B) Can't be true because "Kramer and O'Neill both review Tamerlan"
C) Can't be true because "Neither Lopez nor Megregian reviews any play Jiang reviews."
D) Can't be true because "Kramer and O'Neill both review Tamerlan"
E) Can't be true because "Kramer and Lopez each review fewer of the plays than Megregrian. " (Meaning Megregrian reviewed more than 1 play)

Therefore the answer is A.

 

r6ashih

Senior member
May 29, 2003
667
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0
Originally posted by: bahbahbooey
19) can NOT be A. If it is A, then J is the only one that does S, and that means M must be at T and U(because M must be at LEAST 2 review cuz of rule 1)...HOWEVER, J must be at a T or U, and J cannot be with M

the Answer is E. --if u want me to go thru why all the others are wrong, PM me


This is hwo i Set it up...i did not draw out hte last rule, so u gotta remember the last rule in ur head ----> http://img161.imageshack.us/my.php?image=lsat19ny5.jpg

19. which one of the following could be an accurate and complete list of the students who review only Sunset?

C is the answer because J can only go once, and L can only go once. so in one instance, J reviews only sunset, and in another instance, L reviews ONLY sunset and no other play.
 

bahbahbooey

Member
Nov 18, 2006
83
0
0
Originally posted by: r6ashih
Originally posted by: bahbahbooey
19) can NOT be A. If it is A, then J is the only one that does S, and that means M must be at T and U(because M must be at LEAST 2 review cuz of rule 1)...HOWEVER, J must be at a T or U, and J cannot be with M

the Answer is E. --if u want me to go thru why all the others are wrong, PM me


This is hwo i Set it up...i did not draw out hte last rule, so u gotta remember the last rule in ur head ----> http://img161.imageshack.us/my.php?image=lsat19ny5.jpg

19. which one of the following could be an accurate and complete list of the students who review only Sunset?

C is the answer because J can only go once, and L can only go once. so in one instance, J reviews only sunset, and in another instance, L reviews ONLY sunset and no other play.

J and L cant review together. Rule 2
 

dullard

Elite Member
May 21, 2001
25,543
4,038
126
Originally posted by: bahbahbooey
Well, what is the final setup when 19 is solved if you use A ??..cuz i cant figure it out
J: U
K: T
L: S
M: S and T
O: (T) or (S and T)
 

Buttsmacker

Member
Jun 2, 2005
42
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0
I think you all are just going to confuse this poor soul.

They asked HOW TO DIAGRAM, not what the answer was.

I set it up with STU along the top of a table, and then fit in each person's innitials below it.

To illustrate each restriction I used:
M = 2
K = 1
L = 1
J = 1
No MJ
No LJ

this is the kind of problem you need to make THREE tables exactly like each other to come up with all possible answers (Based on J, as it is the most restricted item. ) In the first table, place J under S; second place J under T; third place J under U and go from there.
If you know where J is, you can place M - anywhere J isn't. Based on J's position, L can go in ONE of the two other plays, thus the only unknown variable... and that would be how you solve the "could be true" or "must be false" by using these choices. O always has to be under T, and to satisfy the fourth condition an additional O can be placed in the same category as M (in the sequence where J is located under T, O would then review all three plays).
so here's my best illustration:
First
S........T.......U
(O).....O....(O)
. .......K...... J
M ...... M
(L)<->(L)


Second
S........T.......U
(O).....O....(O)
J ........K
..........M ...... M
..........(L)<->(L)

Third
S........T.......U
(O).....O....(O)
..........K
...........J
M .............. M
(L)....<->.....(L)
I really just need to show you a written version...

(_) indicates interchangable positioning with L and "and/or" positioning for O. Excuse how I put that togehter.. i am SURE there was a better way. Some of these problems just HAVE to be written out, unless you are a frickin genious, which I'm not. I'll be RIGHT THERE with you on Saturday. Best of luck to both of us!
This should be right, but forgive me if I've screwed anything up, I am tired and at work!
 

r6ashih

Senior member
May 29, 2003
667
0
0
Originally posted by: bahbahbooey
Originally posted by: r6ashih
Originally posted by: bahbahbooey
19) can NOT be A. If it is A, then J is the only one that does S, and that means M must be at T and U(because M must be at LEAST 2 review cuz of rule 1)...HOWEVER, J must be at a T or U, and J cannot be with M

the Answer is E. --if u want me to go thru why all the others are wrong, PM me


This is hwo i Set it up...i did not draw out hte last rule, so u gotta remember the last rule in ur head ----> http://img161.imageshack.us/my.php?image=lsat19ny5.jpg

19. which one of the following could be an accurate and complete list of the students who review only Sunset?

C is the answer because J can only go once, and L can only go once. so in one instance, J reviews only sunset, and in another instance, L reviews ONLY sunset and no other play.

J and L cant review together. Rule 2

The question is if the the person reviews ONLY SUNSET and no other play. Not if They are alone in reviewing.
 

bahbahbooey

Member
Nov 18, 2006
83
0
0
Originally posted by: r6ashih
Originally posted by: bahbahbooey
Originally posted by: r6ashih
Originally posted by: bahbahbooey
19) can NOT be A. If it is A, then J is the only one that does S, and that means M must be at T and U(because M must be at LEAST 2 review cuz of rule 1)...HOWEVER, J must be at a T or U, and J cannot be with M

the Answer is E. --if u want me to go thru why all the others are wrong, PM me


This is hwo i Set it up...i did not draw out hte last rule, so u gotta remember the last rule in ur head ----> http://img161.imageshack.us/my.php?image=lsat19ny5.jpg

19. which one of the following could be an accurate and complete list of the students who review only Sunset?

C is the answer because J can only go once, and L can only go once. so in one instance, J reviews only sunset, and in another instance, L reviews ONLY sunset and no other play.

J and L cant review together. Rule 2

The question is if the the person reviews ONLY SUNSET and no other play. Not if They are alone in reviewing.

hahaha.. ONLY sunset. fcuk