I did this for PT56 and it was a good exercise. I'm not sure if anyone is going to ever be helped by these being online, but there is a chance.

Overall, I thought the LGs on PT53 were some of the easier I've seen. Maybe it's just my style though.

Questions 1-5:

See diagram: http://www.psych.ubc.ca/~jchin/diagrams/Image18.jpg

This was a very easy game for me. That rule 1 places X and the rule 2 tells you that Y cannot be with X limits the game greatly. Then the ZY rule pretty much places 3/5 variables, telling you it should be one to breeze through to build up some extra time.

The 2-2-1 or 3-1-1 distributions are good to keep in mind but are never integral to a specific solution.

1. Apply a rule

-start with the X placed on F rule because it's easiest to see

2. Honestly, when there are these few elements, I think you can just eyeball. Starting with A, you can see that if W is at S, ZY goes to P and then T can go either at F or P. Bingo A is correct.

3. Same strategy

A: Can W be with Z? Sure, place ZY at S, and add W. T goes to P and X is at F of coruse

B: ZY can't go at F. So that leaves W and T. If you place one of them there, then one group will go empty. This is the correct answer

4. Look right to the most restricted variables. I thought about ZY first:

E: if Z goes to P, then Y does as well. This seems pretty restricted but it's easy to see that both TW could then go to S or just W. -wrong

Next I thought about T because that sets of some conditionals

A: T goes to F. This leaves all kinds of options

B: T goes to S. Seems like the obvious answer, but sometimes those are right. T brings W with him to S. ZY can only go to P. Correct!

5. This one was actually also very easy. Overall such a great way to start a game. I really hope we get some like these on Monday. We are looking that the answer that cannot be true. 4 could be true

Make a quick diagram with ZY at S, and of course X at F. We can see that T will be either at P or F.

A. Our diagram says yes, it could just be T here

B. Yes, W could go to Z

C. No our diagram clearly shows that T can't be at S. Correct choice

## PT53 Logic games explained

- jmc8y
**Posts:**29**Joined:**Wed May 12, 2010 1:16 am

### Re: PT53 Logic games explained

Questions 6-11

Another straightforward game! Essentially, it's just a balanced linear game. The one curveball is that two of the three rules are worded in an unusual way. But the questions are actually easier than they could have been given this bit of complexity.

Diagram: http://www.psych.ubc.ca/~jchin/diagrams/Image19.jpg

Note that with the second and third rules, you could just write them out as G<J <--/--> L<G but the way I have it is actually easier to comprehend because we see all the options right away without having to reason out that if G<J then we know G>L

6. Grab a rule game. Start with the not-laws from the diagram, then the P<M<L rule, then with the somewhat trickier last two rules. This way you've eliminated stuff already so you don't have to use those tricky rules as much. Answer: C

7. This is just asking about a simple not law from a simple rule! A!

8. This is pretty easy to see as well. If L is last then we are definitely talking about the L<G scenario. Like I said, these questions could have easily been harder. Anyway in that scenario we know J<L which is exactly what A says.

9. To diagram the new rule: P & J < M < L and we are looking for a couldbetrueX or cannot be true. Right now I was thinking this can't be this easy but I'll just see if any of these try to place P J or M very late in the order because that would be impossible. And shit, it turned out to be right. D tries to put J fifth which is impossible.

10. What the hell, this one just asks about the initial not law from the P<M<L rule. At this point I started getting worried that I was doing it wrong because it was too easy. but I guess not. C is right here

11. Finally, a somewhat difficult one. I guess we all knew this was coming. No real way to eyeball this to find the right answer, but we can eliminate some without too much of a stink.

A: notlaw says M can't be first. Eliminate

E. this one is easy to see because it places the last three variables. VGL. Look at the last rule, this can't be, because if V is earlier than G it also has to be earlier than P and that's not possible here

D. This also places the last 3 as PLJ but we know M has to be between P and L so this is wrong.

Now we can just test B or C and that will tell us which one is right.

Let's do C and draw it out.

it places __ M __ ___ G J

That can't be right because if G is ahead of J it also has to be ahead of L. Eliminate

B is right.

Another straightforward game! Essentially, it's just a balanced linear game. The one curveball is that two of the three rules are worded in an unusual way. But the questions are actually easier than they could have been given this bit of complexity.

Diagram: http://www.psych.ubc.ca/~jchin/diagrams/Image19.jpg

Note that with the second and third rules, you could just write them out as G<J <--/--> L<G but the way I have it is actually easier to comprehend because we see all the options right away without having to reason out that if G<J then we know G>L

6. Grab a rule game. Start with the not-laws from the diagram, then the P<M<L rule, then with the somewhat trickier last two rules. This way you've eliminated stuff already so you don't have to use those tricky rules as much. Answer: C

7. This is just asking about a simple not law from a simple rule! A!

8. This is pretty easy to see as well. If L is last then we are definitely talking about the L<G scenario. Like I said, these questions could have easily been harder. Anyway in that scenario we know J<L which is exactly what A says.

9. To diagram the new rule: P & J < M < L and we are looking for a couldbetrueX or cannot be true. Right now I was thinking this can't be this easy but I'll just see if any of these try to place P J or M very late in the order because that would be impossible. And shit, it turned out to be right. D tries to put J fifth which is impossible.

10. What the hell, this one just asks about the initial not law from the P<M<L rule. At this point I started getting worried that I was doing it wrong because it was too easy. but I guess not. C is right here

11. Finally, a somewhat difficult one. I guess we all knew this was coming. No real way to eyeball this to find the right answer, but we can eliminate some without too much of a stink.

A: notlaw says M can't be first. Eliminate

E. this one is easy to see because it places the last three variables. VGL. Look at the last rule, this can't be, because if V is earlier than G it also has to be earlier than P and that's not possible here

D. This also places the last 3 as PLJ but we know M has to be between P and L so this is wrong.

Now we can just test B or C and that will tell us which one is right.

Let's do C and draw it out.

it places __ M __ ___ G J

That can't be right because if G is ahead of J it also has to be ahead of L. Eliminate

B is right.

- jmc8y
**Posts:**29**Joined:**Wed May 12, 2010 1:16 am

### Re: PT53 Logic games explained

Questions 12-17

This is actually one of my favorite games of all time. I think it's because it contains an inference that if you get it, makes the game a breeze. And for once, I did!

It's a multilevel linear game that is balanced. There is no set numerical distribution for the confessions.

Diagram:

Pretty straightforward at first. Until you start thinking about the two confessions after T and none after W. That tells you right there that W has to be after T, otherwise W couldn't have all non-confessions after him. But how far after? Far enough to pass the two confessions. And then you see that W<S law and realize you can place W and S exactly. After that it becomes the easiest third game I've ever done.

Questions

12. Probably best to just run through the nearly complete diagram.

A. Nope - not law for that

B. nothing wrong with this - correct

C. Not law

D. We know he is questioned on 6!

E. Again, we know S is on 7

13. I made a minidiagram here but most of you could probably just do this in your head. If Z was the second to confess he clearly can't be first. So he's second, Y is first and they both confessed

It's a could be true except, so the correct answer cannot be true

A. We never know a lot about T's status so could be true

B. Same

C. V could be either

D. Same

E. Our initial inference shows that yes Y must have confessed.

14. Makes a rule: V<Y<X which forces Z first. Diagram or keep in your head

A. correct. V could be either.

15. This just follows from the diagram

E

16. If X and Y both confessed then one of them must be in slot B, because that is one of the two only spots that leaves open the possibility of confessing. This forces V into 4 and not confessing. So A cannot be true and is right.

17. Same idea. but Y has to be in 5 here so she confessed. D.

Great game.

This is actually one of my favorite games of all time. I think it's because it contains an inference that if you get it, makes the game a breeze. And for once, I did!

It's a multilevel linear game that is balanced. There is no set numerical distribution for the confessions.

Diagram:

Pretty straightforward at first. Until you start thinking about the two confessions after T and none after W. That tells you right there that W has to be after T, otherwise W couldn't have all non-confessions after him. But how far after? Far enough to pass the two confessions. And then you see that W<S law and realize you can place W and S exactly. After that it becomes the easiest third game I've ever done.

Questions

12. Probably best to just run through the nearly complete diagram.

A. Nope - not law for that

B. nothing wrong with this - correct

C. Not law

D. We know he is questioned on 6!

E. Again, we know S is on 7

13. I made a minidiagram here but most of you could probably just do this in your head. If Z was the second to confess he clearly can't be first. So he's second, Y is first and they both confessed

It's a could be true except, so the correct answer cannot be true

A. We never know a lot about T's status so could be true

B. Same

C. V could be either

D. Same

E. Our initial inference shows that yes Y must have confessed.

14. Makes a rule: V<Y<X which forces Z first. Diagram or keep in your head

A. correct. V could be either.

15. This just follows from the diagram

E

16. If X and Y both confessed then one of them must be in slot B, because that is one of the two only spots that leaves open the possibility of confessing. This forces V into 4 and not confessing. So A cannot be true and is right.

17. Same idea. but Y has to be in 5 here so she confessed. D.

Great game.

- jmc8y
**Posts:**29**Joined:**Wed May 12, 2010 1:16 am

### Re: PT53 Logic games explained

Questions 18-23

I guess this is the hardest of the bunch, but it isn't so bad. Honestly, as far as LGs come, this is the best case scenario for a section. I guess I'm good at linear ones, with apparent inferences...so easy ones?

Diagram:

Everything follows pretty straightforward from the rules. Keep in mind

-O is a floater

-Since S is yoked to G and G can't be 3rd, neither can S

Didn't see any huge inferences here and from what I see online, there isn't one.

18. Grab a rule. I got tied up on this a bit because I forgot to grab the t goes 2nd rule because it was just in my diagram. Good to have a set pattern on these: go to diagram first maybe?

E.

19. Mini-diagram this. If P is from (H)illtop and hilltop can't be 1st then he can't be there. But he is before N so we know he's not last. So he's 2nd. P is 3rd then. S forced to 1st. O/M can be put in the final two spaces. Leads directly to A

20. Mini-diagram. O in 2 forces S to 1st. The p<n rule comes into play such that p then has to bee first and n last. only one spot left for m, so we can place all the kids. We can also place G because we know s goes to G. I think /M can flip here so let's just see if this is enough to answer the questions.

A. nope G is full and M isn't there. cannot be true.

B. seems possible

C. impossible for same reason as A

D. impossible, p is already assigned to G

E. same as A and C

So without really proving it, we know B must be true

21. This would be hard, except t is placed so we can diagram this out. p and t are in the middle. n must be third and s first because that's the last place Gillom school can go. The answer is four. C.

22. This is actually a pretty complicated one. I got it before but the second time it threw me. That's because I forgot about the pm notblock rule.

Here it goes: if m is < Hilltop then m is first or second. I just diagrammed both out. If she's second S and P go first then N and O go last. There is another scenario but let's just run through it real quick since it's a could be true question and thus maybe this template has the answer. and it does - E

23. Go back through old templates to find who was S's teammates.

A - eliminate by hypo for 19

B

C - eliminate by hypo for 19

D - eliminate by q18 grab a rule

E

just draw out the last one to see that S and T is possible.

B is right

I guess this is the hardest of the bunch, but it isn't so bad. Honestly, as far as LGs come, this is the best case scenario for a section. I guess I'm good at linear ones, with apparent inferences...so easy ones?

Diagram:

Everything follows pretty straightforward from the rules. Keep in mind

-O is a floater

-Since S is yoked to G and G can't be 3rd, neither can S

Didn't see any huge inferences here and from what I see online, there isn't one.

18. Grab a rule. I got tied up on this a bit because I forgot to grab the t goes 2nd rule because it was just in my diagram. Good to have a set pattern on these: go to diagram first maybe?

E.

19. Mini-diagram this. If P is from (H)illtop and hilltop can't be 1st then he can't be there. But he is before N so we know he's not last. So he's 2nd. P is 3rd then. S forced to 1st. O/M can be put in the final two spaces. Leads directly to A

20. Mini-diagram. O in 2 forces S to 1st. The p<n rule comes into play such that p then has to bee first and n last. only one spot left for m, so we can place all the kids. We can also place G because we know s goes to G. I think /M can flip here so let's just see if this is enough to answer the questions.

A. nope G is full and M isn't there. cannot be true.

B. seems possible

C. impossible for same reason as A

D. impossible, p is already assigned to G

E. same as A and C

So without really proving it, we know B must be true

21. This would be hard, except t is placed so we can diagram this out. p and t are in the middle. n must be third and s first because that's the last place Gillom school can go. The answer is four. C.

22. This is actually a pretty complicated one. I got it before but the second time it threw me. That's because I forgot about the pm notblock rule.

Here it goes: if m is < Hilltop then m is first or second. I just diagrammed both out. If she's second S and P go first then N and O go last. There is another scenario but let's just run through it real quick since it's a could be true question and thus maybe this template has the answer. and it does - E

23. Go back through old templates to find who was S's teammates.

A - eliminate by hypo for 19

B

C - eliminate by hypo for 19

D - eliminate by q18 grab a rule

E

just draw out the last one to see that S and T is possible.

B is right

Return to “LSAT Prep and Discussion Forum”

### Who is online

Users browsing this forum: AJordan, Baidu [Spider], JazzyMac, klaudiaxo, Majestic-12 [Bot], QT1994, Slippin' Jimmy, Yahoo [Bot] and 8 guests