## LSAT PrepTest #9

Prepare for the LSAT or discuss it with others in this forum.
Rheastoria

Posts: 55
Joined: Thu May 26, 2011 1:26 pm

### LSAT PrepTest #9

Hey all,

I'm in the process of working on Logic Games, and I started the LG section (3rd section) of PrepTest #9 (October 1993). I have the older, green version of 10 Actual, Official LSAT PrepTests, and I was wondering if there is a typo on the answer key for question 13 of this section: the answer I arrived at was A (J and L) and the answer key says E (M and Q). I may just be overlooking something, but I am almost positive that my answer is correct. Any help would be appreciated!

I did a search and couldn't find any discussion of this, so I apologize if this has already been discussed.

Thanks!

tmon

Posts: 1236
Joined: Sun Apr 17, 2011 10:52 pm

### Re: LSAT PrepTest #9

Rheastoria

Posts: 55
Joined: Thu May 26, 2011 1:26 pm

### Re: LSAT PrepTest #9

Hope that helps

The only image that doesn't work is 1-10, and I need #9. Just my luck, haha. Thank you so much anyway!

Bobeo

Posts: 108
Joined: Tue Apr 05, 2011 7:51 pm

### Re: LSAT PrepTest #9

The answer is most definitely E.

If M and Q are selected, then you know K is selected. The only other option is the N/P choice. N can only be selected if L is, which it is not. Therefore, P is selected. MQKP is the only option.

If J and L is selected, either N or P could be selected making the answer choice wrong.

Rheastoria

Posts: 55
Joined: Thu May 26, 2011 1:26 pm

### Re: LSAT PrepTest #9

Bobeo wrote:The answer is most definitely E.

If M and Q are selected, then you know K is selected. The only other option is the N/P choice. N can only be selected if L is, which it is not. Therefore, P is selected. MQKP is the only option.

If J and L is selected, either N or P could be selected making the answer choice wrong.

Okay, I see where I made my initial mistake (I reversed the dependency of N and L) so I do see that A is wrong.

I do have a small logical problem though - if N cannot be selected unless L is selected, doesn't this imply that L can be selected independently of N? So if N is selected over P, then it must also include L, but can you choose P and L since L does not need N to be chosen?

Sorry if I'm being anal, but this is just one of two questions I answered incorrectly, so I want to completely understand this!

Bobeo

Posts: 108
Joined: Tue Apr 05, 2011 7:51 pm

### Re: LSAT PrepTest #9

Your intital setup should look something like this

(J/K) --- (N/P) --- (2 of MLQ)

We know that M and Q are selected, per the AC. So that leaves L not selected. Q requires K, which means K is selected and J is not. So the only choice is the N/P spot. If N is chosen, then L needs to be, which it is not. Therefore, P must be chosen. This is why there is only one possible solution when M and Q are selected.

You are correct that L can be chosen with or without N, but if we choose M and Q (per the AC), then there is no room for L.

Rheastoria

Posts: 55
Joined: Thu May 26, 2011 1:26 pm

### Re: LSAT PrepTest #9

Bobeo wrote:Your intital setup should look something like this

(J/K) --- (N/P) --- (2 of MLQ)

We know that M and Q are selected, per the AC. So that leaves L not selected. Q requires K, which means K is selected and J is not. So the only choice is the N/P spot. If N is chosen, then L needs to be, which it is not. Therefore, P must be chosen. This is why there is only one possible solution when M and Q are selected.

You are correct that L can be chosen with or without N, but if we choose M and Q (per the AC), then there is no room for L.

I see now - thank you so much for explaining it for me!