June 2017 LSAT
Section 4
Question 8

# If Otero performs earlier than Miller, then the fifth performer could be any of the following EXCEPT:

Reply

Ravi on January 5, 2019

@jack515,Happy to diagram this for you.

The first rule is that L must go before N

1) L-N

The second rule is that M must go before T

2) M-T

The third rule states that there must be exactly one performance

between the performances of L and O, whether or not L performs earlier

than O

3) L_O or O_L

The fourth rule states that there must be exactly one performance

between the performances of M and P, whether or not M performs earlier

than P

4) M_P or P_M

The fifth rule states that P must perform either first or seventh

5) P 1/7

The fifth rule gives us a great opportunity to split this logic game

into two hypotheticals/sub game boards. One has P in 1, and the other

has P in 7.

P_ _ _ _ _ _

_ _ _ _ _ _ P

Let's apply our other rules to these hypotheticals to draw more

inferences. The next rule we should look at is rule four since it

deals with P and we've already put P on our boards.

P_ M_ _ _ _

_ _ _ _ M _ P

Now, let's look at our second rule, since it deals with M and we've

now put M on our boards

P_ M _ _ _ _ (T must go after M here)

_ _ _ _ M T P

At this point, the other rules deal with L, O, and N, and we haven't

put them on our boards yet, so it's time for us to go to the

questions.

Question 7 asks us to identify which answer could be true.

A is out, as we know that M is either third or fifth.

B is out since we know that L has to go before N.

C is out because O can only go in 5 in our first board. If O goes in 5

in this board, then L must go in 7 since M is in 3. If L is in 7, then

N can't follow it, so this answer is wrong.

D is correct, as there are no rules that govern where S goes. We can

place it in 7 just fine. If we do, we can still put L and O into 2 and

4, respectively, on our first game board. We can then put T and N in 5

or 6 (they can switch)

P L M O T N S

P L M O N T S

E is out; T must go after M in the first board, so it has to be in 4

or lower. In the second board, we've already placed T in 6.

Question 8 says if O is before M, then the fifth performer could be

any of the following except...

In using our two boards, if we put O before M in our second game

board, then M is in the fifth spot since it's already been placed. We

can eliminate answer B. If we put O before M in the first game board,

then L must go into four because of rule 3. This means that L cannot

go into the fifth spot, so answer A is our correct choice.

Question 9 says which of the following cannot be the third performer...

We know M can be in the third spot from our first game board, so we

can eliminate B. Using our second game board, if we put L and O into

the first and third spots we can put N and S into 2 and 4, and this

works. We can also put O and L into the first and third spots and put

S and N into 2 and 4, and this works, too. We can eliminate A and D as

a result.

If we put N into our second game board in the third spot, we can put L

into 2, O into 4, and S into one with no problems. This eliminates

answer C. Lastly, if we put S into 3 in the second board, then L and O

must go into 2 and 4 since those are the only available spots that

have one space between them to satisfy rule 3. This means N has to go

in 1, but this violates rule 1, which states L must go before N. Our

answer is E.

Question 10 gives us an additional premise (that S is immediately

before T) and asks us to identify which of the answer choices cannot

be true.

ST

We know from this that we can only use our first board since T is

sandwiched between M and P in the second board. Looking at the board,

it first appears that ST can go in 4 and 5, 5 and 6, or 6 and 7.

However, let's try each out. If we put ST into 4 and 5, then there is

nowhere to put L and O since they need one space in between them. This

means that S can't go in 4 and T can't go in 5. Does either of these

appear as an answer choice? Yes, answer choice D says T in 5, and we

know that this cannot be true, so D is our answer.

Question 11 says that the order in which the musicians perform is

completely determined if which one of the following is true...

Let's start by running through and testing the answers.

For A, if we put L into 4 in the first game board (we can't in the

second because there's no space for N after it), If L is in 4 in the

first board, then we could put O into 2 and still have lots of options

for our other pieces. If M is in 5, that's our second game board, and

we know that our second game board contains many options and is not

completely determined, so B is out. If N is in 4, which we could do in

the second game board, then L and O could switch between 1 and 3, so

this does not completely determine all game pieces. We can get rid of

C. For D, if O is in 3 (which we can do in the second game board) then

L must go in 1, but then N and S can switch between 2 and 4, so this

doesn't fully determine the order for us. D is out.

Lastly for E, if S is in 1, then we must use our second game board. L

and O must go into 2 and 4 since those are the only remaining spaces

with one space in between them. L must go into 2 and not 4 since N has

to go after it and spots 5 through 7 are full. This means L is in 2, O

is in 4, and N must be in 3. This fully determines the order, so it's

our correct answer.

Does this help? Let us know if you have more questions!