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June 2017 LSAT
If Otero performs earlier than Miller, then the fifth performer could be any of the following EXCEPT:
on January 4, 2019
question 7-11 june 17
could you diagram this for me?
on January 5, 2019
Happy to diagram this for you.
The first rule is that L must go before N
The second rule is that M must go before 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
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
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
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
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
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!
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