June 2010 LSAT
Section 5
Question 18

# Based on the passage, it can be concluded that the author and Broyles-González hold essentially the same attitude toward

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Skylar on February 13, 2020

@rinavaleriano, happy to help!Let's work through a couple of examples to try to clarify these drills.

(#1) Argument Completion Drills example:

P: not B - most -X

P: X -> A

C: ?

We should start by finding any other statements that can be deduced from our given statements. For a most statement, we can reverse and change the "most" to "some." For a some statement, we can reverse. For an S->N statement, we can reverse and negate to find the contrapositive. This gives us the following:

P: not B - most - X

X - some - not B

P: X -> A

not A -> not X

C: ?

Now we should look for a conclusion that combines our two premises. The variable X is our only mutual variable, so we should look here. It's also important to note that X is in the Sufficient condition, so the arrow is pointing away from the quantifier. This means we can combine the two premises as follows:

not B - most - X -> A

We can then shorten this to:

not B - most - A

We can reverse and change to a some statement as follows:

A - some - not B

This gives us the answer of:

P: not B - most - X

X - some - not B

P: X -> A

not A -> not X

C: not B - most - A

A - some - not B

(#2) Missing Premise example:

P: not D - some - A

P: ?

C: not D - some - X

Again, we can start by deducing other logical statements from those we're given. Since we have two some statements, the only possibility is to reverse the statements. This gives us:

P: not D - some - A

A - some - not D

P: ?

C: not D - some - X

X - some - not D

Now let's try to find the missing logical part that would fill the gap between the premise and the conclusion we are given. We should notice that "not D - some - __" is consistent, which means we need to focus on the part that changes. In this case, A changes to X. We can make this happen with an S->N statement, but remember that A must be Sufficient condition so that the arrow points away from the quantifier.

This gives us: A -> X

We can reverse and negate to find that the contrapositive of this is: not X -> not A

Altogether, this provides us the following logic: not D - some - A -> X

We can simply this to: not D - some - X

This fills the logical gap and gives us an answer of:

P: not D - some - A

A - some - not D

P: A -> X

not X -> not A

C: not D - some - X

X - some - not D

Does that make sense? Please let us know if you have any other questions!