Q23 - Every brick house on River Street

[mshinners]

What does the Question Stem tell us?
The question stem is slightly unusual, but this is still a Match the Flaw question. We're given a flawed argument in the stimulus, and asked to find the answer choice that contains matching flawed reasoning.

Break down the Stimulus:
"We're given two premises and a conclusion:

P1: brick house —> front yard
P2: front yard —most—> two stories
------------------------------------------------------
C: brick house —most—> two stories

It's important to recognize that only the first premise is a conditional statement. The second premise and conclusion are quantified statements. They don't guarantee anything. We can't link conditional statements with quantified statements in the same way that we can link conditional statements.

Suppose you live in one of the brick houses. The first statement tells us that you have a front yard. But there could be other, non-brick houses that also have front yards. And those non-brick houses with front yards might be the only ones that have two stories. None of the brick houses need to be among those included in the second premise. This is why the conclusion is flawed."

Any prephrase?
To be a match, the correct answer has to contain the same type of reasoning and the same flaw. We can eliminate answers choices if either the logical structure or the flaw doesn't match.

Answer choice analysis:
A) "The conclusion in this answer choice could be a match, but the first premise is reversed, so this isn't an analogous argument:
L —> P
L —most—> R
-----------------------
P —most—> R"

B) This answer choice has two quantified ("most") premises, and no conditional premise, so it can't be a match.

C) The conclusion in this answer choice is a "some" statement (some public servants have not run for office), not a "most" statement, and so is the second premise. This should make us suspicious of answer choice (C), but there's another problem. In the original argument, the sufficient condition of the conditional premise (brick house) was included in the conclusion. The sufficient condition of the conditional premise in this answer choice (legislator) does not appear in the conclusion. It's not a match.

D) "Correct. This answer choice has the same structure, and therefore the same flaw:

legislator —> pub. servant
pub. servant —most—> ~ run for office
------------------------------------------------------
legislator —most—> ~ run for office"

E) Like answer choice (B), this answer choice has two quantified ("most") premises, and no conditional premise, so it can't be a match.

Takeaway/Pattern: Matching questions are often based on conditional logic. You can often eliminate several answer choices based on obvious mismatches. More difficult questions will present two answer choices with similar logical structures. You'll need to compare them carefully to the stimulus to determine which one is correct.

#officialexplanation

[toodusolu]

Hi! could someone please explain how D is flawed in the same way as the stimulus? I'm not seeing it...

thanks!

[ManhattanPrepLSAT1]

Sure. This one is testing conditional logic. If you're strong on this concept, this question becomes much easier.

Stimulus:

Brick House ---> Front Yard
Front Yard most 2 Stories
----------------------------
Brick House most 2 Stories

This is flawed reasoning in that no inference can be drawn from the two premises. A valid argument would have read:

Brick House most Front Yard
Front Yard ---> 2 Stories
--------------------------------
Brick House most 2 Stories

So we're trying to match the structural flaw committed in the stimulus and answer choice (D) reads:

Legislator ---> Public Servant
Public Servant most ~Run for Office
------------------------------------
Legislator most ~Run for Office

That's a perfect match!

Remember that the order in which the statements are presented is irrelevant. We simply care about the logical structure of the claims.

Does that answer your question?

[toodusolu]

yes it does, thanks!

[randitect]

I had trouble diagramming and finding the appropriate flaw on this one because of the relationship between first and second statements: every brick house on River St. has a front yard, and most houses on River St. that have front yards also have two stories.
Had the second sentence said "most BRICK houses on river street with front yards also have two stories" then I would not have had a problem. However, since it did not, I couldn't figure out how to diagram and I was searching the answer choices for a flaw that paralleled the above: one group (brick houses on R St.) with X quality (front yard) and larger group (houses on R st. in general with X quality) with Y quality (two stories).

Given the above difference between the two groups (brick houses vs. houses generally) I'd really appreciate help with diagramming, as well as with thinking the problem through. Thank you.

[alex.cheng.2012]

Could someone please elaborate why answer (A) is incorrect?

P: all L are P (L --> P)
P: most L run office (most L --> run)
C: most P run office (most P --> run)

If we translate that into A/B terms that would be:
A --> B
most A --> R
most B --> R

The original reasoning is
A --> B
most B --> R
most A --> R

So is (A) incorrect because the original reasoning uses most B as a premise for the conclusion most A, while in answer (A), it uses most A as a premise for most B?

Additionally, is the reasoning itself in answer (A) is incorrect?
P: all L are P (L --> P)
P: most L run office (most L --> run)
C: most P run office (most P --> run)

So if there are 100 L, then we know there are at least 100 P.
If 60 L run, then most L run.
So at least 60 P run.
But we don't know if 60 P running = most P run, because we don't know if all P are L.
There could be other Ps that aren't L, perhaps the total P is 200, and then 60 P out of 200 P would not be most.

Is that why the reasoning in answer (A) itself is incorrect?

[ManhattanPrepLSAT1]

The reasoning is answer choice (A) is not valid.

(A)
L --> P
L -m-> RO
------------
P -m-> RO

** the valid inference would have been P <-s-> RO

The error of reasoning in the stimulus however contains a more serious error: no inference at all can be made from the stated premises.

Stimulus
BH --> FY
FY -m-> 2S
-------------
BH -m-> 2S

Remember to connect "most" + "all" statements, the linking term (in the stimulus the linking term in the premises is FY) needs to sit in the sufficient condition of the "all" statement.

For example, the premises

A -m-> B
B ----> C

lead to the valid inference

A -m-> C

While the premises

A -m-> B
A ----> C

lead to the inference

B <-s-> C

Any combination that puts the linking term in the necessary condition of the "all" statement leads to no inference.

[ttunden]

Here is my process. its a little easier to understand and not convoluted like the above responses.


so basically it is a conditional reasoning flaw.

BH --> FY
Most FY--> 2 stories
__________

Most BH --> 2 stories


1.) make sure answer choice has similar conditional reasoning structure
2.) all + most premise and most conclusion
3.) conclusion sufficient indicator is same term as all statement in premise

I was mostly looking for item 1 and 3. I am not really that technical and you don't have to be when you're doing this time and you at the end of the section with a proctor giving you a 5 minute warning while dong this question.

A - Conclusion is different, it doesn't match with sufficient of premise (Legislators) and the premises are different as well. Just have to spend 10-12 seconds writing it out to see that it is wrong and doesn't match.

B- Most + Most premise. get rid. different CR flaw
C - CR structure is different. Conclusion sufficient doesn't match up with sufficient of the All statement. Not the same CR flaw as the stimulus. Get rid

D- left with this. if you spend 10 seconds diagramming this you will see it utilizes same CR flaw as the stimulus.

E- Different CR flaw. also has Most Most premise.

[benjamin.w.richmond]

I had trouble diagramming and finding the appropriate flaw on this one because of the relationship between first and second statements: every brick house on River St. has a front yard, and most houses on River St. that have front yards also have two stories.
Had the second sentence said "most BRICK houses on river street with front yards also have two stories" then I would not have had a problem. However, since it did not, I couldn't figure out how to diagram and I was searching the answer choices for a flaw that paralleled the above: one group (brick houses on R St.) with X quality (front yard) and larger group (houses on R st. in general with X quality) with Y quality (two stories).

Given the above difference between the two groups (brick houses vs. houses generally) I'd really appreciate help with diagramming, as well as with thinking the problem through. Thank you.

Hey, I completely agree that this question does not make logical sense because of the Brick house/front yard parsing:

Brick House -> Front Yard
Front Yard -most-> 2 stories
----------------------------------------
brick house -most-> 2 stories

There are 2 parts of the argument that look like errors
1. you cannot link a most statement after a conditional statement, and conclude a conditional chain
2. What if most of the houses on river street are not brick, but all have front yards? There is incomplete information about the set in question to make a conclusion.

This question turned into a big time sink for me b/c D only seemed to parallel flaw 1. Ultimately though I think 2 is unique to this type of argument and doesnt take away from 1 being the bigger error.

Thoughts?

[zen]

I usually do pretty well on parallel questions but this one tripped me up.

Arg: All Brickhouse ---> Front Yard
Most House ---> 2 Stories
--------------------------------------------
Conc: Most BH ---> 2 Stories


Answer Choices:

We can whittle the choices down to (A) and (D) fairly quickly.

(B)- Contains three "mosts"; our original argument contained an "all", "most", and had a "most" for its conclusion. Does not match.

(C)- Contains a "some" which our original did not. Does not Match

(E)- Contains three "mosts". Does not match.

(A) Has a different flaw than the original.. The original had an issue where we concluded something about brick houses based on info we had about most houses in general which is invalid reasoning. (A) has a separate issue where it does not switch terms in the sufficient conditions of the premises like the original, but instead, makes a mistaken reversal, putting information we had only received as a necessary assumption into the sufficient assumption of the conclusion when we are not given premises to make the link.

For more visual people:

All L---> P
Most L---> O
----------------------
Conc: Most P--->O

(D) Matches because it contains the same types of modifiers( most, all, etc.) and has the same type of flaw as the original-where the premises have different sufficient conditions and are mistakenly linked to conclude something about the "ALL" premise's necessary condition based upon a "Most" of that ALL premises necessary condition located in the sufficient condition of the MOST premise.
I bet that sucks to read. I'm sorry. Here's a visual approach:

All L--->P
Most P---> -O
-----------------------
Conc: Most L---> -O

Cheers.

(D)

[cacrv]

bumping this question because I also struggled with this aspect:

I had trouble diagramming and finding the appropriate flaw on this one because of the relationship between first and second statements: every brick house on River St. has a front yard, and most houses on River St. that have front yards also have two stories.
Had the second sentence said "most BRICK houses on river street with front yards also have two stories" then I would not have had a problem. However, since it did not, I couldn't figure out how to diagram and I was searching the answer choices for a flaw that paralleled the above: one group (brick houses on R St.) with X quality (front yard) and larger group (houses on R st. in general with X quality) with Y quality (two stories).

Given the above difference between the two groups (brick houses vs. houses generally) I'd really appreciate help with diagramming, as well as with thinking the problem through. Thank you.

how did those who got this correct know to link up the two conditionals and ignore the subset aspect of the conditionals?

[Laura Damone]

Full Explanation

What does the stem tell us: This is a unique one, but since an analogous argument is an argument that is a structural match, we know this is a Matching Family question. We’re told the original argument is flawed, so that makes this a Match Flaw question.

Breakdown the stimulus: We have an “every” statement and 2 “most” statements, so we’re looking at a quantified logic problem. And since we’re going to have to match that structure to the answers, diagramming is going to be the right call for most folks.

Brick House → Front Yard
Front Yard –most-> 2 Stories
----------------------------------------------
Brick House –most-> 2 Stories

Any prephrase? Since this is Match Flaw, we want to try to ID the flaw first, but we also can use other kinds of mismatches to make eliminations elsewhere. Since there are quantifiers, we should be on the lookout for quantifier mismatches. The Flaw that’s exhibited here can be hard to spot, because if those “most” statements were “all” statements, we’d be looking at a basic A→B→C transitive argument. But remember that when you combine quantified statements, the “all” statement’s sufficient condition has to be the shared term. In this argument, it’s the necessary condition “Front Yard” that is shared, which means nothing can be concluded from the combination of these statements. Structurally, we are looking for an argument that proceeds thusly:

A→ B
B-most->C
----------------------
A –most-> C

A) is the first example of a tricky move the test writers employed: putting the conclusion first. Remember, the order of statements in answer choice doesn’t impact an argument’s logical structure, so this is not a mismatch. This argument can be diagrammed:
L → P
L –m→ RO
-------------------
P –most-> RO

This argument is also flawed, but not in the same way as the original. From this argument we could rightly conclude that some politicians have run for office, because the shared term is the “all” statement’s sufficient condition, whereas our original argument allows us to conclude nothing.

B) has 3 “most” statements and no “all” statements. This quantifier mismatch allows us to eliminate without diagramming.

C) “not every … has” is equivalent to “some …have not,” so this too can be eliminated via a quantifier mismatch without diagramming.

D) has all the right quantifiers, so proceed with the diagram:
PS –most-> -RO
L → PS
------------------------
L –most-> -RO

It’s a match, even though the premises come in a different order!

E) has 3 “most” statements and can be eliminated without diagramming.

Takeaway/Pattern: Don’t be afraid to diagram Match questions because you think it will be too time consuming to do 6 diagrams: other types of mismatches, particularly quantifier mismatches, can eliminate wrong answers on a first pass, and diagramming the remaining choices gives you concrete proof that one is a match and the others are not.