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26 Oct 2022, 21:06
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
34% (02:31) correct
66%
(02:34)
wrong
based on 338
sessions
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Every brick house on River Street has a front yard. Most of the houses on River Street that have front yards also have two stories. So most of the brick houses on River Street have two stories.
Which one of the following is most appropriate as an analogy demonstrating that the reasoning in the argument above is flawed?
(A) By that line of reasoning, we could concluden that most politicians have run for office, since all legislators are politicians and most legislators have run for office.
(B) By that line of reasoning, we could conclude that most public servants are legislators, since most legislators have run for office and most politicians who have run for office are public servants.
(C) By that line of reasoning, we could conclude that not every public servant has run for office, since every legislator is a public servant but some public servants are not legislators.
(D) By that line of reasoning, we could conclude that most legislators have never run for office, since most public servants have never run for office and all legislators are public servants.
(E) By that line of reasoning, we could conclude that most legislators are not public servants, since most public servants have not run for
office and most legislators have run for office.
Which one of the following is most appropriate as an analogy demonstrating that the reasoning in the argument above is flawed?
(A) By that line of reasoning, we could concluden that most politicians have run for office, since all legislators are politicians and most legislators have run for office.
(B) By that line of reasoning, we could conclude that most public servants are legislators, since most legislators have run for office and most politicians who have run for office are public servants.
(C) By that line of reasoning, we could conclude that not every public servant has run for office, since every legislator is a public servant but some public servants are not legislators.
(D) By that line of reasoning, we could conclude that most legislators have never run for office, since most public servants have never run for office and all legislators are public servants.
(E) By that line of reasoning, we could conclude that most legislators are not public servants, since most public servants have not run for
office and most legislators have run for office.
30 Oct 2022, 00:26
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Every brick house on River Street has a front yard. Most of the houses on River Street that have front yards also have two stories. So most of the brick houses on River Street have two stories.
Since it is an analogy, we have to break down the main reasoning and the options in certain form. Even Venn diagram would help.
Every brick house on River Street (A) has a front yard (condition for B).
Most of the houses on River Street that have front yards (B) also have two stories (C).
So most of the brick houses on River Street (most of A) have two stories (C).
The breakdown is
Every A has B. Most houses having B have C.
Conclusion: Most of A have C.
Flaw:- Every A may be part of those some houses within B that do not have C OR every A may have C.
Which one of the following is most appropriate as an analogy demonstrating that the reasoning in the argument above is flawed?
Conclusion: Most B run for C
Premise and conclusion have interchanged when compared to original argument
Conclusion: Not every B has C.
Not connected as no C in the premise.
Conclusion: Most A have ‘no C’.
Exactly as per our original argument.
CORRECT
D
Since it is an analogy, we have to break down the main reasoning and the options in certain form. Even Venn diagram would help.
Every brick house on River Street (A) has a front yard (condition for B).
Most of the houses on River Street that have front yards (B) also have two stories (C).
So most of the brick houses on River Street (most of A) have two stories (C).
The breakdown is
Every A has B. Most houses having B have C.
Conclusion: Most of A have C.
Flaw:- Every A may be part of those some houses within B that do not have C OR every A may have C.
Which one of the following is most appropriate as an analogy demonstrating that the reasoning in the argument above is flawed?
Quote:
A are B. Most A run for C.
Conclusion: Most B run for C
Premise and conclusion have interchanged when compared to original argument
Quote:
Can be discarded immediately as it does not connect A and C, and we have variables A, B, C and D in our argument.
Quote:
All A are B. Some B are not A.
Conclusion: Not every B has C.
Not connected as no C in the premise.
Quote:
A are B. Most B have ‘no C’.
Conclusion: Most A have ‘no C’.
Exactly as per our original argument.
CORRECT
Quote:
Not connected.
D
21 Dec 2022, 07:22
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This one was hard ..... hope not too many of this on gmat or will be working in Mcdonalds kitchen.
