When 100 people who have not used cocaine are tested for : GMAT Critical Reasoning (CR)
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# When 100 people who have not used cocaine are tested for

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When 100 people who have not used cocaine are tested for [#permalink]

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05 Feb 2008, 04:22
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When 100 people who have not used cocaine are tested for cocaine use, on average only 5 will test positive. By contrast, of every 100 people who have used cocaine 99 will test positive. Thus, when a randomly chosen group of people is tested for cocaine use, the vast majority of those who test positive will be people who have used cocaine.

A reasoning error in the argument is that the argument

(A) attempts to infer a value judgment from purely factual premises
(B) attributes to every member of the population the properties of the average member of the population
(C) fails to take into account what proportion of the population have used cocaine
(D) ignores the fact that some cocaine users do not test positive
(E) advocates testing people for cocaine use when there is no reason to suspect that they have used cocaine
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Neelabh Mahesh

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05 Feb 2008, 05:00
B

When 100 people who have not used cocaine are tested for cocaine use, on average only 5 will test positive. By contrast, of every 100 people who have used cocaine 99 will test positive. Thus, when a randomly chosen group of people is tested for cocaine use, the vast majority of those who test positive will be people who have used cocaine.

A reasoning error in the argument is that the argument

(A) attempts to infer a value judgment from purely factual premises - too common
(B) attributes to every member of the population the properties of the average member of the population - the best. positive test = usage of cocaine.
(C) fails to take into account what proportion of the population have used cocaine - irrelevant
(D) ignores the fact that some cocaine users do not test positive - vice-versa. ignores the fact that some test positive users have not used cocaine
(E) advocates testing people for cocaine use when there is no reason to suspect that they have used cocaine - irrelevant
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05 Feb 2008, 05:11
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Walker. The answer is not B
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05 Feb 2008, 05:35
It is C
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05 Feb 2008, 05:43
walker wrote:
It is C

Walker, Even not C.
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05 Feb 2008, 05:46
neelabhmahesh wrote:
Walker, Even not C.

I give up! I definitely need a cup of coffee.
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05 Feb 2008, 05:52
But I used "search" and found that OA is C

for example, only 1% of all population have used cocaine.

therefore, among 100 people we have 5 test positive persons but who have not used cocaine and ~1 who have used.
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05 Feb 2008, 05:53
walker wrote:
neelabhmahesh wrote:
Walker, Even not C.

I give up! I definitely need a cup of coffee.

Pls think over your cup of coffee why E is OA . The answer puzzles me.
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05 Feb 2008, 05:54
neelabhmahesh wrote:
walker wrote:
neelabhmahesh wrote:
Walker, Even not C.

I give up! I definitely need a cup of coffee.

Let me check again.
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05 Feb 2008, 05:56
The answer in my stuff says E is the answer. If it wrong pls tell me.
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05 Feb 2008, 06:15
Test 3, Section 1, #23 in 1000CR: OA is C
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06 Feb 2008, 04:09
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Let us consider following give samples:
Sample A: Not Used Cocaine Users: Number of people reported positive: 5% (on average)
Sample B: Cocaine User: Number of people reported positive: 99%
As per argument, if you chosen randomly, the vast majority of those who test positive will be people who have used cocaine. Definitely the author would have taken Cocaine when he concluded this (Sorry for being rude - ). Nevertheless, author faultily reasoning all the samples will be from “Sample B”. That is, he didn’t take complete count of the two groups.

(A) attempts to infer a value judgment from purely factual premises (If so, author reasoning would have been a correct one – eliminate it)

(B) attributes to every member of the population the properties of the average member of the population (Enticing! Yes. This could be a potential winner (out of 200, 104 are Cocaine Users. ) However, the members were chosen randomly – Oops – Eliminate it).

(C) fails to take into account what proportion of the population have used cocaine(This is a good candidate too. But is this valid under random selection? Consider, the probability of picking a Cocaine User is 99 / 100 and picking non-Cocaine User is as Cocaine User = 5/100.So, no-way that we end up have same vast majority of Cocaine Users.
Let twist the argument by saying 99% of non-Cocaine users will be tested positive. Then both probabilities will match. In short, if we have proper proportions, then we can accurately find the numbers that tested positive. – Hold it
)

(D) ignores the fact that some cocaine users do not test positive (may be, but does not provide vast majority – eliminate it)

(E) advocates testing people for cocaine use when there is no reason to suspect that they have used cocaine (correct – but won’t address logical error – eliminate it)

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Re: When 100 people who have not used cocaine are tested for [#permalink]

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30 Oct 2013, 21:11
This is my understand ding of the passage. Say there are 100 people if u choose say 50 people from them randomly. there will two groups 1) who test positive 2) who don't test positive
the argument says that of those who test positive (group 1), majority are cocaine users.
what you guys are saying is that argument says that majority of the people randomly picked from the crowd are cocaine users. In that case D stands true but not otherwise is what I feel. Please correct me if I am wrong.
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Re: When 100 people who have not used cocaine are tested for [#permalink]

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23 Nov 2015, 23:02
neelabhmahesh wrote:
The answer in my stuff says E is the answer. If it wrong pls tell me.

The answer is C, not E.
In the argument, we make a conclusion about the majority who might be taking cocaine. E talks about the advocating the use of the tests. So, E is irrelevant.
Re: When 100 people who have not used cocaine are tested for   [#permalink] 23 Nov 2015, 23:02
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