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Did more than half of the men on the committee wear blue?

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Did more than half of the men on the committee wear blue?  [#permalink]

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New post 20 Dec 2016, 01:23
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A
B
C
D
E

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  25% (medium)

Question Stats:

80% (00:49) correct 20% (00:35) wrong based on 97 sessions

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Re: Did more than half of the men on the committee wear blue?  [#permalink]

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New post 20 Dec 2016, 11:13
Bunuel wrote:
Did more than half of the men on the committee wear blue?

(1) Fifty percent of the committee members are male.
(2) Fifty percent of the committee members wore blue



suppose there are 100 members(50 male+50 female) on the committee

Q:- is men wearing blue>25

(1) no info about dress they wear....not suff..
(2) 50 members wearing blue...it may be all females wearing blue...NO
or all men wearing blue...YES..........
Not suff...

Combined
may be 26 men and 24females wearing blue...YES..
OR 10 men wearing blue. & 40 females wear blue..NO
not suff

Ans E
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Re: Did more than half of the men on the committee wear blue?  [#permalink]

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New post 21 Dec 2016, 04:06
1
Hey everyone,

PFB the detailed solution :)


Steps 1 & 2: Understand Question and Draw Inferences

    • If M is the number of Men in the committee, then is Mwearing blue > M/2 ?


Step 3: Analyze Statement 1 independently

    • It is given that fifty percent of the committee members are male
    • If the total members in the committee is taken as C then
      o \(M = 50\)% of \(C\)
      o \(M = \frac{C}{2}\)
    • But we don’t have any information on the number of men wearing blue(Mwearing blue).

So statement 1 is not sufficient to arrive at a unique answer.

Step 4: Analyze Statement 2 independently

    • It is given that fifty percent of the committee members wore blue
    • If the total members in the committee is C then
      o \((M+W)\)wearing blue = \(50\)% of \(C\)
      o \((M+W)\)wearing blue \(= \frac{C}{2}\)
    • We don’t know out of these \(\frac{C}{2}\) people how many are men and how many are women.
    • Also, we don’t have any information about the total number of men in the committee.

Therefore, statement 2 is not sufficient to arrive at a unique answer.

Step 5: Analyze Statement 1 and Statement 2 together


    • From the first statement we know that the total number of M in the committee is

      o \(M = \frac{C}{2}\) ……..(i)
    • From the second statement, we know that
      o \((M+W)\)wearing blue \(= \frac{C}{2}\) ……….(ii)
      o Combining equation (i) and (ii) we can write
      o \((M+W)\)wearing blue \(= M\)
      o whic means \(M\)wearing blue = \(M - W\)wearing blue
      o Since we don’t have the value for Wwearing blue, we cannot find the number of men wearing blue.

Hence, even after combining both statement 1 and statement 2 we cannot answer the question.

Therefore, the correct Answer is E


Thanks,
Saquib
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Re: Did more than half of the men on the committee wear blue?  [#permalink]

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New post 27 Dec 2016, 11:00
Bunuel wrote:
Did more than half of the men on the committee wear blue?

(1) Fifty percent of the committee members are male.
(2) Fifty percent of the committee members wore blue


1 st statement is clearly Insuff...
2nd statement says 50% wear blue, hence "MORE" than 50% are definitely not wearing blue, even if all 50% are assumed to be men...hence answer to the question in No...SUFFICIENT.

Hence-B 2nd statement alone.
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Re: Did more than half of the men on the committee wear blue?  [#permalink]

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New post 06 Feb 2017, 06:26
If we combine both statements, we have two extreme cases, either all the men wore blue or none of them wore blue. In both cases, the number of men wearing blue is equal to or below 50%. Hence, the answer should be C.
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Re: Did more than half of the men on the committee wear blue?  [#permalink]

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New post 06 Feb 2017, 21:44
SaumitD wrote:
If we combine both statements, we have two extreme cases, either all the men wore blue or none of them wore blue. In both cases, the number of men wearing blue is equal to or below 50%. Hence, the answer should be C.


Hi, I can feel where u r heading to...a nice thought. However, what u may b trying to say can b deduced from B alone....

Thanx
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Re: Did more than half of the men on the committee wear blue?  [#permalink]

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New post 05 Jul 2017, 10:36
saurabhsavant wrote:
Bunuel wrote:
Did more than half of the men on the committee wear blue?

(1) Fifty percent of the committee members are male.
(2) Fifty percent of the committee members wore blue


1 st statement is clearly Insuff...
2nd statement says 50% wear blue, hence "MORE" than 50% are definitely not wearing blue, even if all 50% are assumed to be men...hence answer to the question in No...SUFFICIENT.

Hence-B 2nd statement alone.


Hi,

Your reasoning is on the right track. However, you are missing the point that 50% of the committee members and 50% of the male members on the committee are two different groups.

Second statement says that 50% of the committee members wore blue. That means if there are 100 members, 50 of them wore blue. But if there are 40 male members in the committee and 30 of them wore blue then more than 50% of the male members wore blue but for the committee still only 50% are wearing blue.

Hence the correct answer here would be E.

Hope all of the above makes sense.
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Re: Did more than half of the men on the committee wear blue?  [#permalink]

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New post 05 Jul 2017, 10:42
Asoka wrote:
If we combine both statements, we have two extreme cases, either all the men wore blue or none of them wore blue. In both cases, the number of men wearing blue is equal to or below 50%. Hence, the answer should be C.


Hi,

I do not think you are applying the right course of reasoning. There are no extremes here. Sure, the group of men is a subset of members of the committee. However, even if all the men on the committee are wearing blue, it could be less than 50% of the committee which is wearing blue.

For ex, if there are 30 men and all of them are wearing blue, and there are 100 members on the committee. Then, 100% of the male members are wearing blue, whereas not even 50% of the committee members are wearing blue. (Assuming that only the male members are wearing blue for the sake of this example)
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Re: Did more than half of the men on the committee wear blue?  [#permalink]

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New post 05 Jul 2017, 11:09
Bunuel wrote:
Did more than half of the men on the committee wear blue?

(1) Fifty percent of the committee members are male.
(2) Fifty percent of the committee members wore blue


Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.


Committee | Male | Female
-------------------------------
....Blue.....|.. \(a\)...|... \(b\)..
-------------------------------
....Other....|...\(c\)...|...\(d\)..
-------------------------------
\(a + b + c + d = 100\)

There are 4 variables and 1 equation. Thus the answer E is most likely.

1) \(a + c = 50\)
2) \(a + b = 50\)

The questions asks if \(\frac{a}{a+c} > \frac{1}{2}\).

When we consider both conditions 1) and 2), we can't identify the value of \(a\) from two equations \(a + c = 50\) and \(a + b = 50\).
Therefore, the answer is E.

For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80 % chance that E is the answer, while C has 15% chance and A, B or D has 5% chance. Since E is most likely to be the answer using 1) and 2) together according to DS definition. Obviously there may be cases where the answer is A, B, C or D.
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Re: Did more than half of the men on the committee wear blue?   [#permalink] 05 Jul 2017, 11:09
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