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Last year 3/5 of the members of a certain club were males.

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Last year 3/5 of the members of a certain club were males. [#permalink]

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Last year \(\frac{3}{5}\) of the members of a certain club were males. This year the members of the club include all the members from last year plus some new members. Is the fraction of the members of the club who are males greater this year than last year?

(1) More than half of the new members are male.

(2) The number of members of the club this year is \(\frac{6}{5}\) the number of members last year.
[Reveal] Spoiler: OA

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Last year 3/5 of the members of a certain club were males. [#permalink]

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New post 27 Jul 2017, 08:55
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carcass wrote:
Last year \(\frac{3}{5}\) of the members of a certain club were males. This year the members of the club include all the members from last year plus some new members. Is the fraction of the members of the club who are males greater this year than last year?

(1) More than half of the new members are male.

(2) The number of members of the club this year is \(\frac{6}{5}\) the number of members last year.




Hi..
total =T
Male = M
so \(\frac{3}{5}*T=M\)
new member = x..
Ratio of M will be GREATER if the male in new group >\(\frac{3}{5}*x\) otherwise No

lets see the statements
(1) More than half of the new members are male.
as can be seen slightly more than the 3/5 of new members should be male..
here if new males are between \(\frac{1}{2}..&..\frac{3}{5}\) including ans is NO
if > \(\frac{3}{5}\), ans is YES
Insuff

(2) The number of members of the club this year is \(\frac{6}{5}\) the number of members last year.
Nothing about the number of males in new addition
Insuff

combined..
nothing new

E
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Re: Last year 3/5 of the members of a certain club were males. [#permalink]

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New post 02 Aug 2017, 08:48
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this is hard
suppose there are 50 member, 30 of them are male.

1. not sufficient
2. not sufficient

combined
total number is 60
new member is 10
more than half is men, so, there are more than 6 is men
if 6 are men
we have 30+6=36
fraction of men is 36/60=3/5
if 7 are men
we have 30+7=37 men

fraction is 37/60>3/5

so not sufficient.

E

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Re: Last year 3/5 of the members of a certain club were males. [#permalink]

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New post 07 Aug 2017, 07:09
this question can very well explained by considering it to be like adding 2 solutions, a solution with 3/5 male concentration and another one with 1/2 male concentration

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Re: Last year 3/5 of the members of a certain club were males. [#permalink]

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New post 12 Aug 2017, 13:06
My 2 cents:
Attachment:
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Re: Last year 3/5 of the members of a certain club were males. [#permalink]

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New post 14 Aug 2017, 05:18
carcass wrote:
Last year \(\frac{3}{5}\) of the members of a certain club were males. This year the members of the club include all the members from last year plus some new members. Is the fraction of the members of the club who are males greater this year than last year?


Last Year
Total Members -> T
Male Members -> \(\frac{3}{5}\)*T => In other words M : T = 3 : 5

This Year
Total Members -> T + N
New Male Members -> m

Is \(\frac{(M + m)}{(T+N)}\) > \(\frac{3}{5}\) ?

Quote:
(1) More than half of the new members are male.
(2) The number of members of the club this year is \(\frac{6}{5}\) the number of members last year.


1) m > 0.5N
N = 10, m = 6,7,8..
Ratio will change according to the number of males added
=> Let's say T = 10 => M = 6
and N = 10 => m = 6,7,8..
If m = 6
=> \(\frac{(M + m)}{(T+N)}\) = \(\frac{3}{5}\) (Ratio is the same)

If m = 7
=> \(\frac{(M + m)}{(T+N)}\) = \(\frac{7}{10}\) (Ratio is greater)

Hence, Insufficient.

2) T + N = 1.2T
=> N = 0.2T
We don't know the value of N or T. Plus, we cannot derive the values of M and m.
Insufficient.

1+2)
We gain no additional information, so we still don't know the value of N,T,M or m.
Insufficient.

E is the answer.
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Re: Last year 3/5 of the members of a certain club were males. [#permalink]

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New post 15 Aug 2017, 20:38
Say last year total number of members be = Old
Given, # of males = \(3Old/5\)

This year, new members are added, say \(New\) and let the number of males within the new group be \(x\).

Question stem asks, if \((x+3Old/5)/(New+Old) > \frac{3}{5}\)
Simplifying this, we get, if \(x > \frac{3}{5}New\) or \(x > 0.6 New\)

Stmt1: Gives us \(x>0.5 New\) but we are not sure if \(x\) is greater than \(0.6 New\). Hence this statement is insufficient.
Stmt2: Gives \(New = \frac{1}{5} Old\). This gives us nothing about \(x\) and \(New\) relationship. Hence it is insufficient as well.

Combining 1 and 2, we get nothing fresh about the relationship of x and new members 'New' and hence E is the answer.

Hope this helps.
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Re: Last year 3/5 of the members of a certain club were males. [#permalink]

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New post 15 Aug 2017, 22:16
The answer is E for sure

Explanation : let there be 100 members in the club at starting
So as given 3/5 of them are male
So 60 would be male

Now new members get in and no of males in the new members is more than 1/2

Now 2 bf statement says that now the no of members after addition of new members is 6/5 of the earlier
Earlier there were 100 members
Now 6/5 *100 = 120
So the no of new members who have joined is

120-100=20

Given more than half are male so no of males >10

It can be 11,12,13 ......20

So
Now if the no of males are 11 then this time males are less in fraction
If 12 men are there then fraction of males would remain the same
If the no is 13 or greater than 13 the fraction will we greater this time

So we don't the actual no of males in the new members

So the answers is E



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Re: Last year 3/5 of the members of a certain club were males.   [#permalink] 15 Aug 2017, 22:16
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