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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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27 Jul 2017, 07:17
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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.

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

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17 Dec 2017, 22:08
25
14
Bunuel Can you please share your number picking approach? would be really helpful.

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. Not sufficient.

(2) The number of members of the club this year is $$\frac{6}{5}$$ the number of members last year. Not sufficient.

(1)+(2):
The number of the members last year = 100 (assume).
The number of males last year = 3/5*100 = 60 (from the stem).
The number of the members this year = 100*6/5 = 120 (from 2). So, 20 new members joined this year.
More than 10, out of those 20, are males (from 1).

The question asks whether the number of males this year is more than 3/5*120 = 72. So, basically whether more than 72 - 60 = 12 males joined this year.

We know that the number of males who joined this year was more than 10, so 11, 12, 13, ..., 20. If that number is 11 or 12, the answer would be NO but of that number is more than 12 the answer would be YES. Not sufficient.

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

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27 Jul 2017, 07:55
6
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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15 Aug 2017, 19:38
7
3
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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02 Aug 2017, 07:48
3
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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14 Aug 2017, 04:18
3
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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13 Dec 2017, 20:44
1
Erjan_S wrote:
Bunuel wrote:
Erjan_S wrote:
Honestly, i found this problem to be confusing because of 2) The number of members of the club this year is 3/5 the number of members last year.
Why? Because it contradicts this statement: "This year the members of the club include all the members from last year plus some new members".
How can it be? If all members of the last year are still in the club and we have an addition of new members, why the then the total number of members is less than the total number of last year? How it could be?

I would appreciate your comment chetan2u

(2) says: The number of members of the club this year is 6/5 the number of members last year, NOT 3/5.

That is what I have in my official quant online version of Wiley!

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

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15 Sep 2018, 12:18
1
3/5 = 60%. So, There are 60% males in the group.

The fraction or the percentage of the male this year will be greater only when there will be more than 60% male in the new group.

St 1: Says more than 50%. But we are not sure whether it is more than 60%. NS

St 2: Tell nothing about the new male member. NS

Combining we are not getting anything new. NS

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

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07 Aug 2017, 06:09
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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12 Aug 2017, 12:06
My 2 cents:
Attachment:

my 2 cents.png [ 6.37 MiB | Viewed 14380 times ]
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Re: Last year 3/5 of the members of a certain club were males.  [#permalink]

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15 Aug 2017, 21:16
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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Last year 3/5 of the members of a certain club were males.  [#permalink]

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13 Dec 2017, 19:52
Honestly, i found this problem to be confusing because of 2) The number of members of the club this year is 3/5 the number of members last year.
Why? Because it contradicts this statement: "This year the members of the club include all the members from last year plus some new members".
How can it be? If all members of the last year are still in the club and we have an addition of new members, why the then the total number of members is less than the total number of last year? How it could be?

I would appreciate your comment chetan2u
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Posts: 62455
Last year 3/5 of the members of a certain club were males.  [#permalink]

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13 Dec 2017, 20:16
Erjan_S wrote:
Honestly, i found this problem to be confusing because of 2) The number of members of the club this year is 3/5 the number of members last year.
Why? Because it contradicts this statement: "This year the members of the club include all the members from last year plus some new members".
How can it be? If all members of the last year are still in the club and we have an addition of new members, why the then the total number of members is less than the total number of last year? How it could be?

I would appreciate your comment chetan2u

(2) says: The number of members of the club this year is 6/5 the number of members last year, NOT 3/5.
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Re: Last year 3/5 of the members of a certain club were males.  [#permalink]

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13 Dec 2017, 20:41
Bunuel wrote:
Erjan_S wrote:
Honestly, i found this problem to be confusing because of 2) The number of members of the club this year is 3/5 the number of members last year.
Why? Because it contradicts this statement: "This year the members of the club include all the members from last year plus some new members".
How can it be? If all members of the last year are still in the club and we have an addition of new members, why the then the total number of members is less than the total number of last year? How it could be?

I would appreciate your comment chetan2u

(2) says: The number of members of the club this year is 6/5 the number of members last year, NOT 3/5.

That is what I have in my official quant online version of Wiley!
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File comment: OG Quant - efficient learning by Wiley

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

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17 Dec 2017, 21:05
Bunuel Can you please share your number picking approach? would be really helpful.
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Re: Last year 3/5 of the members of a certain club were males.  [#permalink]

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15 Sep 2018, 16:38
Last year, club has X members
3/5X are males

This year, club has Y new members
n is the proportion that are male
Is (3/5X + nY)/(X+Y) > 3/5 ?

(1) n > 1/2
n must be greater than 3/5 so this is not sufficient

(2) Y = 6/5 X - X = 1/5 X
Is (3/5X + n/5X)/(6/5X) > 3/5?
Is 3+n > 18/5?
Is n > 3/5?
Not sufficient

(1) and (2) n must be greater than 3/5 but we are only told n is greater than 1/2

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

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07 Apr 2019, 12:37
Working with the stem, we can come up with an equation: is 3+"new male"/ 5 + "new total" >3/5?
-> (3+m)/(5+t)>3/5? -> m/t>3/5? So, if m/t <=3/5 gives "no", m/t>3/5 gives "yes"

Then, we could do a bit of guessing should it come to this on the test day.
St. 2 looks easier to eliminate - we know nothing about the total and hence cannot judge about the ratio. SO B, D are gone.
We will be choosing between A, C or E.
St. 1 tells us that more than half are male, so it could still be 3/5, and thus will give us "yes" or "no". A is gone.
Between C or E, E should be the answer as there is not enough information to find the answer.

OR we can use smart numbers.
Lets say we have 5 people in total, with 3 men. We need to understand if 3 men + new men is greater that 5 + newcomers.

St. 1 tells that more than half are men. If 3 men out of 5 come, we'll keep the ratio. If 4 out of 5 come, it will exceed the ratio.
St. 2 tells that more men came than before. If 4 men and 0 women came, we will have a bigger ratio, if 4 men and 100 women come, we will have a smaller ratio.
E it is
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Re: Last year 3/5 of the members of a certain club were males.  [#permalink]

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24 Jun 2019, 19:34
Hi bb, I tried searching this question in the PS forum by putting all the necessary filters but this question didn't show up there. And I have faced similar issue in all the searches. I can't post link in this message otherwise I would have shared the link where I am trying this. Can you please help why that search is not working, will help save me a lot of time in filtering!!!!
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Re: Last year 3/5 of the members of a certain club were males.  [#permalink]

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30 Nov 2019, 22:56
The fraction of male members this year would be the weighted average of old Male ratio and the Male ratio of new member,

Statement 1) the male ratio in new members > 1/2,

But 1) if the male ratio in new members is more than 3/5, the new ratio would be more than 3/5.
2) if the male ratio in new members is less than 3/5, the new ratio would be less than 3/5.
3)if the male ratio in new members is equal to 3/5, the new ratio would be equal to 3/5.

So, Not Sufficient

Statement2: The number of members of the club this year is $$\frac{6}{5}$$ the number of members last year.
Number of new total members/ new members is inconclusive without the Male ratio in new members.
NOT Sufficient.

Now, after combining St1 & 2, we cant be sure of the new ratio, as we dont know the male ratio in new members as explained above.
Hence, the answer is E

if where
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.

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Re: Last year 3/5 of the members of a certain club were males.   [#permalink] 30 Nov 2019, 22:56
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