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# In a 200 member association consisting of men and women

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Intern
Joined: 18 Feb 2010
Posts: 27
In a 200 member association consisting of men and women [#permalink]

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12 Dec 2010, 06:21
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In a 200 member association consisting of men and women, exactly 20% of men and exactly 25 % women are homeowners. What is the least number of members who are homeowners?

A. 49
B. 47
C. 45
D. 43
E. 41
Math Expert
Joined: 02 Sep 2009
Posts: 46284
Re: Least number of homeowners [#permalink]

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12 Dec 2010, 06:59
10
16
In a 200 member association consisting of men and women, exactly 20% of men and exactly 25 % women are homeowners. What is the least number of members who are homeowners?

A) 49
B) 47
c) 45
D) 43
E) 41

Thanks

Let the # of women be $$w$$, then # of men will be $$200-w$$. We want to minimize $$0.25w+0.2(200-w)$$ --> $$0.25w+0.2(200-w)=0.05w+40=\frac{w}{20}+40$$, so we should minimize $$w$$, but also we should make sure that $$\frac{w}{20}+40$$ remains an integer (as it represent # of people). Min value of $$w$$ for which w/20 is an integer is for $$w=20$$ --> $$\frac{w}{20}+40=1+40=41$$.

Or: as there are bigger percentage of homeowner women then we should minimize # of women, but we should ensure that $$\frac{1}{4}*w$$ and $$\frac{1}{5}*(200-w)$$ are integers. So $$w$$ should be min multiple of 4 for which $$200-w$$ is a multiple of 5 (basically w should be min positive multiple of 20), which turns out to be for $$w=20$$.

Hope it's clear.
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Intern
Joined: 18 Feb 2010
Posts: 27
Re: Least number of homeowners [#permalink]

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12 Dec 2010, 11:10
Great! Thanks for the quick reply. Did not think it would be so straightforward

You rock!
Intern
Joined: 30 Nov 2010
Posts: 25
Location: Boston
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Re: Least number of homeowners [#permalink]

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12 Dec 2010, 13:57
awesome explanation. thanks!
Senior Manager
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Re: Least number of homeowners [#permalink]

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13 Dec 2010, 07:34
Great explanation.
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Re: Least number of homeowners [#permalink]

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14 Dec 2010, 00:32
Nice Question and Great explaination by Bunuel .. Thanks a lot
Intern
Joined: 20 Jun 2011
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Re: Least number of homeowners [#permalink]

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27 Jun 2012, 07:50
Bunuel wrote:
In a 200 member association consisting of men and women, exactly 20% of men and exactly 25 % women are homeowners. What is the least number of members who are homeowners?

A) 49
B) 47
c) 45
D) 43
E) 41

Thanks

Let the # of women be $$w$$, then # of men will be $$200-w$$. We want to minimize $$0.25w+0.2(200-w)$$ --> $$0.25w+0.2(200-w)=0.05w+40=\frac{w}{20}+40$$, so we should minimize $$w$$, but also we should make sure that $$\frac{w}{20}+40$$ remains an integer (as it represent # of people). Min value of $$w$$ for which w/20 is an integer is for $$w=20$$ --> $$\frac{w}{20}+40=1+40=41$$.

Or: as there are bigger percentage of homeowner women then we should minimize # of women, but we should ensure that $$\frac{1}{4}*w$$ and $$\frac{1}{5}*(200-w)$$ are integers. So $$w$$ should be min multiple of 4 for which $$200-w$$ is a multiple of 5 (basically w should be min positive multiple of 20), which turns out to be for $$w=20$$.

Hope it's clear.

Could you explain this one; $$=0.05w+40=\frac{w}{20}+40$$ ? How did you get $$\frac{w}{20}$$ ?
Math Expert
Joined: 02 Sep 2009
Posts: 46284
Re: Least number of homeowners [#permalink]

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27 Jun 2012, 08:00
2
superpus07 wrote:
Bunuel wrote:
In a 200 member association consisting of men and women, exactly 20% of men and exactly 25 % women are homeowners. What is the least number of members who are homeowners?

A) 49
B) 47
c) 45
D) 43
E) 41

Thanks

Let the # of women be $$w$$, then # of men will be $$200-w$$. We want to minimize $$0.25w+0.2(200-w)$$ --> $$0.25w+0.2(200-w)=0.05w+40=\frac{w}{20}+40$$, so we should minimize $$w$$, but also we should make sure that $$\frac{w}{20}+40$$ remains an integer (as it represent # of people). Min value of $$w$$ for which w/20 is an integer is for $$w=20$$ --> $$\frac{w}{20}+40=1+40=41$$.

Or: as there are bigger percentage of homeowner women then we should minimize # of women, but we should ensure that $$\frac{1}{4}*w$$ and $$\frac{1}{5}*(200-w)$$ are integers. So $$w$$ should be min multiple of 4 for which $$200-w$$ is a multiple of 5 (basically w should be min positive multiple of 20), which turns out to be for $$w=20$$.

Hope it's clear.

Could you explain this one; $$=0.05w+40=\frac{w}{20}+40$$ ? How did you get $$\frac{w}{20}$$ ?

$$0.05w=\frac{5}{100}*w=\frac{1}{20}*w=\frac{w}{20}$$.

Hope it's clear.
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Re: In a 200 member association consisting of men and women [#permalink]

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05 Nov 2012, 13:20
1
I got E too (41)
X: number of men and 200-X: number of women
0.20*X +0.25*(200-X) = 50 - 0.05*X = 50 - (1/20)*X To minimize this, we should maximize X (i.e X = 180, since there must be some women and X should be a multiple of 20)
50 - (1/20)*180 = 41
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Re: In a 200 member association consisting of men and women [#permalink]

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09 Jun 2013, 05:42
Hi Bunuel

I have reached to the answer but the solution is little weird after I saw yours.
Please guide me whether we can go by this

Question
In a 200 member association consisting of men and women, exactly 20% of men and exactly 25 % women are homeowners. What is the least number of members who are homeowners?

A. 49
B. 47
C. 45
D. 43
E. 41

Solution simple

Out of 200 20% are male i.e 40 and 25% are female i.e 50 , so total homeowner is 90.

Now min number homeowner is 40 and max is 90 so question ask us to find least and 41 has least value among all option.

So ans is 41.

Rgds
Prasannajeet
Math Expert
Joined: 02 Sep 2009
Posts: 46284
Re: In a 200 member association consisting of men and women [#permalink]

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09 Jun 2013, 08:06
prasannajeet wrote:
Hi Bunuel

I have reached to the answer but the solution is little weird after I saw yours.
Please guide me whether we can go by this

Question
In a 200 member association consisting of men and women, exactly 20% of men and exactly 25 % women are homeowners. What is the least number of members who are homeowners?

A. 49
B. 47
C. 45
D. 43
E. 41

Solution simple

Out of 200 20% are male i.e 40 and 25% are female i.e 50 , so total homeowner is 90.

Now min number homeowner is 40 and max is 90 so question ask us to find least and 41 has least value among all option.

So ans is 41.

Rgds
Prasannajeet

We are told that exactly 20% of men and exactly 25 % women are homeowners, not that 20% of 200 are males (homeowners)...

If there are 40 males and 50 females who are the remaining 200-(90)=110 people?
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Re: In a 200 member association consisting of men and women [#permalink]

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09 Jun 2013, 19:32
6
1
1. The percentage of women who are homemakers is higher than the percentage of men who are homemakers. So the least number of homemakers would be, when the number of women is the lowest possible
2. The number of women should be a multiple of 4 and the number of men should be a multiple of 5
3. Given our objective in (1) and the constraint in (2), the minimum number of women is 20 and the corresponding number of men is 180
4. So 25% of 20 is 5 and 20% of 180 is 36. The least number of homemakers is 5+36= 41.

Therefore the answer is choice E.
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Intern
Joined: 28 Dec 2012
Posts: 9
Re: In a 200 member association consisting of men and women [#permalink]

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08 Aug 2013, 10:17
Total numbers = 200. Men = X, Women = Y.
We are told that exactly 20% of men are homeowner => X is a multiple of 5 => X=5a (a>=1). Similarly, Y = 4b (b>=1). Find min(a+b)

X + Y = 200 <=> 5a + 4b = 200. a and b are all integers.
When a = 36, b=5 => (a+b)--->min.

E.
Intern
Joined: 29 Aug 2012
Posts: 29
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Re: In a 200 member association consisting of men and women [#permalink]

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27 Sep 2013, 01:25
I didn't set up any algebra equations.

Basically, we know that we have 1/5 of the men and 1/4 of the women. So both, men and women should be a multiple of 20 (lcm of 5 and 4).

Now, in order to minimize, we want the maximum multiple for men (because is the least fraction) and the minimun (dintinct to zero) for women.

So we have 180 men and 20 women. Now the solution is clear. 180*(1/5)+20(1/4)=41
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Re: Least number of homeowners [#permalink]

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05 Feb 2014, 06:34
Bunuel wrote:
In a 200 member association consisting of men and women, exactly 20% of men and exactly 25 % women are homeowners. What is the least number of members who are homeowners?

A) 49
B) 47
c) 45
D) 43
E) 41

Thanks

Let the # of women be $$w$$, then # of men will be $$200-w$$. We want to minimize $$0.25w+0.2(200-w)$$ --> $$0.25w+0.2(200-w)=0.05w+40=\frac{w}{20}+40$$, so we should minimize $$w$$, but also we should make sure that $$\frac{w}{20}+40$$ remains an integer (as it represent # of people). Min value of $$w$$ for which w/20 is an integer is for $$w=20$$ --> $$\frac{w}{20}+40=1+40=41$$.

Or: as there are bigger percentage of homeowner women then we should minimize # of women, but we should ensure that $$\frac{1}{4}*w$$ and $$\frac{1}{5}*(200-w)$$ are integers. So $$w$$ should be min multiple of 4 for which $$200-w$$ is a multiple of 5 (basically w should be min positive multiple of 20), which turns out to be for $$w=20$$.

Hope it's clear.

Hi Bunuel,

I solved it with a similar approach but came out with the option A. Can you please explain where am I wrong:

Let males be = m , then women = 200-m

0.2m +0.25(200-m)
we need to minimize m

so solving the equation gives us
50-m/20

For m to be an integer, the least value it can have is 20. So the ans: 50-1= 49.

Where am I wrong please guide.

Thanks.
Senior Manager
Joined: 15 Aug 2013
Posts: 272
Re: Least number of homeowners [#permalink]

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25 May 2014, 18:43
Bunuel wrote:
In a 200 member association consisting of men and women, exactly 20% of men and exactly 25 % women are homeowners. What is the least number of members who are homeowners?

A) 49
B) 47
c) 45
D) 43
E) 41

Thanks

Let the # of women be $$w$$, then # of men will be $$200-w$$. We want to minimize $$0.25w+0.2(200-w)$$ --> $$0.25w+0.2(200-w)=0.05w+40=\frac{w}{20}+40$$, so we should minimize $$w$$, but also we should make sure that $$\frac{w}{20}+40$$ remains an integer (as it represent # of people). Min value of $$w$$ for which w/20 is an integer is for $$w=20$$ --> $$\frac{w}{20}+40=1+40=41$$.

Or: as there are bigger percentage of homeowner women then we should minimize # of women, but we should ensure that $$\frac{1}{4}*w$$ and $$\frac{1}{5}*(200-w)$$ are integers. So $$w$$ should be min multiple of 4 for which $$200-w$$ is a multiple of 5 (basically w should be min positive multiple of 20), which turns out to be for $$w=20$$.

Hope it's clear.

Hi Bunuel, while solving this, I substituted for males instead of female and my equation looked like this:

.2m+.25(200-m) === 50 - 0.05m

Now, since it's asking me to minimize this value, would I choose the GREATEST possible value for m? How would I go about choose a value? At first, i thought it would be 196 but that doesn't lend itself well to being an integer.

Am I using the right though process here?
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Re: In a 200 member association consisting of men and women [#permalink]

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22 Jun 2015, 00:11
good easy solution
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Joined: 08 Mar 2015
Posts: 6
Re: In a 200 member association consisting of men and women [#permalink]

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14 Oct 2015, 06:05
Bunuel wrote:
In a 200 member association consisting of men and women, exactly 20% of men and exactly 25 % women are homeowners. What is the least number of members who are homeowners?

A) 49
B) 47
c) 45
D) 43
E) 41

Thanks

brunel:

Let the # of women be $$w$$, then # of men will be $$200-w$$. We want to minimize $$0.25w+0.2(200-w)$$ --> $$0.25w+0.2(200-w)=0.05w+40=\frac{w}{20}+40$$, so we should minimize $$w$$, but also we should make sure that $$\frac{w}{20}+40$$ remains an integer (as it represent # of people). Min value of $$w$$ for which w/20 is an integer is for $$w=20$$ --> $$\frac{w}{20}+40=1+40=41$$.

Or: as there are bigger percentage of homeowner women then we should minimize # of women, but we should ensure that $$\frac{1}{4}*w$$ and $$\frac{1}{5}*(200-w)$$ are integers. So $$w$$ should be min multiple of 4 for which $$200-w$$ is a multiple of 5 (basically w should be min positive multiple of 20), which turns out to be for $$w=20$$.

Hope it's clear.

brunel: If we want to minimize w, why cant we just assume that w =0. Ie. The number of women is 0. The question does not state anywhere that the number of women CANNOT be 0.
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Posts: 12
Re: In a 200 member association consisting of men and women [#permalink]

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13 Apr 2016, 23:07
Bunuel wrote:
In a 200 member association consisting of men and women, exactly 20% of men and exactly 25 % women are homeowners. What is the least number of members who are homeowners?

A) 49
B) 47
c) 45
D) 43
E) 41

Thanks

i notice if i put w = 200 -m
i get
-m/20 + 50

then it would be completely different and il get min/max wrong way around?
Math Expert
Joined: 02 Aug 2009
Posts: 5928
In a 200 member association consisting of men and women [#permalink]

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13 Apr 2016, 23:39
2
sabxu1 wrote:
Bunuel wrote:
In a 200 member association consisting of men and women, exactly 20% of men and exactly 25 % women are homeowners. What is the least number of members who are homeowners?

A) 49
B) 47
c) 45
D) 43
E) 41

Thanks

i notice if i put w = 200 -m
i get
-m/20 + 50

then it would be completely different and il get min/max wrong way around?

Hi,
no you will still get the correct answer...
Two cases--
1) we maximize m, as the % of m is lower
2) If the answer is in w, we minimize w..

here you have taken your answer in m and you get it as $$50 - \frac{m}{20}$$..
so you have to maximize m..
m has to be a multiple of 20 to get an integer value for m/20..
m cannot be 200, as there are some w...

Next can be 200-20=180..
substitue m as 180, So number =$$50 - \frac{180}{20} = 50 - 9 = 41$$
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In a 200 member association consisting of men and women   [#permalink] 13 Apr 2016, 23:39

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