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Re: In a 200 member association consisting of men and women, exactly 20% [#permalink]
Great! Thanks for the quick reply. Did not think it would be so straightforward

You rock!
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Re: In a 200 member association consisting of men and women, exactly 20% [#permalink]
awesome explanation. thanks!
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Re: In a 200 member association consisting of men and women, exactly 20% [#permalink]
Great explanation.
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Re: In a 200 member association consisting of men and women, exactly 20% [#permalink]
Nice Question and Great explaination by Bunuel .. Thanks a lot
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Re: In a 200 member association consisting of men and women, exactly 20% [#permalink]
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Bunuel wrote:
nades09 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

Please explain

Thanks
NAD



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 \(w=20\) --> \(\frac{w}{20}+40=1+40=41\).

Answer: E.

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 \(\frac{w}{20}\) ?
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Re: In a 200 member association consisting of men and women, exactly 20% [#permalink]
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superpus07 wrote:
Bunuel wrote:
nades09 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

Please explain

Thanks
NAD


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 \(w=20\) --> \(\frac{w}{20}+40=1+40=41\).

Answer: E.

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, exactly 20% [#permalink]
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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
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Re: In a 200 member association consisting of men and women, exactly 20% [#permalink]
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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, exactly 20% [#permalink]
Bunuel wrote:
nades09 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

Please explain

Thanks
NAD


I have a question about this method
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?
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Re: In a 200 member association consisting of men and women, exactly 20% [#permalink]
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sabxu1 wrote:
Bunuel wrote:
nades09 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

Please explain

Thanks
NAD


I have a question about this method
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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Re: In a 200 member association consisting of men and women, exactly 20% [#permalink]
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 =\(50 - \frac{180}{20} = 50 - 9 = 41\)[/quote]

Thanks for replying to me I apologise for asking another question about it. WHy is it maximum m as % is lower? Sorry I'm kind of confused now is there a reason. I'm trying to understand this concept.
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Re: In a 200 member association consisting of men and women, exactly 20% [#permalink]
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sabxu1 wrote:
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 =\(50 - \frac{180}{20} = 50 - 9 = 41\)


Thanks for replying to me I apologise for asking another question about it. WHy is it maximum m as % is lower? Sorry I'm kind of confused now is there a reason. I'm trying to understand this concept.[/quote]

Hi,

the Q gives us few details which cannot be changed..


1) total = m + w= 200
2) exactly 20% of men are homeowners = \(\frac{20}{100}* m = \frac{m}{5}\)
3) exactly 25 % women are homeowners = \(\frac{25}{100}* w = \frac{w}{4}\)
4) homeowners = H =\(\frac{m}{5}+\frac{w}{4}\)

Q is " What is the least number of members who are homeowners?" or "least H"

Now H depends on m and w, as these two numbers can be played around with, that is no fixed value is given..
Rest there is nothing we can do with 200 , 20% and 25%...

If you look at the Equation - H =\(\frac{m}{5}+\frac{w}{4}\)
w/4 which means every 4th women is home owners and m/5 means every 5th men is a homeowner...
so for H to be least, we have to maximize people who are least likely to possess HOME and here it is 'm', so we try to maximize 'm'..
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Re: In a 200 member association consisting of men and women, exactly 20% [#permalink]
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nades09 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


Letting m = the number of men in the association and w = the number of women in the association, we know that:

m + w = 200

m = 200 - w

Thus:

0.2(200 - w) + 0.25w = homeowners

40 - 0.2w + 0.25w = homeowners

40 + 0.05w = homeowners

40 + w/20 = homeowners

We see that w must be a multiple of 20 and since we want the least number of homeowners, w = 20. So the least number of homeowners is 40 + 20/20 = 40 + 1 = 41.

Answer: E
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Re: In a 200 member association consisting of men and women, exactly 20% [#permalink]
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nades09 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


We can PLUG IN THE ANSWERS, which represent the least number of homeowners.
Since the correct answer must be the least possible option, start with the smallest answer choice.
To test the answers, use ALLIGATION.

E: 41, implying that the percentage who are homeowners \(= \frac{41}{200} = \frac{20.5}{100} = 20.5\)%

Step 1: Plot the 3 percentages on a number line, with the percentages for M and W on the ends and the percentage for the mixture in the middle.
M 20------------20.5------------25 W

Step 2: Calculate the distances between the values on the number line.
M 20----0.5----20.5----4.5----25 W

Step 3: Determine the ratio of men to women.
The ratio of M to W is equal to the RECIPROCAL of the distances in red.
\(\frac{M}{W} = \frac{4.5}{0.5} = \frac{9}{1} = \frac{90}{10} =\) \(\frac{180}{20}\)

The ratio in green indicates that the percentage of homeowners will be 20.5% -- yielding a total of 41 homeowners -- if M=180 and W=20, for a total of 200 people.

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Re: In a 200 member association consisting of men and women, exactly 20% [#permalink]
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nades09 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


In order to minimize the number of homeowners, we must MAXIMIZE the number of men in the group, since the proportion of male homeowners (20%) is less than the proportion of female homeowners (25%)
So, let's see what happens if there are 199 men and 1 woman.
Well, if 20% (aka 1/5) of the men are homeowners, then the number of male homeowners = 20% of 199 = 39.8. This makes no sense, since we can't have 39.8 men.
Likewise, if 25% (aka 1/4) of the women are homeowners, then the number of female homeowners = 25% of 1 = 0.25. This makes no sense either.

Let's now focus on the women. We know that, in order to have an INTEGER number of female homeowners, the number of females must be divisible by 4.
Likewise, in order to have an INTEGER number of male homeowners, the number of females must be divisible by 5.

So, the first pair of values that meet the above conditions are: 180 men and 20 women.
20% of 180 = 36, so there are 36 male homeowners.
25% of 20 = 5, so there are 5 female homeowners.
MINIMUM number of homeowners = 36 + 5 = 41

Answer: E

Cheers,
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Re: In a 200 member association consisting of men and women, exactly 20% [#permalink]
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nades09 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

Excellent opportunity for the grid (aka double-matrix)!

Important: we have chosen the blue expression wisely and the others follow from it. (Justify all of them!)



\(? = {\left( {50 - M} \right)_{\min }}\)


\(\left( {50 - M} \right)\,\,{\mathop{\rm int}} \,\,\,\, \Rightarrow \,\,\,\,\,M\,\,{\mathop{\rm int}}\)

\(\left( {10 - M} \right) \ge 1\,\,{\mathop{\rm int}} \,\,\,\, \Rightarrow \,\,\,M \le 9\,\,{\mathop{\rm int}}\)


\(? = {\left( {50 - M} \right)_{\min }}\,\,\, = 41\,\,\,\,\left( {M = 9} \right)\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: In a 200 member association consisting of men and women, exactly 20% [#permalink]
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Hi All,

We're told that in a 200 member association consisting of men and women, exactly 20% of men and exactly 25 % women are homeowners. We're asked for the LEAST number of members who are homeowners. This question is built around a couple of Number Properties - and to MINIMIZE the number of people who are homeowners, we have to MAXIMIZE the number of men in the group (since a smaller percentage of men are homeowners).

To start, since 20% of men are homeowners, we know that the number of men MUST be a multiple of 5. In that same way, since 25% of women are homeowners, we know that the number of women MUST be a multiple of 4. Thus, we need to add the largest possible multiple of 5 to a multiple of 4 and get a total of 200, while accounting for the fact that there MUST be some men and some women. Logically, we can 'work down' from 200 to find those numbers.

IF there were...
4 women, then there'd be 196 men (not valid; number of men needs to be a multiple of 5)
8 women, then there'd be 192 men (not valid; number of men needs to be a multiple of 5)
Etc.

With a little more work, you'll find that 20 women and 180 men is situation that is needed. From there the LEAST possible number of homeowners would be...
(.25)(20) + (.2)(180) =
5 + 36 =
41 homeowners

Final Answer:

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