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In a 200 member association consisting of men and women

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Joined: 26 Feb 2016
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Re: In a 200 member association consisting of men and women [#permalink]

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New post 13 Apr 2016, 23:50
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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In a 200 member association consisting of men and women [#permalink]

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New post 14 Apr 2016, 00:02
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 [#permalink]

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New post 12 Aug 2017, 08:02
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


There are men and women in 200 member association.
20% of men are homeowners and 25% of women are homeowners.

Let n be the number of men and 200-n be the number of women
So, 0.2 * n + .25 * (200-n) = n/5 + (200-n)/4 are homeowners
Now both the parameters must be an integer. Also we need to find the least number of homeowners so we need to maximize the number of men i.e. n as only 20% of men are homeowners.

So, Lets tart taking such values
First such value is (200,0) but number of women can't be 0
2nd such value is (180,20)

So, no. of homeowners = 180/5 +20/4 = 36 +5 = 41.

If we further find the values which satisfies the expression n/5 + (200-n)/4, we will find the value n/5 + (200-n)/4 increasing since the percentage of females who are homeowners > Percentage of male who are homeowners. And hence increase in no. of women will increase the number of homeowners.


Answer E
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Re: In a 200 member association consisting of men and women   [#permalink] 12 Aug 2017, 08:02

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