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In a 200 member association consisting of men and women, exa

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In a 200 member association consisting of men and women, exa

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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

OA E

Source: Official Guide

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BTGmoderatorDC 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

OA E

Source: Official Guide
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.
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,
Brent

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BTGmoderatorDC 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
Source: Official Guide
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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Easiest approach here is trial and error. Pick the lowest number (41) and put it into equation

(1) M + W = 200
(2) $$\frac{1}{5}M+\frac{1}{4}W\ =\ 41$$

Thus you get W = 20 and M = 160; which are valid numbers. Since this is the lowest answer...

Answer = 41

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