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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
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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scott@targettestprep.com



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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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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: E

GMAT assassins aren't born, they're made,
Rich

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Contact Rich at Rich.C@empowergmat.com

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