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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



Last edited by BTGmoderatorDC on Wed Jan 16, 2019 6:58 am; edited 2 times in total

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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
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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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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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
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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Consider the following equation:
7x + 5y = 70.

If x and y are nonnegative integers, the following solutions are possible:
x=10, y=0
x=5, y=7
x=0, y=14.

Notice the following:
The value of x changes in increments of 5 (the coefficient for y).
The value of y changes in increments of 7 (the coefficient for x).
This pattern will be exhibited by any fully reduced equation that has two variables constrained to nonnegative integers.

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
.
Since only 1/5 of men own homes -- versus 1/4 of women -- the number of homeowners will be minimized if we MAXIMIZE THE NUMBER OF MEN.

Let x = the number of male homeowners.
Since 1/5 of the men own homes, the total number of men must be 5 times the number of male homeowners = 5x.
Let y = the number of female homeowners.
Since 1/4 of the women own homes, the total number of women must be 4 times the number of female homeowners = 4y.
Since there are 200 members in total, we get:
5x + 4y = 200.

Nonnegative integral solutions for the equation in blue, in accordance with the pattern discussed above:
x=40, y=0
x=36, y=5
x=32, y=10
And so on.

Since x and y must both be positive, the option in red is not viable.
Implication:
To maximize the value of x and thus the total number of men, we must assign x and y the combination in green:
x=36 and y=5, with the result that the total number of homeowners = 36+5 = 41.

The correct answer is E.

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OP, are you sure this is an OG question? Which edition of the OG did you get this problem from? It's not in any edition I own. A google search yields only 1 post from gmatclub, and 3 posts on this forum - all posted by you in the last month.

The wording of this problem does not sound like an OG problem. Any OG problem asking us to minimize or maximize would stipulate "what is the least POSSIBLE number..." It is also missing "25% of women."

This unfortunately has the effect of casting doubt on the veracity of your sources.

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

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EdM in Mind, Brain, and Education
Harvard Graduate School of Education


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