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In a 200 member association [#permalink]
18 May 2020, 06:31
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Question Stats:
66% (02:37) correct
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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 What is the best strategy to use here ?




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Re: In a 200 member association [#permalink]
18 May 2020, 21:37
It is given that there are 200 members and 20% men and 25% female are homeowners. To find the least number of members, we need to know that when the lesser the number, more lesser the resulting percent will be. Thus, if we take a large number for 20% and a small number for 25%, we can get the least amount possible. In this case, 180*0.2 = 36 and 20*0.25 = 5, thus 36+5=41, option E. Hence, always take a smaller number for larger percentage.



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Re: In a 200 member association [#permalink]
18 May 2020, 21:56
sukrut96 wrote: It is given that there are 200 members and 20% men and 25% female are homeowners. To find the least number of members, we need to know that when the lesser the number, more lesser the resulting percent will be. Thus, if we take a large number for 20% and a small number for 25%, we can get the least amount possible. In this case, 180*0.2 = 36 and 20*0.25 = 5, thus 36+5=41, option E. Hence, always take a smaller number for larger percentage. Is there any thinking behind dividing 200 into 180 and 20??



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Re: In a 200 member association [#permalink]
18 May 2020, 22:38
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vndnjn wrote: sukrut96 wrote: It is given that there are 200 members and 20% men and 25% female are homeowners. To find the least number of members, we need to know that when the lesser the number, more lesser the resulting percent will be. Thus, if we take a large number for 20% and a small number for 25%, we can get the least amount possible. In this case, 180*0.2 = 36 and 20*0.25 = 5, thus 36+5=41, option E. Hence, always take a smaller number for larger percentage. Is there any thinking behind dividing 200 into 180 and 20?? By using trial and error method, I reached this number. Started with 140, 60 and tried till 180 and 20. Also we need an integer value, so even integers will work well.



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Re: In a 200 member association [#permalink]
29 Jun 2020, 11:49
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The way I would do it is also trail and error, but not random trail and error. As sukrut96 points out we want the 25% group (women) to be as small as possible. As we know that at least one person in each group is a home owner because there are both men and women in the association and the percentages are positive. Starting with one woman who is a home owner > 3 women are not > there are 2004=196 men left in the group. We also know that 0.2x = x/5 > the number of remaining men should be divisable by 5, and 196 is not. Then: 2 female homeowners > 8 women in total > 192 not divisible by 5 3 > 12 > 188 4 > 16 > 184 5 > 20 > 180 180 is divisible by 5 so here we stop. 180/5 + 5 = 41




Re: In a 200 member association
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29 Jun 2020, 11:49





