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Re: In a 200 member association consisting of men and women
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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 20020=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
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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 20020=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
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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 200n be the number of women So, 0.2 * n + .25 * (200n) = n/5 + (200n)/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 + (200n)/4, we will find the value n/5 + (200n)/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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In a 200 member association consisting of men and women
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13 May 2018, 08:06
Bunuel wrote: 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
Please explain
Thanks NAD Let the # of women be \(w\), then # of men will be \(200w\). We want to minimize \(0.25w+0.2(200w)\) > \(0.25w+0.2(200w)=0.05w+40=\frac{w}{20}+40\), so we should minimize \(w\), but also we should make sure that \(\frac{w}{20}+40\) remains an integer (as it represent # of people). Min value of \(w\) for which w/20 is an integer is for \(w=20\) > \(\frac{w}{20}+40=1+40=41\). Answer: E. Or: as there are bigger percentage of homeowner women then we should minimize # of women, but we should ensure that \(\frac{1}{4}*w\) and \(\frac{1}{5}*(200w)\) are integers. So \(w\) should be min multiple of 4 for which \(200w\) is a multiple of 5 (basically w should be min positive multiple of 20), which turns out to be for \(w=20\). Hope it's clear. Hi Bunuel generis, pushpitkc hope you are having nice weekend please advise normally in MIN/MAX problems to find minimal value we need to MAXimize another value, right ? you say " We want to minimize \(0.25w+0.2(200w)\) " why are you saying this " we need to minimize" instead of this " In order to minimize the number of homeowners we need to maximize number of non homeowners" shouldnt we maximize number of non homeowners in order to minimize number of homeowners ? What do you minimize here and why are you subtracting number of women from 200 and not number of men ? Also it says "exactly 20% of men and exactly 25 % women are homeowners" this phrase does it mean tat there are total 25 % woman and all of them are homeowners ? what is the diference between exactly 20% of men and exactly 25 % women what if the question were "In a 200 member association consisting of men and women, exactly 20% of men and exactly 25 % of women are homeowners. What is the least number of members who are homeowners? what would be the equation ? And one more thing, let me copy and paste some MIN/MAX guidlines from this link, see below https://gmatclub.com/forum/gmattipof ... l#p2058670Minimizing the highest value by maximizing the others
Maximizing the highest value by minimizing the othersMinimizing the lowest value by maximizing the others
Maximizing the lowest value by minimizing the othersaren't the above mentioned rules the same ? I mean two rules marked in red are the same as two rules marked in green i would appreciate your explanation thank you UPDATE:) i think today i figured out answer to one of my questions:) human brain cosists of two parts VERBAL and QUANT, so i yesterday turmed on the wrong part of my brain to solve quant so let me answer this question you say "We want to minimize \(0.25w+0.2(200w)\) " why are you saying this " we need to minimize" instead of this " In order to minimize the number of homeowners we need to maximize number of non homeowners"
so we duduct number of woman from 200 because we need to maximize number of nonhomeowners and minimize number of homeowners Now why do we deduct # of woman from 200 and not number of men ? well, because we know that 25% are woman and this will give us upportunity maximize # of non homeowners and minimize number of homeowners 0.2(200w) here maximize # of non homeowners and minimize number of men homeowners YAY! EUREKA ! Guys how about this method if 25 % are woman and answer choices are within 40  49 range AND 20% of men are menhomeowners than i could say that 25% of 200 is 40 and since i need to minimize number of men and only 20% of men are menhomeowners. then total #of men is 5 and out of them 0.20*5 =1 hence 40+1 = 41 is it valid approach ? Bunuel, generis , pushpitkc



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In a 200 member association consisting of men and women
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14 May 2018, 07:01
dave13 , before I check thoroughly (I skimmed), I'm a little unclear about something in your answer. What do you mean by this phrase: Quote: if 25 % are woman 25% of what?
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Re: In a 200 member association consisting of men and women
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14 May 2018, 07:33
generis wrote: dave13 , before I check thoroughly (I skimmed), I'm a little unclear about something in your answer. What do you mean by this phrase: Quote: if 25 % are woman 25% of what? generis thank you I again did mistake:) sorry I thought 25 of 200 is 40, it should be vice versa 20 of men are homeowners ...but it cant be vice versa lol ...but yeah looks like my method is incorrect:) thanks for pointing out its called gmat salad with two many ingredients ok just forget my solution I figured out it is incorrect



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Re: In a 200 member association consisting of men and women
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14 May 2018, 08:11
dave13 wrote: generis wrote: dave13 , before I check thoroughly (I skimmed), I'm a little unclear about something in your answer. What do you mean by this phrase: Quote: if 25 % are woman 25% of what? generis thank you I again did mistake:) sorry I thought 25 of 200 is 40, it should be vice versa 20 of men are homeowners ...but it cant be vice versa lol ...but yeah looks like my method is incorrect:) thanks for pointing out its called gmat salad with two many ingredients ok just forget my solution I figured out it is incorrect dave13  I think you are headed in the right direction, with an orientation similar to this post, above . If you have already figured out the answer, ignore it. If not, use those CR skills of yours with that solution. The solution you are focused on is also similarly oriented, but sometimes if we see the same idea written a different way, the "AHA" hits. Just a thought.
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Re: In a 200 member association consisting of men and women
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14 Jul 2018, 19:14
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 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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Re: In a 200 member association consisting of men and women
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14 Jul 2018, 19:23
Good question.Thanks for posting!!
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Re: In a 200 member association consisting of men and women
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14 Jul 2018, 23:32
Bunuel wrote: 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
Please explain
Thanks NAD Let the # of women be \(w\), then # of men will be \(200w\). We want to minimize \(0.25w+0.2(200w)\) > \(0.25w+0.2(200w)=0.05w+40=\frac{w}{20}+40\), so we should minimize \(w\), but also we should make sure that \(\frac{w}{20}+40\) remains an integer (as it represent # of people). Min value of \(w\) for which w/20 is an integer is for \(w=20\) > \(\frac{w}{20}+40=1+40=41\). Answer: E. Or: as there are bigger percentage of homeowner women then we should minimize # of women, but we should ensure that \(\frac{1}{4}*w\) and \(\frac{1}{5}*(200w)\) are integers. So \(w\) should be min multiple of 4 for which \(200w\) is a multiple of 5 (basically w should be min positive multiple of 20), which turns out to be for \(w=20\). Hope it's clear. Or we can have no women, and the total number of homeowners would be 40.



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Re: In a 200 member association consisting of men and women
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15 Jul 2018, 01:44
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 m+w = 200 m = 200w minimize (200w)/5 + w/4 or minimize (800+w)/20 or minimize 40 + w/20 w=20 gives min. so 41.
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In a 200 member association consisting of men and women
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15 Jul 2018, 03:23
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 We can PLUG IN THE ANSWERS, which represent the least number of homeowners. Since the correct answer must be the least possible option, start with the smallest answer choice. To test the answers, use ALLIGATION. E: 41, implying that the percentage who are homeowners \(= \frac{41}{200} = \frac{20.5}{100} = 20.5\)% Step 1: Plot the 3 percentages on a number line, with the percentages for M and W on the ends and the percentage for the mixture in the middle.M 2020.525 W Step 2: Calculate the distances between the values on the number line.M 20 0.520.5 4.525 W Step 3: Determine the ratio of men to women.The ratio of M to W is equal to the RECIPROCAL of the distances in red. \(\frac{M}{W} = \frac{4.5}{0.5} = \frac{9}{1} = \frac{90}{10} =\) \(\frac{180}{20}\)The ratio in green indicates that the percentage of homeowners will be 20.5%  yielding a total of 41 homeowners  if M=180 and W=20, for a total of 200 people.
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Re: In a 200 member association consisting of men and women
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20 Oct 2018, 09:03
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 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. Well, 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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Re: In a 200 member association consisting of men and women
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06 Dec 2018, 18:37
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
Excellent opportunity for the grid (aka doublematrix)! 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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Re: In a 200 member association consisting of men and women
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07 Dec 2018, 01:05
I solved this using venn diagram. Suggest if smthing wrong in this approach
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Re: In a 200 member association consisting of men and women
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01 Jan 2019, 12:08
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: GMAT assassins aren't born, they're made, Rich
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Re: In a 200 member association consisting of men and women
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17 Jan 2019, 19:10
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 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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Re: In a 200 member association consisting of men and women
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08 Oct 2019, 17:14
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 m + w = homeowners 5m + 4w = 200 1) 5m + 4*0 = 200 gives m=40 5*0 + 4w = 200 gives w=50 Now we know we want to minimize the number of women and maximize the number of men. 2) 5m + 4w = 200 > m must be even (since 4w will always be e) and e+e = e 3) Try m=38, 5m gives 190, 10 not div by 4 Try m=36, 5m gives 180, 20 div by 4 = 5 women So 36 + 5 = 41




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