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A certain high school has 5,000 students. Of these students, x are
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14 Sep 2015, 07:50
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A certain high school has 5,000 students. Of these students, x are taking music, y are taking art, and z are taking both music and art. How many students are taking neither music nor art? (A) 5,000 – z (B) 5,000 – x – y (C) 5,000 – x + z (D) 5,000 – x – y – z (E) 5,000 – x – y + z og2016
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Re: A certain high school has 5,000 students. Of these students, x are
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14 Sep 2015, 16:46
Easier formula to remember is Group1+ Group2 Both +Neither = Total (Source: Kaplan) So, solution is x + y  z + Neither = 5000 Neither = 5000  x  y + z ..Choice E!




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Re: A certain high school has 5,000 students. Of these students, x are
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14 Sep 2015, 08:02
ske wrote: A certain high school has 5,000 students. Of these students, x are taking music, y are taking art, and z are taking both music and art. How many students are taking neither music nor art?
(A) 5,000 – z (B) 5,000 – x – y (C) 5,000 – x + z (D) 5,000 – x – y – z (E) 5,000 – x – y + z
og2016 I believe the wording of the question is incorrect (for the options given)  when you say "take music", that means take either music or music and art. In that case here's the answer: Music U Art = Music + Art  Music n Art = (xz) + (yz)  z = x+y3z Total = 5000 So, people not taking any of these is: 5000  (x+y3z). Now, from the options, I understand that the question asked has the assumption that "take music" means "music only". In that case, the calculation becomes: 5000  (x+yz) = 5000  x  y + z So, (E) is the answer.
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Re: A certain high school has 5,000 students. Of these students, x are
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16 Sep 2015, 07:36
HardWorkBeatsAll wrote: I believe the wording of the question is incorrect (for the options given)  when you say "take music", that means take either music or music and art. In that case here's the answer: Music U Art = Music + Art  Music n Art = (xz) + (yz)  z = x+y3z
Total = 5000 So, people not taking any of these is: 5000  (x+y3z)
Thanks for submitting an explanation. It is correct to say that: Music U Art = Music + Art  Music n Art. However, your logic falls apart when you attribute Music = xz and Art = yz. This is wrong because (xz) and (yz) refers to the number of students who ONLY take music or who ONLY take art. So in your equation, you've subtracted the number of students who take both music and art three times over. It should be: x + y  z. That represents the number of students who took music OR art. Then, 5,000  (x + y  z) would give you the number of students who took neither. And you would have answer choice (E). Alternatively, if you want to follow your methodology of accounting for the students who took music ONLY and Art ONLY, use the following equation (which would still arrive at the same original equation for union of two sets): (x  z) + (y  z) + z = x + y  z.



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Re: A certain high school has 5,000 students. Of these students, x are
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07 Oct 2015, 09:51
pdxyj wrote: HardWorkBeatsAll wrote: I believe the wording of the question is incorrect (for the options given)  when you say "take music", that means take either music or music and art. In that case here's the answer: Music U Art = Music + Art  Music n Art = (xz) + (yz)  z = x+y3z
Total = 5000 So, people not taking any of these is: 5000  (x+y3z)
Thanks for submitting an explanation. It is correct to say that: Music U Art = Music + Art  Music n Art. However, your logic falls apart when you attribute Music = xz and Art = yz. This is wrong because (xz) and (yz) refers to the number of students who ONLY take music or who ONLY take art. So in your equation, you've subtracted the number of students who take both music and art three times over. It should be: x + y  z. That represents the number of students who took music OR art. Then, 5,000  (x + y  z) would give you the number of students who took neither. And you would have answer choice (E). Alternatively, if you want to follow your methodology of accounting for the students who took music ONLY and Art ONLY, use the following equation (which would still arrive at the same original equation for union of two sets): (x  z) + (y  z) + z = x + y  z. from the options, I understand that the question asked has the assumption that "take music" means "music only". In that case, the calculation becomes: 5000  (x+yz) = 5000  x  y + z So, (E) is the answer.



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A certain high school has 5,000 students. Of these students, x are
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25 Apr 2017, 21:42
The items in the yellow boxes are the ones given to us (the 'y' is also given) and the green one is what we're looking for: 5000x(yz) 5000xy+zOA
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Re: A certain high school has 5,000 students. Of these students, x are
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20 Oct 2015, 09:46
Bunuel wrote: A certain high school has 5,000 students. Of these students, x are taking music, y are taking art, and z are taking both music and art. How many students are taking neither music nor art?
(A) 5,000 − z (B) 5,000 − x − y (C) 5,000 − x + z (D) 5,000 − x − y − z (E) 5,000 − x − y + z
Kudos for a correct solution. The simple formula for overlapping sets is as follows: Total = Group A + Group B  Both + NeitherIn this case we have 5,000 = x + y  z + n Solving for N equals 5,000 − x − y + z Answer E
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Re: A certain high school has 5,000 students. Of these students, x are
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29 Apr 2017, 09:05
ske wrote: A certain high school has 5,000 students. Of these students, x are taking music, y are taking art, and z are taking both music and art. How many students are taking neither music nor art?
(A) 5,000 – z (B) 5,000 – x – y (C) 5,000 – x + z (D) 5,000 – x – y – z (E) 5,000 – x – y + z We can use the following formula: Total students = # taking music + # taking art  # taking both + # taking neither 5,000 = x + y  z + neither 5,000  x  y + z = neither Answer: E
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Re: A certain high school has 5,000 students. Of these students, x are
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20 Oct 2015, 09:55
Bunuel wrote: A certain high school has 5,000 students. Of these students, x are taking music, y are taking art, and z are taking both music and art. How many students are taking neither music nor art?
(A) 5,000 − z (B) 5,000 − x − y (C) 5,000 − x + z (D) 5,000 − x − y − z (E) 5,000 − x − y + z
Kudos for a correct solution. Students taking Music and art both = z Students taking Music only = x  z Students taking Art only = y  z Students taking Music and/or art = (xz) + (yz) + z = x + y  z Students without Music and Art = 5000  (x + y  z) Students without Music and Art = 5000  x  y + z Answer: option E
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Re: A certain high school has 5,000 students. Of these students, x are
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08 Aug 2016, 16:12
Can someone show me how to solve this using the Matrix Method?
For some reason I cannot find E, I find D as the answer.



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A certain high school has 5,000 students. Of these students, x are
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11 Aug 2016, 11:51
g3lo18 wrote: Can someone show me how to solve this using the Matrix Method?
For some reason I cannot find E, I find D as the answer. Hope this is helpful. 5000=x+yz+n As z is counted twice both in x and also in y, we should subtract once to get the actual count. Let n be the number of students taking neither music nor art so n=5000xy+z answer option (E)
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Re: A certain high school has 5,000 students. Of these students, x are
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03 Sep 2019, 23:34
This is a very simple question on overlapping sets where we need to draw a twoset Venn diagram and fill out the regions to obtain the answer. It’s as simple as that. Here's how the Venn diagram should look: Attachment:
04th Sept 2019  Reply 2.JPG [ 18.09 KiB  Viewed 1578 times ]
From the Venn diagram, we can say that the sum of all the regions should be equal to 5000. So, xz + y – z + z + n = 5000. Here, n represents the number of people who have taken up neither music nor art. Therefore, n = 5000 – x – y +z. The correct answer option is E. Hope this helps!
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Re: A certain high school has 5,000 students. Of these students, x are
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20 Oct 2015, 04:29
Bunuel wrote: A certain high school has 5,000 students. Of these students, x are taking music, y are taking art, and z are taking both music and art. How many students are taking neither music nor art?
(A) 5,000 − z (B) 5,000 − x − y (C) 5,000 − x + z (D) 5,000 − x − y − z (E) 5,000 − x − y + z
Kudos for a correct solution. My Solution:
Total = 5000
Total = X + Y Z+Neither
Neither = Total  X Y +Z
Neither = 5000XY+Z Option E Answer
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Re: A certain high school has 5,000 students. Of these students, x are
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21 Oct 2015, 11:07
Hi All, While this question can be solved Algebraically, it can also be solved by TESTing VALUES and taking some basic notes: We're given a series of facts to work with: 1) A certain high school has 5,000 students. 2) Of these students: X are taking music, Y are taking art, and Z are taking BOTH music and art. We're asked how many students are taking NEITHER music nor art? Let's TEST X = 2 Y = 2 Z = 1 So, we have 2 students taking music, 2 taking art and 1 taking BOTH music and art. That 1 person has been counted TWICE though (once in the music 'group' and once in the art 'group'), so what we really have is... 1 student taking JUST music 1 student taking JUST art 1 student taking BOTH music and art Total = 3 students We're asked for the total number of students who are taking NEITHER Course. That is 5000  3 = 4997. So that's the answer that we're looking for when X=2, Y=2 and Z=1. There's only one answer that matches... Final Answer: GMAT assassins aren't born, they're made, Rich
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Re: A certain high school has 5,000 students. Of these students, x are
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22 Oct 2015, 02:00
Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer. A certain high school has 5,000 students. Of these students, x are taking music, y are taking art, and z are taking both music and art. How many students are taking neither music nor art? (A) 5,000 − z (B) 5,000 − x − y (C) 5,000 − x + z (D) 5,000 − x − y − z (E) 5,000 − x − y + z The number of students who take music or art is X+YZ( Z students take also art among the students who take music and the same Z students are included in the students who take art, so the number of students who take music or art is X+YZ). The number of students who take neither music nor art is 5000(X+YZ) = 5000XY+Z. The answer is (E)
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Re: A certain high school has 5,000 students. Of these students, x are
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