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# Of the 400 members at a health club, 260 use the weight room and 300

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Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7869
Location: Pune, India
Of the 400 members at a health club, 260 use the weight room and 300 [#permalink]

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23 Nov 2011, 21:13
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Difficulty:

35% (medium)

Question Stats:

69% (01:34) correct 31% (01:43) wrong based on 569 sessions

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Here is the question I promised (Sets from Veritas Book: Maximize and Minimize):

Of the 400 members at a health club, 260 use the weight room and 300 use the pool. If at least 60 of the members do not use either, then the number of members using both the weight room and the pool must be between:

(A) 40 to 100
(B) 80 to 140
(C) 160 to 260
(D) 220 to 260
(E) 220 to 300
[Reveal] Spoiler: OA

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Karishma
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Director Joined: 28 Jul 2011 Posts: 514 Location: United States Concentration: International Business, General Management GPA: 3.86 WE: Accounting (Commercial Banking) Re: Of the 400 members at a health club, 260 use the weight room and 300 [#permalink] ### Show Tags 23 Nov 2011, 22:33 2 This post received KUDOS Its D because the keyword here is "atleast" which says initially 60 but it can be a maximum of 100 as Minimun of 300 are use the something so we are lest with 100 so it ranges from 340=260+300-x x=220 and 300=260+300-x x=260 so the range is from 220 to 260 _________________ +1 Kudos If found helpful.. Intern Joined: 06 Sep 2011 Posts: 20 GMAT 1: 670 Q42 V40 GMAT 2: 750 Q50 V42 Re: Of the 400 members at a health club, 260 use the weight room and 300 [#permalink] ### Show Tags 24 Nov 2011, 06:43 AuB=L (AuB)'=K K+L=400 => L=400 - K L=A+B-A@B => A@B = 560 - L => A@B = 160 + K, but K>=60 => A@B>=220 But A@B <=MIN(A,B) A@B<=260, hence D. (@ means intersection) _________________ Persistence. Manager Joined: 23 Oct 2011 Posts: 83 Re: Of the 400 members at a health club, 260 use the weight room and 300 [#permalink] ### Show Tags 24 Nov 2011, 10:36 2 This post received KUDOS 1 This post was BOOKMARKED VeritasPrepKarishma wrote: Here is the question I promised: Of the 400 members at a health club, 260 use the weight room and 300 use the pool. If at least 60 of the members do not use either, then the number of members using both the weight room and the pool must be between: (A) 40 to 100 (B) 80 to 140 (C) 160 to 260 (D) 220 to 260 (E) 220 to 300 1)We understand that its a min/max question regarding 2 overlapping sets. 2)We know that $$(Total Members)=(Use Weight Room)+(Use Pool)-(Use Both)+(Use Neither)$$ ---> $$(Use Both)=160+(Use Neither)$$ and $$(Use Neither)\geq 60$$ If we didn't have a restriction on $$(Use Neither)$$ how could we find the lowest value it can take? For $$(Use Neither)=60$$ we have the lowest value of people who use both ---> $$(Use Both)=160+60=220$$. So we know its either D or E. To find the maximum value of $$(Use Both)$$ we have know that $$(Use Both)\leq 260$$. It is common sense that the maximum value can be 260 but is there a mathematical way to prove it? If we had x and y what would be the inequality? So: $$220 \leq (Use Both)\leq 260$$ ---> D. Karishma could you help me with the red phrases please... Intern Joined: 29 Aug 2011 Posts: 22 Re: Of the 400 members at a health club, 260 use the weight room and 300 [#permalink] ### Show Tags 30 Jan 2012, 12:41 3 This post received KUDOS 1 This post was BOOKMARKED VeritasPrepKarishma wrote: Here is the question I promised: Of the 400 members at a health club, 260 use the weight room and 300 use the pool. If at least 60 of the members do not use either, then the number of members using both the weight room and the pool must be between: (A) 40 to 100 (B) 80 to 140 (C) 160 to 260 (D) 220 to 260 (E) 220 to 300 To reduce the options, I started looking for the max value first. The max value will be the lowest of the two, which is 260. Hence I started looking at options (C) & (D). Then I calculated the 'both' value as 400 = 260 +300 - 'both' + 60 = 220. Thus I reached (D). For this question I did not worry about the atleast part because I already found the max range. Hope my approach is fine in solving the question. Regards Math Expert Joined: 02 Sep 2009 Posts: 43348 Of the 400 members at a health club, 260 use the weight room and 300 [#permalink] ### Show Tags 30 Jan 2012, 15:31 4 This post received KUDOS Expert's post 4 This post was BOOKMARKED VeritasPrepKarishma wrote: Here is the question I promised: Of the 400 members at a health club, 260 use the weight room and 300 use the pool. If at least 60 of the members do not use either, then the number of members using both the weight room and the pool must be between: (A) 40 to 100 (B) 80 to 140 (C) 160 to 260 (D) 220 to 260 (E) 220 to 300 One can also solve this problem with double set matrix quite easily: Numbers in black are given. To minimize the number of members using both we should maximize yellow boxes and for this we should minimize the number of members who do not use either, so make it 60. The rest of the boxes can be filled with simple arithmetic; To maximize the number of members using both we should minimize blue boxes and for this we should maximize the number of members who do not use either, so make it 100 (it cannot be more as the number to the right of it is its upper limit). The rest of the boxes can be filled with simple arithmetic Answer: D. Hope it's clear. [Reveal] Spoiler: Attachment: Matrix.PNG [ 7.88 KiB | Viewed 6122 times ] _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7869 Location: Pune, India Re: Of the 400 members at a health club, 260 use the weight room and 300 [#permalink] ### Show Tags 30 Jan 2012, 21:39 Expert's post 2 This post was BOOKMARKED saxenaashi wrote: VeritasPrepKarishma wrote: Here is the question I promised: Of the 400 members at a health club, 260 use the weight room and 300 use the pool. If at least 60 of the members do not use either, then the number of members using both the weight room and the pool must be between: (A) 40 to 100 (B) 80 to 140 (C) 160 to 260 (D) 220 to 260 (E) 220 to 300 To reduce the options, I started looking for the max value first. The max value will be the lowest of the two, which is 260. Hence I started looking at options (C) & (D). Then I calculated the 'both' value as 400 = 260 +300 - 'both' + 60 = 220. Thus I reached (D). For this question I did not worry about the atleast part because I already found the max range. Hope my approach is fine in solving the question. Regards This approach is fine as long as you understand the following about this question: When we minimize "members who do not use either", we are minimizing the "both" number as well. Look at the equation above, 400 = 260 +300 - 'both' + 60 Since 400 is the constant sum, if you keep increasing the '60' term, you will have to keep increasing the 'both' term too to maintain the sum of 400. Least value of "members who do not use either" is 60 so here we will get the least value of "both". Just to give you the complement of this concept, On the same lines, if you maximize the "members who do not use either" value, you will maximize "both" value. 400 = 260 +300 - 'both' + 100 (since there are 400 people and 300 use the pool) "both" will be 260 here. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Joined: 23 Oct 2010
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Location: Azerbaijan
Concentration: Finance
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Re: Of the 400 members at a health club, 260 use the weight room and 300 [#permalink]

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30 Jan 2012, 22:26
W=260
P=300
ALL=400
N(NEITHER)=AT LEAST 60
BOTH=?

W+P-BOTH+N=ALL
260+300-BOTH+60=400
BOTH=220

now pay attention to the fact, that both at most can be 260, since W=260

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Re: Of the 400 members at a health club, 260 use the weight room and 300 [#permalink]

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17 Aug 2013, 12:42
2
KUDOS
In the figure attached:

writing eqn of venn diagram:

260-X+X+300-X+GT60 = 400

=>X = GT620-400
=> X = GT220
=> X >= 220 (Here GT means Greater than)

So Min value of X is 220 coming to options D/E

X's maximum value cannot exceed 260 as per the venn diagramsince 'W' circle cannot accomodate 'X' More than 260

Hence (D) !
Attachments

soln.JPG [ 16.4 KiB | Viewed 5204 times ]

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Director
Joined: 23 Jan 2013
Posts: 603
Schools: Cambridge'16
Re: Of the 400 members at a health club, 260 use the weight room and 300 [#permalink]

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12 Nov 2015, 22:18
400=260+300-x+neither

400-560+x>=60

x>=220

only D fits
Intern
Joined: 10 Dec 2017
Posts: 3
Re: Of the 400 members at a health club, 260 use the weight room and 300 [#permalink]

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10 Dec 2017, 07:12
VeritasPrepKarishma wrote:
Here is the question I promised (Sets from Veritas Book: Maximize and Minimize):

Of the 400 members at a health club, 260 use the weight room and 300 use the pool. If at least 60 of the members do not use either, then the number of members using both the weight room and the pool must be between:

(A) 40 to 100
(B) 80 to 140
(C) 160 to 260
(D) 220 to 260
(E) 220 to 300

Solution : Let 'x' people use both weight room and pool. Then 260 -x use only weight room and 300-x use only pool. Since atleast 60 people use neither,
(260-x) +x+(300-x) - 400 >=60
560 - x-400 >=60
Solving the above equation, x <= 100
Therefore the number of members using both the weight room and the pool must be less than or equal to 100.
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Re: Of the 400 members at a health club, 260 use the weight room and 300   [#permalink] 10 Dec 2017, 07:12
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