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Of the 400 members at a health club, 260 use the weight room and 300
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VeritasKarishma
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Of the 400 members at a health club, 260 use the weight room and 300 [#permalink] New post  23 Nov 2011, 21:13
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A
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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
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Bunuel
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Of the 400 members at a health club, 260 use the weight room and 300 [#permalink] New post  30 Jan 2012, 15:31
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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:

Image

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.

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Attachment:
Matrix.PNG
Matrix.PNG [ 7.88 KiB | Viewed 11980 times ]

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Re: Of the 400 members at a health club, 260 use the weight room and 300 [#permalink] New post  23 Nov 2011, 22:33
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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
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Re: Of the 400 members at a health club, 260 use the weight room and 300 [#permalink] New post  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)
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Re: Of the 400 members at a health club, 260 use the weight room and 300 [#permalink] New post  24 Nov 2011, 10:36
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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...
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Re: Of the 400 members at a health club, 260 use the weight room and 300 [#permalink] New post  30 Jan 2012, 12:41
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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.
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Re: Of the 400 members at a health club, 260 use the weight room and 300 [#permalink] New post  30 Jan 2012, 21:39
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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.
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Re: Of the 400 members at a health club, 260 use the weight room and 300 [#permalink] New post  30 Jan 2012, 22:26
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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

so, the answer is D
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Re: Of the 400 members at a health club, 260 use the weight room and 300 [#permalink] New post  17 Aug 2013, 12:42
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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) !
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Re: Of the 400 members at a health club, 260 use the weight room and 300 [#permalink] New post  12 Nov 2015, 22:18
400=260+300-x+neither

400-560+x>=60

x>=220

only D fits
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Re: Of the 400 members at a health club, 260 use the weight room and 300 [#permalink] New post  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.
Hence Answer is A
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