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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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)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...

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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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) !

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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