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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
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
30 Jan 2012, 16:31
Kudos
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VeritasPrepKarishma
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
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.
Attachment:
Matrix.PNG [ 7.88 KiB | Viewed 24845 times ]
30 Jan 2012, 22:39
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saxenaashi
VeritasPrepKarishma
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
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.
