A survey was conducted to find out how many people in a housing colony of 144 residents could swim, dance and drive a car. It was found that the number of people who could not swim was 89, the number of people who could not dance was 100 and that the number of people who could not drive a car was 91. If the number of people who could do at least two of these things, was found to be 37 and the number of people who could do all these things was found to be 6, how many people could not do any of these things?
A) 17
B) 23
C) 29
D) 35
E) 50
OA is D
swim, dance and drive
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- cans
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total 144.
not swim=89 or swim = 144-89 = 55
dance: 44
car: 53
atleast 2 =37
all 3:6
thus exactly 2 = 37-6 = 31
thus 144 = 55+44+53 + none -31 -2*6
none = 35
IMO D
not swim=89 or swim = 144-89 = 55
dance: 44
car: 53
atleast 2 =37
all 3:6
thus exactly 2 = 37-6 = 31
thus 144 = 55+44+53 + none -31 -2*6
none = 35
IMO D
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- sl750
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@razorback
Each of 55, 44, 53 represents an overlap with more than one group.
Example The number who can swim, includes, people who can ONLY swim, people who can swim and dance, people who can swim and drive, and people who can swim dance and drive.
This is the case for the other two groups as well. Since we consider the swim,drive,dance category three times and we only need one instance of it, we subtract it twice. Hence the number who can do all three activities is subtracted twice
Hope it is clear
Each of 55, 44, 53 represents an overlap with more than one group.
Example The number who can swim, includes, people who can ONLY swim, people who can swim and dance, people who can swim and drive, and people who can swim dance and drive.
This is the case for the other two groups as well. Since we consider the swim,drive,dance category three times and we only need one instance of it, we subtract it twice. Hence the number who can do all three activities is subtracted twice
Hope it is clear
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This is a useful formula in an overlapping set of 3 categories. The formula is:razorback wrote:why did you multiply 2*6 in your last equation?
Those that can do 1 - Those that can do 2 - 2*(Those that can do all 3) + Those that can't do any = Total Set
The reason you multiply by two is that by counting all that can do 1, you are double counting those that can do two skills and you are triple counting those that can do all skills. These must be subtracted out.
As cans said, you can find those that can do 1 (Swim = 55, Dance = 44, Car = 53) as 152.
We can find those that can do 2 by subtracting those that can do all 3 by those that can do at least two (37-6 = 31)
Those that can do 3 are given - 6.
Plugging this all in:
152 - 31 - 2*6 + None = 144
None = 35
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Given:LalaB wrote:A survey was conducted to find out how many people in a housing colony of 144 residents could swim, dance and drive a car. It was found that the number of people who could not swim was 89, the number of people who could not dance was 100 and that the number of people who could not drive a car was 91. If the number of people who could do at least two of these things, was found to be 37 and the number of people who could do all these things was found to be 6, how many people could not do any of these things?
A) 17
B) 23
C) 29
D) 35
E) 50
OA is D
# of swimmers = 144 - 89 = 55
# of dancers = 144 - 100 = 44
# of drivers = 144 - 91 = 53
Let's sketch the overlapping sets and use whatever numbers satisfy the given information. To do this, I start at the very middle (person does all 3) and place a 6.
From here, we know that the areas where there are exactly two overlaps must add to 31 since there are 37 people who do 2 or more activities. So, I used the values 20, 10 and 1 to satisfy that given condition. Note: I could have used any 3 numbers that add to 31; the result will still be the same - try it.
At that point, we need to fill in the remaining parts of the circles based on the number of people who do each activity. When we do so, we get this.
When we add up the different sections, we get:
17+10+6+20+17+1+38 = 109
So, we've now accounted for 109 people out of 144 people.
So, the remaining 35 people must not do any of these activities.
Answer = D
Cheers,
Brent
Someone pls help i'm going bonkers as I dont knw which formula shld be used
P(A u B u C) = P(A) + P(B) + P(C) - P(A n B) - P(A n C) - P(B n C) + P(A n B n C)
P(A u B u C) = P(A) + P(B) + P(C) - P(A n B) - P(A n C) - P(B n C) - 2P(A n B n C)
Experts advice pls.
P(A u B u C) = P(A) + P(B) + P(C) - P(A n B) - P(A n C) - P(B n C) + P(A n B n C)
P(A u B u C) = P(A) + P(B) + P(C) - P(A n B) - P(A n C) - P(B n C) - 2P(A n B n C)
Experts advice pls.