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# A survey was conducted to find out how many people in a hous

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A survey was conducted to find out how many people in a hous [#permalink]

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19 Sep 2011, 09:00
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Question Stats:

52% (04:01) correct 48% (03:27) wrong based on 351 sessions

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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
[Reveal] Spoiler: OA

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19 Sep 2011, 09:21
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people who could swim =55

People who clould dance =44

people who could drive = 53

total number of driving,swiming and dancing events = 55 +44 +53= 152

6 people have been triple counted for their expertise, same people extra counted 12 times.

(37 - 6) = 31 people have been double counted for their expertise.

total number of different people who could do these different activities = 152 -(12 +31) =109

so people who could do nothing 144 -109 = 35

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20 Sep 2011, 00:17
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people who could swim =55

People who clould dance =44

people who could drive = 53

total number of driving,swiming and dancing events = 55 +44 +53= 152

6 people have been triple counted for their expertise, same people extra counted 12 times.

(37 - 6) = 31 people have been double counted for their expertise.

total number of different people who could do these different activities = 152 -(12 +31) =109

so people who could do nothing 144 -109 = 35

please could you elaborate as to how did you get the individual nos
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20 Sep 2011, 01:25
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gsd85 wrote:
How can we do this Q using the formula....

T=n(A)+n(B)+n(C)-n(Exactly two of the events)-2*n(All 3 Events)+n(None of the events)

T=144
n(A)=T-n(A')=144-89=55
n(B)=T-n(B')=144-100=44
n(C)=T-n(C')=144-91=53
n(Exactly two of the events)=n(At least 2 Events)-n(All 3 Events)=37-6=31
n(All 3 Events)=6

144=55+44+53-31-2*6+n(None of the events)
n(None of the events)=144-55-44-53+31+12=35

Ans: "D"
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23 Sep 2011, 20:22
I used to take more time to solve this type of question. Fluke's method is awesome and will help to reduce the time to solve.
Thanks Fluke!!!!!!!!
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If you find my posts useful, Appreciate me with the kudos!! +1

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24 Sep 2011, 00:27
chawlavinu wrote:
I used to take more time to solve this type of question. Fluke's method is awesome and will help to reduce the time to solve.
Thanks Fluke!!!!!!!!

You're welcome!!! The appreciation goes to the contributors of this thread:

formulae-for-3-overlapping-sets-69014.html
esp.
formulae-for-3-overlapping-sets-69014.html#p729340
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25 Sep 2011, 12:45
#not swim = 89 => #swim = 144-89 = 55
#not dance = 100 => #dance = 144 - 100 = 44
#not drive = 91 => #drive = 144 - 91 = 53

at least two = sum of two's + sum of three's = 37
sum of three's = 6
=> sum of two's = 31

total = S + D + Dr - (sum of two's ) - 2*(sum of three's) + none

144 = 55 + 44 + 53 -31 - 2*6 + none

=> none = 35

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22 Oct 2011, 10:50
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+1 for D.

Use 3 set formula:

Total = Group1 + Group2 + Group3 - (sum of 2-group overlaps) - 2*(all three) + Neither

Found individual numbers as explained in the earlier posts.
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09 Nov 2011, 08:42
Hey liftoff,

I used to be confused with such kind of problems and what helped me is drawing Venn. If you follow this there is no need to remember any formula. One just need ot remember simple +/- and also take less than 2 min.

Cheers
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03 Jan 2013, 06:21
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

I am not sure that stated official answer is correct one. or I am tricked by the language (which generally is the case in these kind of question). IMHO, the number of people who could do at least two of the things also includes 18 ( 3 *6 where 6 are the number of people who could all these things). My answer is coming as option B - 23. Can anybody help.
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03 Jan 2013, 09:41
saxenaashi 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

I am not sure that stated official answer is correct one. or I am tricked by the language (which generally is the case in these kind of question). IMHO, the number of people who could do at least two of the things also includes 18 ( 3 *6 where 6 are the number of people who could all these things). My answer is coming as option B - 23. Can anybody help.

The number of people who can do at least two things includes 6 (number of people who can do all three), not 6*3.

Understand here that 37 is the number of people, not the number of instances. Hence 6 is not counted 3 times in 37. Out of 37 people, 18 people cannot do all three. Only 6 can do all three. So 31 can do exactly 2 things.

I have discussed this concept in my blog post given below:
http://www.veritasprep.com/blog/2012/09 ... ping-sets/

Question 1 on the diagram is this.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Status: Waiting for Decisions Joined: 23 Dec 2012 Posts: 42 Location: India Sahil: Bansal GMAT 1: 570 Q49 V20 GMAT 2: 690 Q49 V34 GPA: 3 WE: Information Technology (Computer Software) Followers: 0 Kudos [?]: 50 [0], given: 42 Re: A survey was conducted to find out how many people in a hous [#permalink] ### Show Tags 10 Nov 2013, 05:29 Hello All, I started as Da, Dr, Sw, be the number of people who could only dance, Drive and swim respectively X is number of people who could dance and drive Y is number of people who could dance and swim Z is number of people who could drive and swim. so X+ y+ z=37-6=31 Da + Dr + X = 89 Da + Sw + y = 91 Dr + Sw + z = 100 2(Da + Dr + Sw) +x+y+z = 89+91+100=280 2 (Da + Dr + Sw) + 31 = 280 2 (Da + Dr + Sw) = 280-31 = 249 where did i go wrong here??? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7119 Location: Pune, India Followers: 2132 Kudos [?]: 13631 [1] , given: 222 Re: A survey was conducted to find out how many people in a hous [#permalink] ### Show Tags 11 Nov 2013, 06:32 1 This post received KUDOS Expert's post bsahil wrote: Da + Dr + X = 89 Da + Sw + y = 91 Dr + Sw + z = 100 Not correct. 89 includes the people who cannot do any of the three things. So does 91 and 100. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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11 Nov 2013, 18:05
fluke wrote:
gsd85 wrote:
How can we do this Q using the formula....

T=n(A)+n(B)+n(C)-n(Exactly two of the events)-2*n(All 3 Events)+n(None of the events)

T=144
n(A)=T-n(A')=144-89=55
n(B)=T-n(B')=144-100=44
n(C)=T-n(C')=144-91=53
n(Exactly two of the events)=n(At least 2 Events)-n(All 3 Events)=37-6=31
n(All 3 Events)=6

144=55+44+53-31-2*6+n(None of the events)
n(None of the events)=144-55-44-53+31+12=35

Ans: "D"

But the stem says "at least 2", not exactly. so it should be the formula as follows:

T=n(A)+n(B)+n(C)-n(at least two of the events)+n(All 3 Events)+n(None of the events)

so 144=55+44+53-37+6+X

144=152-37+6+X
144=121+X
X=23

OA is incorrect,
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11 Nov 2013, 18:06
VeritasPrepKarishma wrote:
saxenaashi 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

I am not sure that stated official answer is correct one. or I am tricked by the language (which generally is the case in these kind of question). IMHO, the number of people who could do at least two of the things also includes 18 ( 3 *6 where 6 are the number of people who could all these things). My answer is coming as option B - 23. Can anybody help.

The number of people who can do at least two things includes 6 (number of people who can do all three), not 6*3.

Understand here that 37 is the number of people, not the number of instances. Hence 6 is not counted 3 times in 37. Out of 37 people, 18 people cannot do all three. Only 6 can do all three. So 31 can do exactly 2 things.

I have discussed this concept in my blog post given below:
http://www.veritasprep.com/blog/2012/09 ... ping-sets/

Question 1 on the diagram is this.

How do you know that's true though; to me the question read like every other overlapping data set question. It didn't specify 'exactly' two, and seemed to be worded as other questions which mention people being in at least two sets...
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11 Nov 2013, 20:55
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AccipiterQ wrote:
VeritasPrepKarishma wrote:
Understand here that 37 is the number of people, not the number of instances. Hence 6 is not counted 3 times in 37. Out of 37 people, 18 people cannot do all three. Only 6 can do all three. So 31 can do exactly 2 things.

I have discussed this concept in my blog post given below:
http://www.veritasprep.com/blog/2012/09 ... ping-sets/

Question 1 on the diagram is this.

How do you know that's true though; to me the question read like every other overlapping data set question. It didn't specify 'exactly' two, and seemed to be worded as other questions which mention people being in at least two sets...

The statement given in the question is this:
"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"

Say, I have 37 people in front of me and I say that these are the people who can do at least two of the three things - say these people are P1, P2, ...P37.
I also know that exactly 6 people can do all three things. These 6 are P1, P4, P8, P9, P10, P12
Tell me, how many people can do exactly 2 of the three things? 31 or 19? The answer here is 31.

Note that this situation is different from the usual: 10 people can swim and dance, 20 people can dance and drive and 7 people can swim and drive. In this case, each of the 10, 20 and 7 includes the people who can do all 3 things and hence 10 + 20 + 7 - 6*3 = 19 people can do exactly two things.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7119 Location: Pune, India Followers: 2132 Kudos [?]: 13631 [0], given: 222 Re: Venn Time [#permalink] ### Show Tags 12 Nov 2013, 05:18 AccipiterQ wrote: But the stem says "at least 2", not exactly. so it should be the formula as follows: T=n(A)+n(B)+n(C)-n(at least two of the events)+n(All 3 Events)+n(None of the events) so 144=55+44+53-37+6+X 144=152-37+6+X 144=121+X X=23 OA is incorrect, Using the formula is a bad idea if you don't understand exactly when and how to use it. If you understand exactly when and how to use the formula, then you would find it too cumbersome to use it and will anyway prefer to reason out the answer! In your formula, n(at least two of the events) is the sum of the intersection of the circles. This means each intersection includes the area where only two overlap and where all 3 overlap. To check out the two formulas, check out this link: a-school-has-3-classes-math-class-has-14-students-150221.html#p1207266 After you check out the link, note that in your formula, n(at least two of the events) = (d + g) + (e + g) + (f + g) whereas the 37 given to you in this question is (d + e + f + g) _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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