Author 
Message 
TAGS:

Hide Tags

VP
Joined: 30 Jun 2008
Posts: 1007

Formulae for 3 overlapping sets [#permalink]
Show Tags
Updated on: 14 Oct 2008, 01:20
Formulae to solve problems with 3 Overlapping setsu = union and n = intersection1. For 3 sets A, B, and 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) + P(A n B n C) 2. To determine the No of persons in exactly one set : P(A) + P(B) + P(C) – 2P(A n B) – 2P(A n C) – 2P(B n C) + 3P(A n B n C) 3. To determine the No of persons in exactly two of the sets : P(A n B) + P(A n C) + P(B n C) – 3P(A n B n C) 4. To determine the No of persons in exactly three of the sets : P(A n B n C) 5. To determine the No of persons in two or more sets (at least 2 sets) : P(A n B) + P(A n C) + P(B n C) – 2P(A n B n C) (SEE THE ATTACHED PICTURE TO UNDERSTAND THE BELOW MORE CLEARLY) 1. For three sets A, B, and C, P(A u B u C): (A + B + C + X + Y + Z + O) 2. Number of people in exactly one set: ( A+B+C) 3. Number of people in exactly two of the sets: (X+Y+Z) 4. Number of people in exactly three of the sets: O 5. Number of people in two or more sets: ( X+Y+Z+O) 6. Number of people only in set A: A 7. P(A): A+X+Y+O 8. P(A n B): X+O Hope this helps. Please add if i have missed something.
Attachments
SET_THEORY.jpg [ 5.6 KiB  Viewed 193009 times ]
_________________
"You have to find it. No one else can find it for you."  Bjorn Borg
Originally posted by amitdgr on 12 Sep 2008, 03:41.
Last edited by amitdgr on 14 Oct 2008, 01:20, edited 1 time in total.



VP
Joined: 05 Jul 2008
Posts: 1334

Re: Formulae for 3 overlapping sets [#permalink]
Show Tags
29 Sep 2008, 10:43
Great post! How ever, it is very difficult to remember all such formulae, at least for me.
We should be able to derive the rest of the formulae based on
For 3 sets A, B, and 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) + P(A n B n C)
lets discuss how we arrived at each such formula
To determine the No of persons in exactly two of the sets : P(A n B) + P(A n C) + P(B n C) – 3P(A n B n C)
The reason why we need to subtract 3 times the (A n B n C) is because we need number of persons in exactly 2 sets. How ever, there are people who are in all the three sets and we need to subtract them. In other words when we counted A n B we already counted A n B n C once. Similarly with B n C and C n A. So we subtract them 3 times again.
To determine the No of persons in exactly one set : P(A) + P(B) + P(C) – 2P(A n B) – 2P(A n C) – 2P(B n C) + 3P(A n B n C)
we need only in one set. that means if we look at the figure we need to find out
only A + only B + only C
Only A = A  { ( A n B) + ( A n C )  (A n B n C) }
The region with only A is the whole region/circle A subtracted from ( A n B) + ( A n C ). In doing so, we accounted for (A n B n C) twice, so subtract (A n B n C) once
Similarly only B and only c
Add them up to get P(A) + P(B) + P(C) – 2P(A n B) – 2P(A n C) – 2P(B n C) + 3P(A n B n C)
Any takers for the next set of formulas.



Intern
Joined: 29 Sep 2008
Posts: 46

Re: Formulae for 3 overlapping sets [#permalink]
Show Tags
29 Sep 2008, 14:22
You have done 1, 2, 3: I can do point 4 and 5 in the first list. I am confused what 6 means.
4. To determine the No of persons in exactly three of the sets : this is clearly the people who are a part of all three sets, look at the picture and it is the shared section of all three circles. that is (A n B n C)
5. To determine the No of persons in two or more sets (at least 2 sets) : this is the set of people who are shared by two sets or by all three sets. Let's start with (A n B) + (B n C) + (C n A). But remember, while adding these three, we added the shared part of all three, three times. thus remove that part 2 times (as we want to keep that part too). hence, (A n B) + (B n C) + (C n A)  2( A n B n C)



Senior Manager
Joined: 17 Mar 2009
Posts: 270

Re: Formulae for 3 overlapping sets [#permalink]
Show Tags
21 Jul 2009, 18:24
seofah wrote: Thanks for putting this thread together, but can we also have some GMAT examples where one can use these concepts? Q1 : In a consumer survey, 85% of those surveyed liked at least one of three products: 1, 2, and 3. 50% of those asked liked product 1, 30% liked product 2, and 20% liked product 3. If 5% of the people in the survey liked all three of the products, what percentage of the survey participants liked more than one of the three products? Answer: 10% like more than 1 product. Q2 : There are 70 students in Math or English or German. Exactly 40 are in Math, 30 in German, 35 in English and 15 in all three courses. How many students are enrolled in exactly two of the courses? Math, English and German. Answer: 5 students are enrolled in exactly 2 of the courses.



Manager
Joined: 29 Oct 2009
Posts: 197

Re: Formulae for 3 overlapping sets [#permalink]
Show Tags
31 Oct 2009, 14:35
understudy wrote: This is great stuff. Any idea how to use the above to solve a max / min problem like the following? I've seen some people on here answer it but none of the explanations were that clear.
In a village of 100 households, 75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x – y is:
A. 65
B. 55
C. 45
D. 35
E. 25
yes.. it is possible to relate it to the formula above. consider the following: Set A  dvd players, Set B  cell phones Set C  mp3 players Now we know that the max value of P (A n B n C) = 55 because C is the limiting factor and its value is 55. Now to find the min value of P (A n B n C), we have: 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) 100 = 75 + 80 + 55  [ P (A n B) + P (A n C) + P (B n C) ] + P (A n B n C) P (A n B n C) = [ P (A n B) + P (A n C) + P (B n C) ]  110 for P (A n B n C) to be min, [ P (A n B) + P (A n C) + P (B n C) ] should be min. this just means that we have to find the min possible overlap between AB, BC and AC and then add them together... how do we do this? simple.. first consider A and B. if the no. of houses having only A = 75 and no. of houses having only B = 80, then the total no of houses having either only A or only B = 155. BUT there are only 100 houses.. therefore a min of 55 houses must have both A and B. another way to understand this is to assume the first 75 houses have A and the next 25 houses have B. this still leaves us with a balance of 55 B. since no house can have more than 1B, a min of 55 houses must have both A and B. Similarly, a min of 35 houses must have both B and C, and a min of 30 houses must have both A and C. therefore, min value of [ P (A n B) + P (A n C) + P (B n C) ] = [55 + 30 + 35] = 120 now substitute this in our above equation and we get min value of P (A n B n C) = 10. thus xy = 5510 = 45 which is choice C. Whats the OA?
_________________
Click below to check out some great tips and tricks to help you deal with problems on Remainders! http://gmatclub.com/forum/compilationoftipsandtrickstodealwithremainders86714.html#p651942
Word Problems Made Easy! 1) Translating the English to Math : http://gmatclub.com/forum/wordproblemsmadeeasy87346.html 2) 'Work' Problems Made Easy : http://gmatclub.com/forum/workwordproblemsmadeeasy87357.html 3) 'Distance/Speed/Time' Word Problems Made Easy : http://gmatclub.com/forum/distancespeedtimewordproblemsmadeeasy87481.html



Intern
Joined: 17 Nov 2009
Posts: 35
Schools: University of Toronto, Mcgill, Queens

Re: Formulae for 3 overlapping sets [#permalink]
Show Tags
05 Dec 2009, 03:55
crejoc wrote: seofah wrote: Thanks for putting this thread together, but can we also have some GMAT examples where one can use these concepts? Q1 : In a consumer survey, 85% of those surveyed liked at least one of three products: 1, 2, and 3. 50% of those asked liked product 1, 30% liked product 2, and 20% liked product 3. If 5% of the people in the survey liked all three of the products, what percentage of the survey participants liked more than one of the three products? Answer: 10% like more than 1 product. Q2 : There are 70 students in Math or English or German. Exactly 40 are in Math, 30 in German, 35 in English and 15 in all three courses. How many students are enrolled in exactly two of the courses? Math, English and German. Answer: 5 students are enrolled in exactly 2 of the courses. For Question No. 1 Total = G1+G2+G3  (sum of two products)  2(all three) + neither 85 = 50 + 30 +20  (sum of two products)  2(5) + 0(neither) Let X = Sum of two products therefore, X = 5 For question No. 2 70 = 40+30+352(15)X+0 X = 5
_________________
Action is the foundational key to all success.



Intern
Joined: 22 Dec 2009
Posts: 19

Re: Formulae for 3 overlapping sets [#permalink]
Show Tags
28 Dec 2009, 13:59
understudy wrote: This is great stuff. Any idea how to use the above to solve a max / min problem like the following? I've seen some people on here answer it but none of the explanations were that clear.
In a village of 100 households, 75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x – y is:
A. 65
B. 55
C. 45
D. 35
E. 25
i did it this way... total households = 100 at least 1 dvd player = 75 at least 1 cell phone = 80 at least 1 mp3 player = 55 now households with atleast 1 dvd player and 1 cell phone = (75+80)100= 55 households with atleast 1 dvd player, 1 cell phone and 1 mp3 player = (55+55)100 = 10 so lowest possible number of households with all 3 = 10 max number of house holds with all 3 is equal to the number of households with atleast 1 mp3 player (since it is the smallest in number of the 3) = 55 xy = 55  10 =45 Answer: C can anyone confirm that my method is correct
_________________
Deserve before you Desire



Intern
Joined: 23 Jan 2010
Posts: 28
Schools: Kellogg, Booth, Harvard, Wharton, Stanford
WE 1: Product Strategy
WE 2: Operations
WE 3: Entrepreneurship

Re: Formulae for 3 overlapping sets [#permalink]
Show Tags
27 Feb 2010, 11:33
meganbaxter1 wrote: I still don't understand the possibility of just deriving all of these in the exam rather than remembering them by heart!
can any one help me with this one:
There are 150 students at Seward High School. 66 students play baseball, 45 play basketball, and 42 play soccer. 27 students play exactly two sports, and three students play all three of the sports. How many of the 150 students play none of the three sports?
A) 0
B) 27
C) 30
D) 99
E) 78 Total Students = GroupA + GroupB +GroupC Exactly Two 2(All Three) + None 150= 66 + 45 + 24 27  2(3) + X X = 30



Director
Joined: 25 Aug 2007
Posts: 837
WE 1: 3.5 yrs IT
WE 2: 2.5 yrs Retail chain

Re: Formulae for 3 overlapping sets [#permalink]
Show Tags
18 Apr 2010, 11:13
Bullet wrote: crejoc wrote: seofah wrote: Thanks for putting this thread together, but can we also have some GMAT examples where one can use these concepts? Q1 : In a consumer survey, 85% of those surveyed liked at least one of three products: 1, 2, and 3. 50% of those asked liked product 1, 30% liked product 2, and 20% liked product 3. If 5% of the people in the survey liked all three of the products, what percentage of the survey participants liked more than one of the three products? Answer: 10% like more than 1 product. Q2 : There are 70 students in Math or English or German. Exactly 40 are in Math, 30 in German, 35 in English and 15 in all three courses. How many students are enrolled in exactly two of the courses? Math, English and German. Answer: 5 students are enrolled in exactly 2 of the courses. For Question No. 1 Total = G1+G2+G3  (sum of two products)  2(all three) + neither 85 = 50 + 30 +20  (sum of two products)  2(5) + 0(neither) Let X = Sum of two products therefore, X = 5 For question No. 2 70 = 40+30+352(15)X+0 X = 5 For Q1, we need to find those who liked more than 1 product that mean people who liked 2 or 3 products: p(1U2U3) = p(1) + p(2) + p(3)  p(1n2)  p(2n3)  p(1n3) + p (1n2n3) 85=50+30+20 [p(1n2) + p(2n3) + p(1n3)] + p (1n2n3) [p(1n2) + p(2n3) + p(1n3)] = 25. Please post OA and OE.
_________________
Want to improve your CR: http://gmatclub.com/forum/crmethodsanapproachtofindthebestanswers93146.html Tricky Quant problems: http://gmatclub.com/forum/50trickyquestions92834.html Important Grammer Fundamentals: http://gmatclub.com/forum/keyfundamentalsofgrammerourcruciallearningsonsc93659.html



Manager
Joined: 04 May 2009
Posts: 119
Location: London
Schools: Haas (WL), Kellogg (matricultating), Stanford (R2, ding), Columbia (ding)
WE 1: 3 years hotel industry sales and marketing France
WE 2: 3 years financial industry marketing UK

Re: Formulae for 3 overlapping sets [#permalink]
Show Tags
24 May 2010, 13:21
Can someone confirm whether this formula is true? Total = Group1 + Group2 + Group3  (sum of 2group overlaps)  2*(all three) + Neither I am getting confused with all different formulas and have seen people adding the 2*(all three) rather than substracting....
_________________
Yes I can!



Math Expert
Joined: 02 Sep 2009
Posts: 46991

Formulae for 3 overlapping sets [#permalink]
Show Tags
25 May 2010, 06:48
nifoui wrote: Can someone confirm whether this formula is true? Total = Group1 + Group2 + Group3  (sum of 2group overlaps)  2*(all three) + Neither
I am getting confused with all different formulas and have seen people adding the 2*(all three) rather than substracting.... I'd advise to understand the concept rather than to memorize formulas. Look at the diagram below: Attachment:
Union_3sets.gif [ 11.63 KiB  Viewed 186248 times ]
FIRST FORMULAWe can write the formula counting the total as: \(Total=A+B+C(AnB+AnC+BnC)+AnBnC+Neither\). When we add three groups A, B, and C some sections are counting more than once. For instance: sections 1, 2, and 3 are counted twice and section 4 thrice. Hence we need to subtract sections 1, 2, and 3 ONCE (to get one) and subtract section 4 TWICE (again to count section 4 only once). Now, in the formula AnB means intersection of A and B (sections 3 and 4 on the diagram). So when we are subtracting \(AnB\) (3 an4), \(AnC\) (1 and 4), and \(BnC\) (2 and 4) from \(A+B+C\), we are subtracting sections 1, 2, and 3 ONCE BUT section 4 THREE TIMES (and we need to subtract section 4 only twice), therefor we should add only section 4, which is intersection of A, B and C (AnBnC) again. That is how thee formula \(Total=A+B+C(AnB+AnC+BnC)+AnBnC+Neither\) is derived. SECOND FORMULAThe second formula you are referring to is: \(Total=A+B+C \){Sum of Exactly 2 groups members} \( 2*AnBnC + Neither\). This formula is often written incorrectly on forums as Exactly 2 is no the same as intersection of 2 and cannot be written as AnB. Members of exactly (only) A and B is section 3 only (without section 4), so members of A and B only means members of A and B and not C. This is the difference of exactly (only) A and B (which can be written, though not needed for GMAT, as \(AnBC\)) from A and B (which can be written as \(AnB\)) Now how this formula is derived? Again: when we add three groups A, B, and C some sections are counting more than once. For instance: sections 1, 2, and 3 are counted twice and section 4 thrice. Hence we need to subtract sections 1, 2, and 3 ONCE (to get one) and subtract section 4 TWICE (again to count section 4 only once). When we subtract {Sum of Exactly 2 groups members} from A+B+C we are subtracting sum of sections 1, 2 and 3 once (so that's good) and next we need to subtracr ONLY section 4 (\(AnBnC\)) twice. That's it. Now, how this concept can be represented in GMAT problem? Example #1:Workers are grouped by their areas of expertise, and are placed on at least one team. 20 are on the marketing team, 30 are on the Sales team, and 40 are on the Vision team. 5 workers are on both the Marketing and Sales teams, 6 workers are on both the Sales and Vision teams, 9 workers are on both the Marketing and Vision teams, and 4 workers are on all three teams. How many workers are there in total?Translating: "are placed on at least one team": members of none =0; "20 are on the marketing team": M=20; "30 are on the Sales team": S=30; "40 are on the Vision team": V=40; "5 workers are on both the Marketing and Sales teams": MnS=5, note here that some from these 5 can be the members of Vision team as well, MnS is sections 3 an 4 on the diagram (assuming Marketing=A, Sales=B and Vision=C); "6 workers are on both the Sales and Vision teams": SnV=6 (the same as above sections 2 and 4); "9 workers are on both the Marketing and Vision teams": MnV=9. "4 workers are on all three teams": MnSnV=4, section 4. Question: Total=? Applying first formula as we have intersections of two groups and not the number of only (exactly) 2 group members. Total=M+S+V(MnS+SnV+SnV)+MnSnV+Neither=20+30+40(5+6+9)+4+0=74. Answer: 74. Example #2:Each of the 59 members in a high school class is required to sign up for a minimum of one and a maximum of three academic clubs. The three clubs to choose from are the poetry club, the history club, and the writing club. A total of 22 students sign up for the poetry club, 27 students for the history club, and 28 students for the writing club. If 6 students sign up for exactly two clubs, how many students sign up for all three clubs?Translating: "Each of the 59 members in a high school class is required to sign up for a minimum of one and a maximum of three academic clubs" Total=59; Neither=0 (as members are required to sign up for a minimum of one); "22 students sign up for the poetry club": P=22; "27 students for the history club": H=27; "28 students for the writing club": W=28; "6 students sign up for exactly two clubs": {Exactly 2 groups members}=6, so sum of sections 1, 2, and 3 is given to be 6, (among these 6 students there are no one who is the member of ALL 3 clubs) "How many students sign up for all three clubs": question is \(PnHnW=x\). Or section 4 =? Apply second formula: \(Total=P+H+W \){Sum of Exactly 2 groups members}\(2*PnHnW + Neither\) > \(59=22+27+2862*x+0\) > \(x=6\). Answer: 6. Similar problem at: psquestion94457.html#p728852Hope it helps.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 04 May 2009
Posts: 119
Location: London
Schools: Haas (WL), Kellogg (matricultating), Stanford (R2, ding), Columbia (ding)
WE 1: 3 years hotel industry sales and marketing France
WE 2: 3 years financial industry marketing UK

Re: Formulae for 3 overlapping sets [#permalink]
Show Tags
25 May 2010, 12:47
Thanks Bunuel for your very clear explanation, this is very helpful, as always!
_________________
Yes I can!



Intern
Joined: 23 May 2010
Posts: 7

Re: Formulae for 3 overlapping sets [#permalink]
Show Tags
04 Jun 2010, 18:48
AWESOME POST Bunuel....I cant thankyou enough



Manager
Joined: 24 Jan 2010
Posts: 110
Location: India
Schools: ISB

Re: Formulae for 3 overlapping sets [#permalink]
Show Tags
24 Jul 2010, 23:33
Thanks Bunuel for your very clear explanation.
All your explanations/tips are really very helpful.
+1 Kudos



Math Expert
Joined: 02 Sep 2009
Posts: 46991

Re: Formulae for 3 overlapping sets [#permalink]
Show Tags
02 Sep 2010, 11:19
harshamadabal wrote: Hi All,
Am a new chap here....
I did not quite understand the below sentence from A+B+C, we are subtracting sections 1, 2, and 3 ONCE BUT section 4 THREE TIMES (and we need to subtract section 4 only twice)
i understood that section 4 is deleted thrice, but why should it be deleted twice..it should be deleted only once right? Hi, and welcome to Gmat Club! Formula is: \(Total=A+B+C(AnB+AnC+BnC)+AnBnC+Neither\). Not to have double counting of section 4 we should count it only once. A+B+C counts it 3 times (so to count section 4 once we should subtract it twice), but then when we subtract (AnB+AnC+BnC) we subtract section 4 three times so at this stage we haven't counted it at all, so we should add it once and that' why we have +AnBnC part. Hope it's clear.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 31 Aug 2010
Posts: 1

Re: Formulae for 3 overlapping sets [#permalink]
Show Tags
02 Sep 2010, 20:04
its clear now ...thanks Bunuel



Manager
Joined: 20 Apr 2010
Posts: 185
Schools: ISB, HEC, Said

Re: Formulae for 3 overlapping sets [#permalink]
Show Tags
16 Dec 2010, 06:19
Nice explaination Thanks



Manager
Joined: 18 Aug 2010
Posts: 83

Re: Formulae for 3 overlapping sets [#permalink]
Show Tags
02 Feb 2011, 01:48
Bunuel wrote: I'd advise to understand the concept rather than to memorize formulas. Look at the diagram below: Attachment: Union_3sets.gif FIRST FORMULAWe can write the formula counting the total as: \(Total=A+B+C(AnB+AnC+BnC)+AnBnC+Neither\). ... +Kudo. very clear explanation .



Manager
Joined: 17 Dec 2010
Posts: 91
Location: Australia
GPA: 3.37
WE: Engineering (Consulting)

Re: Formulae for 3 overlapping sets [#permalink]
Show Tags
13 Feb 2011, 00:01
I've got a curly one  which just isn't working out using the formulas (I'm sure I'm doing something wrong though!) Here's the question: The 38 movies in the video store fall into the following 3 categories: 10 action, 20 drama, and 18 comedy. However, some movies are classified under more than one category: 5 are both action & drama, 3 are both action & comedy, and 4 are both drama and comedy. How many actiondramacomedies are there?Solving it via firstprinciples gives me the correct answer  but it takes me something like 4 mins  which is way too slow. Using the formula, I can solve these ones in just under 90 seconds....but this one just plain isn't working out with me: For this question, I am using the following info: A = 10, B = 18, C = 20 Sum of exactly 2 groups = 5 + 3 + 4 = 12 x = ? T = 38 By using the formula: 38 = 10 + 18 + 20  (12)  2x 38 = 48  12  2x 38 = 36  2x x = 1 ????? Any help?
_________________
Kudos always appreciated if my post helped you



Math Expert
Joined: 02 Sep 2009
Posts: 46991

Re: Formulae for 3 overlapping sets [#permalink]
Show Tags
13 Feb 2011, 03:41
MackyCee wrote: I've got a curly one  which just isn't working out using the formulas (I'm sure I'm doing something wrong though!)
Here's the question:
The 38 movies in the video store fall into the following 3 categories: 10 action, 20 drama, and 18 comedy. However, some movies are classified under more than one category: 5 are both action & drama, 3 are both action & comedy, and 4 are both drama and comedy. How many actiondramacomedies are there?
Solving it via firstprinciples gives me the correct answer  but it takes me something like 4 mins  which is way too slow. Using the formula, I can solve these ones in just under 90 seconds....but this one just plain isn't working out with me:
For this question, I am using the following info: A = 10, B = 18, C = 20 Sum of exactly 2 groups = 5 + 3 + 4 = 12 x = ? T = 38
By using the formula:
38 = 10 + 18 + 20  (12)  2x 38 = 48  12  2x 38 = 36  2x x = 1
?????
Any help? There is a reason for two formulas in my post. 5 are both action and drama DOES NOT mean that 5 movies are ONLY action and drama, among these 5 might be some which are comedy. Similarly for: "3 are both action and comedy, and 4 are both drama and comedy" So you should apply first formula: {Total} = {action} + {drama} + {comedy}  {action&drama+action&comedy+drama&comedy} + {action&drama&comedy} > 38=10+20+18(5+3+4)+{action&drama&comedy} > {action&drama&comedy}=2. You can apply second formula as well but you should make necessary modification: let {action&drama&comedy}=x, then: ONLY action and drama = (5x); ONLY action and comedy = (3x); ONLY drama and comedy = (4x); {Total} = {action} + {drama} + {comedy}  {EXACTLY two movies}  2*{action&drama&comedy} > 38=10+20+18(5x+3x+4x)2x > x=2. Hope it's clear.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Re: Formulae for 3 overlapping sets
[#permalink]
13 Feb 2011, 03:41



Go to page
1 2 3
Next
[ 50 posts ]



