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13 Feb 2011, 05:18
Bunuel  crystal clear mate You are a champion!
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22 May 2011, 08:57
Damn! Amazing explanations Bunuel! +1 every time.



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27 May 2011, 11:24
source gclub3 Q7. http://gmatclub.com/tests/m03#expl7Foodmart customers regularly buy at least one of the following products: milk, chicken, or apples. 60% of shoppers buy milk, 50% buy chicken, and 35% buy apples. If 10% of the customers buy all 3 products, what percentage of Foodmart customers purchase exactly 2 of the products listed above? 5% 10% 15% 25% 30% Alternative Explanation: Assume there are 100 individual buyers. Let ci represent the customers who regularly buy 1, 2, or 3 products, where ci is the number of products. Multiply c2 by 2 and c3 by 3 to accurately represent the number of times these buyers were counted since they have purchased chicken and apples or milk, chicken, and apples but in reality are an individual buyer counted multiple times. Construct the following equations: c1+2c2+3c3 = 145 c1+c2+c3 = 100      c2+2c3 = 45 (Subtract the second equation from the first one:) given c3= 10% hence c2=25% Kindly refer to this link>http://gmatclub.com/tests/m03#expl7 if my solution is not clear(login for gmat club tests.) The correct answer is D. Query Why does the above logic does not hold in all examples discussed in this forum. I found this alternate method a great shortcut and even applied it to a couple questions. 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: this formula worked. 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: this formula worked. 3>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 Answer: could not apply this shortcut 4>ALL of the cofee mixtures sold in a certain store contain either a Colombian, Jamaican or Brazilian cofee or some combinations of these. Of all the mixtures 33 contain Colombian cofee, 43 contain Jamaican cofee and 42 contain Brazilian cofee. Of these, 16 contain atleast Colombian and Jamaican cofees, 18 contain atleast Jamaican and Brazilian cofees, 8 contain atleast Brazilian and colombian cofees and 5 contain all three. How many different mixtures are sold in the store ? Answer: this formula failed where it really looked like it would work. 5>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? formula failed again. Please help. Bunuel ?? anybody on this.
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30 May 2011, 14:48
bblast wrote: source gclub3 Q7. http://gmatclub.com/tests/m03#expl7Foodmart customers regularly buy at least one of the following products: milk, chicken, or apples. 60% of shoppers buy milk, 50% buy chicken, and 35% buy apples. If 10% of the customers buy all 3 products, what percentage of Foodmart customers purchase exactly 2 of the products listed above? 5% 10% 15% 25% 30% Second formula: Total = n(A)+n(B)+n(C)(Sum of Exactly two)2*n(All)+n(None) 100 = 60+50+35SoE22*10+0 SoE2 = 125100=25% Ans: "D"bblast wrote: Query Why does the above logic does not hold in all examples discussed in this forum. I found this alternate method a great shortcut and even applied it to a couple questions.
Yes, you are right!! We can't just apply the formula and get the result in some of these cases. bblast wrote: 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: this formula worked.
Second formula: Total = n(A)+n(B)+n(C)(Sum of Exactly two)2*n(All)+n(None) 85 = 50+30+20SoE22*5 SoE2=9085=5 5% like exactly 2 products. 5% like all 3. 5+5=10% like more than 1 product.
Ans: 10%bblast wrote: 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: this formula worked.
Second formula: 70 = 40+30+35SoE22*15 SoE2 = 5
Ans: 5bblast wrote: 3>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 Answer: could not apply this shortcut
True, I couldn't apply the formula either. Neither am I aware of any formula that can directly solve this. Anyone? I just used some Venn like pictures to understand and get at the result.
Max intersection is easy: 55  Think it as three concentric circles. Innermost circle:55 Intermediate: 75 Outermost: 80
Min intersection was tricky: Make 8020 region as two distinct entity. We need to fit: 55+75=130 in this 8020 disjointed region. We can lay over two products' count on the 20 region: Managed to get rid of 40 that way. Rest is 80 region: We should just layover one product over this 80 region. 13040=90 This is 10 more than 80. Thus, 10 is the minimum intersection within 80 region that we will have to use. Difference: 5510=45
Ans: 45bblast wrote: 4>ALL of the cofee mixtures sold in a certain store contain either a Colombian, Jamaican or Brazilian cofee or some combinations of these. Of all the mixtures 33 contain Colombian cofee, 43 contain Jamaican cofee and 42 contain Brazilian cofee. Of these, 16 contain atleast Colombian and Jamaican cofees, 18 contain atleast Jamaican and Brazilian cofees, 8 contain atleast Brazilian and colombian cofees and 5 contain all three. How many different mixtures are sold in the store ? Answer: this formula failed where it really looked like it would work.
Used top down approach. All=5 C+J=16; Only C+J=165=11 J+B=18; Only J+B=185=13 C+B=8; Only C+B=85=3 C=33; Only C = 33(Only C+J)(Only C+B)All=331135=14 J=43; Only J = 43(Only C+J)(Only J+B)All=4311135=14 B=42; Only B = 42(Only J+B)(Only C+B)All=421335=21
Total = 5+11+13+3+14+14+21=81 Ans: 81bblast wrote: 5>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?
Second formula: 150=66+45+42272*3+None None=150120=30 Ans: 30
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07 Jun 2011, 04:52
can som1 please explain why use the 2nd formula in the next 2 cases: the first demands exactly 2 so it's clear why using the second, but the 2nd question deals with all classes?
In a group of 68 students, each student is registered for at least one of three classes – History, Math and English. Twentyfive students are registered for History, twentyfive students are registered for Math, and thirtyfour students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes? 13 10 9 8 7
4. 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?



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07 Jun 2011, 05:03
thats what i tried to realize: is it true to say that when u look for ALL or EXACTLY 2 use the 2nd formula and when you look for Neither or Total use the 1st.
in the case of the coffee: 43+42+33(16+18+8) +5 gives 81 [quote][/quote]bblast wrote: 4>ALL of the cofee mixtures sold in a certain store contain either a Colombian, Jamaican or Brazilian cofee or some combinations of these. Of all the mixtures 33 contain Colombian cofee, 43 contain Jamaican cofee and 42 contain Brazilian cofee. Of these, 16 contain atleast Colombian and Jamaican cofees, 18 contain atleast Jamaican and Brazilian cofees, 8 contain atleast Brazilian and colombian cofees and 5 contain all three. How many different mixtures are sold in the store ? Answer: this formula failed where it really looked like it would work.
Used top down approach. All=5 C+J=16; Only C+J=165=11 J+B=18; Only J+B=185=13 C+B=8; Only C+B=85=3 C=33; Only C = 33(Only C+J)(Only C+B)All=331135=14 J=43; Only J = 43(Only C+J)(Only J+B)All=4311135=14 B=42; Only B = 42(Only J+B)(Only C+B)All=421335=21
Total = 5+11+13+3+14+14+21=81



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Re: Formulae for 3 overlapping sets [#permalink]
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07 Jun 2011, 05:48
dimri10 wrote: thats what i tried to realize: is it true to say that when u look for ALL or EXACTLY 2 use the 2nd formula and when you look for Neither or Total use the 1st.
Please pick a formula only after fully understanding what the question is trying to ask rather than relying on some keyword. Maybe you can use this methodology as a last resort. This post by Bunuel clears why these formulas should be used: formulaefor3overlappingsets69014.html#p729340Please let us know if you don't understand anything in that.
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Re: Formulae for 3 overlapping sets [#permalink]
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25 Jun 2011, 09:33
Great explanations from Bunuel....Overlapping sets are much clearer now!



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18 Jul 2011, 13:39
Excellent Explanation Bunuel.. Thanks!!



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24 Jul 2011, 02:11
Hi, can anyone help me solve this DS question... thanks..
Of the 200 members of an association, each member who speaks German also speaks English and 70 of the members speak only Spanish. If no member speaks all three languages, how many of the members speak two of the three languages.
1. 60 of the members only speak English. 2. 20 of the members do not speak any of the three languages.
Answer: Formulae  Total = A+B+C  (Exactly 2 members)  2*(Members in all three) + Neither Now, values given Total = 200 German = A English = B Spanish = C = 70 (AnBnC) = 0, According to the Question, each member who speaks german also speaks english , does this indicate A = B ?? I am not sure about this..
According to statement 1, B= 60 ==> not sufficient According to statement 2, Neither = 10 ==> not sufficient Both statements together , if A=B, the correct answer is C..



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07 Aug 2011, 04:26
good explanations



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10 Sep 2011, 05:22
Quote: Of the 200 members of an association, each member who speaks German also speaks English and 70 of the members speak only Spanish. If no member speaks all three languages, how many of the members speak two of the three languages.
1. 60 of the members only speak English. 2. 20 of the members do not speak any of the three languages.
This is how Venn Diagram will look like. Attachment:
Screen Shot 20110910 at 3.05.10 PM.png [ 50.52 KiB  Viewed 12492 times ]
3 things to notice: 1) All those who speak German speak English, represented by smaller circle inside bigger circle and denoted as y. 2) Some who speak English also speak Spanish, represented by x. 3) No one speaks all three language and hence only two circles overlap. Given: Total number of people=200 Only Spanish = 70 German and English=y English and Spanish=x To Find x+y. 1) Only English = 60. This means English circle  x  y = 60 Not sufficient to know x+y. 2) 20 speaks neither. Total=200 Hence 20020=180 speaks one of the language. That is English Circle+Spanish Circle= 180 Not sufficient to answer x+y Together, From 2, Total who speaks atleast one language= 180, Given Only Spanish = 70 Hence Only English + y + x = 18070=110 From 1 Only English = 60 x+y+60=110 x+y=11060 = 50 OA C.
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07 Oct 2011, 08:55
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Bunuel !!! Very Good explanation.



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23 Nov 2011, 09:02
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fluke wrote: bblast wrote: 3>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 Answer: could not apply this shortcut
True, I couldn't apply the formula either. Neither am I aware of any formula that can directly solve this. Anyone?I just used some Venn like pictures to understand and get at the result. Max intersection is easy: 55  Think it as three concentric circles. Innermost circle:55 Intermediate: 75 Outermost: 80 Min intersection was tricky: Make 8020 region as two distinct entity. We need to fit: 55+75=130 in this 8020 disjointed region. We can lay over two products' count on the 20 region: Managed to get rid of 40 that way. Rest is 80 region: We should just layover one product over this 80 region. 13040=90 This is 10 more than 80. Thus, 10 is the minimum intersection within 80 region that we will have to use. Difference: 5510=45 We use the second formula that Bunuel mentioned: \(Total=D+C+M \){Sum of Exactly 2 groups members} \( 2*DnCnM + Neither\)"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" > Everyone has a device so: Neither = 0 "In a village of 100 households" > \(T=100\) "75 have at least one DVD player" > \(D=75\) "80 have at least one cell phone" > \(C=80\) "55 have at least one MP3 player" > \(M=55\) "If x and y are respectively the greatest and lowest possible number of households that have all three of these devices" > \(DnCnM= x or y\) So: \(2*DnCnM=D+C+M Total \) {Sum of Exactly 2 groups members} > \(DnCnM=55\) {Sum of Exactly 2 groups members}/2In order for DnCnM to be maximum {Sum of Exactly 2 groups members} has to be minimum. It will be minimum when it is equal with \(0\). Therefore \(x=55\). In order for \(DnCnM\) to be minimum {Sum of Exactly 2 groups members} has to be Maximum. The value of {Sum of Exactly 2 groups members} will depend on \(DnCnM\). In order to be maximum: {Sum of Exactly 2 groups members}+\(DnCnM=100\) Therefore {Sum of Exactly 2 groups members}\(=100DnCnM\) Therefore, \(DnCnM=10 > y=10\) \(xy=45\)



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Re: Formulae for 3 overlapping sets [#permalink]
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30 Jan 2012, 07:22
Bullet wrote: ALL of the cofee mixtures sold in a certain store contain either a Colombian, Jamaican or Brazilian cofee or some combinations of these. Of all the mixtures 33 contain Colombian cofee, 43 contain Jamaican cofee and 42 contain Brazilian cofee. Of these, 16 contain atleast Colombian and Jamaican cofees, 18 contain atleast Jamaican and Brazilian cofees, 8 contain atleast Brazilian and colombian cofees and 5 contain all three. How many different mixtures are sold in the store ? (A) 71 (B) 81 (C) 109 (D) 118 (E) 165 OA is B
Total = P(A) + P(B) + P(C) (P(AnB) + P(AnC) + P(BnC)) + p(AnBnC) = 33 + 43 + 42  (16+18+8) + 5 = 81 If we replace 'atleast' with 'atmost' then it changes the question quite a lot.  Of these, 16 contain atmost Colombian and Jamaican cofees, 18 contain atmost Jamaican and Brazilian cofees, 8 contain atmost Brazilian and colombian cofees and 5 contain all three.



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Re: Formulae for 3 overlapping sets [#permalink]
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Updated on: 20 Apr 2012, 12:39
Hello Bunuel 
Quick question on the following Overlapping Set Example 
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.
Why didn't you deduct  (2*4). I thought answer would be  62 (20+30+40(5+6+9)2(4)+0 = 62. Please explain.
Originally posted by rpamecha on 20 Apr 2012, 12:28.
Last edited by rpamecha on 20 Apr 2012, 12:39, edited 1 time in total.



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Re: Formulae for 3 overlapping sets [#permalink]
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20 Apr 2012, 12:34
rpamecha wrote: Hello Bunuel 
Quick question on the following Overlapping Set Example 
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 21045 times ]
FIRST FORMULA We can write the formula counting the total as: .
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 (3 an4), (1 and 4), and (2 and 4) from , 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 is derived.
SECOND FORMULA The second formula you are referring to is: {Sum of Exactly 2 groups members} . This formula is often written incorrectly on forums as Exactly 2 is no the same as intersection of 2 and can not 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 ) from A and B (which can be written as )
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 () 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.
Why didn't you dedcut  (2*4). I thought answer would be  62 (20+30+40(5+6+9)2(4)+0 = 62. Please explain. For this question we are applying the first formula. All is explained here: formulaefor3overlappingsets69014.html#p729340P.S. Please format the posts correctly, so that it's clear what are you quoting and asking.
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Re: Formulae for 3 overlapping sets [#permalink]
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01 May 2012, 06:07
Bunuel wrote: 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 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 can not 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. Thanx for the concept and formula ! what about the other formulas given above by amitdgr? it would be nice if your expert comment and guidance could be given. throughout the thread, different formulas are are applied on same questions. Different answers .. OA are not clearly given.



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Re: Formulae for 3 overlapping sets [#permalink]
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06 May 2012, 03:48
Dear Buuel,
please help me with thos question 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?
According to your posts the formaula requred here is FIRST FORMULA We can write the formula counting the total as: .
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 (3 an4), (1 and 4), and (2 and 4) from , 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 is derived.
from that answer derived is 20 and not 10.
Require your help.
Thanks




Re: Formulae for 3 overlapping sets
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06 May 2012, 03:48



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