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# Equations for 3 overlapping sets

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Intern
Joined: 30 May 2013
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Equations for 3 overlapping sets [#permalink]

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21 Jun 2013, 21:28
Q1. In a group of 68 students, each student is registered for at least one of three classes - History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes?

Q2. At a certain dealership, every car on the lot has at least one of three modest options: windows, brakes and radio. 40 cars have windows, 30 cars have brakes, and 50 cars have a radio. 21 cars have B&R, 13 cars have B&W and 17 cars have W&R. If 11 cars have all 3 options, what is the total number of cars on the lot?

For Q1, we are applying the following equation for 3 overlapping sets:

Total = H + M + E - (HM + ME + HE) - 2(HME)
68 = 25 + 25 + 34 - Sum - 6
Sum = 10

While for Q2, we are applying a different equation for 3 overlapping sets:

Total = B + W + R - (BR + BW + WR) + BWR
Total = 40 + 30 + 50 -21 - 13 - 17 + 11
Total = 80

Please explain the different scenarios/equations and when they must be applied...

Thanks,
Intern
Joined: 30 May 2013
Posts: 13
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Re: Equations for 3 overlapping sets [#permalink]

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22 Jun 2013, 16:06
I got it.. found it in another related post and found it extremely easy to follow. I am pasting that post below:

Bunuel wrote:
nifoui wrote:
Can someone confirm whether this formula is true?

Total = Group1 + Group2 + Group3 - (sum of 2-group 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:

FIRST FORMULA
We 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 FORMULA
The 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 $$AnB-C$$) 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.

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);
"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}$$-2*PnHnW + Neither$$ --> $$59=22+27+28-6-2*x+0$$ --> $$x=6$$.

Similar problem at: ps-question-94457.html#p728852

Hope it helps.

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Re: Equations for 3 overlapping sets   [#permalink] 22 Jun 2013, 16:06
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