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Set Theory

Set Theory

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Set Notations
Types of Sets
Set Operations & Venn Diagram
2 Set Venn Diagrams
3 Set Venn Diagrams
4 Set Venn Diagrams
Maximum and Minimum
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Set Theory - 1
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Set Theory : Level 1
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CAT 2025 Lesson : Set Theory - 3 Set Venn Diagrams

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4. Venn Diagrams with 3 Sets

The following is a Venn Diagram with
333 sets - A, B and C. We typically write the name of the set just outside that circle in capital letters. In small letters from aaa to hhh are the numbers of elements in each of the 888 different regions formed. For instance, ddd is the number of elements that are in sets A and B but not in set C.

nnn(A) =a+d+f+g= a + d + f + g=a+d+f+g
nnn(B) =b+d+e+g= b + d + e + g=b+d+e+g
nnn(C) =c+e+f+g= c + e + f + g=c+e+f+g

Let P, Q and R be the number of elements in exactly 1 set, exactly 2 sets and all 3 sets respectively.

P
=a+b+c= a + b + c=a+b+c
Q
=d+e+f= d + e + f=d+e+f
R
=g= g=g



nnn(A ∪\cup∪ B ∪\cup∪ C) =a+b+c+d+e+f+g= a + b + c + d + e + f + g=a+b+c+d+e+f+g
⇒
n\bm{n}n(A  ∪  \ \bm{\cup} \  ∪  B  ∪ \ \bm{\cup} \  ∪  C) === P + Q + R    \space\space\space    (Property 1)

nnn(A) +n+ n+n(B) +n+ n+n(C) =a+b+c+2(d+e+f)+3g= a + b + c + 2(d + e + f) + 3g=a+b+c+2(d+e+f)+3g
⇒
n\bm{n}n(A) +++ n\bm{n}n(B) +++ n\bm{n}n(C) === P + 2Q + 3R    \space\space\space    (Property 2)

Note that
nnn(u) = nnn(A ∪\cup∪ B ∪\cup∪ C) +h+ h+h
⇒
n\bm{n}n(U) === P + Q + R + h    \space\space\space    (Property 3)

Example 7

A survey was conducted in a neighbourhood of 700700700 residents to assess the readership of 333 newspapers - The Hindu (TH), The Indian Express (IE) and The Times of India (TOI). Every resident reads at least one of these papers. At the end of the survey it was found that 240240240 read TH, 400400400 read IE, 500500500 read TOI and 100100100 read all 333 newspapers.

(I) How many read exactly
111 paper?
(II) How many read at least
222 of these newspapers?

Solution

Let P, Q and R be the number of people who read exactly 111, exactly 222 and all 333 newspapers respectively.
As
100100100 residents read all 333 newspapers, R = 100.

As every resident reads at least one of the newspapers,
nnn(U) === P +++ Q +++ R === 700700700
⇒ P
+++ Q =600= 600=600 -----(1)
nnn(TH) +++ nnn(IE) +++ nnn(TOI) =240+400+500= 240 + 400 + 500=240+400+500
⇒ P
+++ 222Q +++ 333R =1140= 1140=1140
⇒ P
+++ 222Q =840= 840=840 -----(2)

Solving (1) and (2), we get P = 360 & Q = 240

Case I: Number of residents who read exactly
111 paper === P =360= 360=360
Case II: Number of people reading at least
222 newspapers === Q +++ R =340= 340=340

Answer: (I)
360360360; (II) 340340340


Example 8

Every student plays at least one of 333 sports - cricket, football and hockey. 40%40 \%40% play cricket, 44%44 \%44% play football and 60%60 \%60% play hockey. 14%14 \%14% play cricket and football, 20%20 \%20% play football and hockey, while 16%16 \%16% hockey and cricket.
(I) What percentage of the students play all sports?
(II) If x is the number of students who play hockey but not football, then what percent of x play cricket?
(III) What percent of students play hockey or cricket but not both?

Solution

Let C, F and H represent the sets of students who play Cricket, Football and Hockey respectively.
Let P, Q and R be the number of people who play exactly 111, exactly 222 and all 333 sports respectively.
Let
100100100 be the total number of students. We can now write the percentages as absolute values.

As every student plays at least one of the sports,
nnn(U) =n= n=n(C ∪\cup∪ F ∪\cup∪ H) =100= 100=100
⇒ P
+++ Q +++ R =100= 100=100 -----(1)

P
+++ 222Q +++ 333R
=n= n=n(C) +++ nnn(F) +++ nnn(H)
=40+44+60= 40 + 44 + 60=40+44+60
⇒ P
+++ 222Q +++ 333R=144 = 144=144 -----(2)

As
141414 play cricket & football ⇒ d+g=14d + g = 14d+g=14 -----(4)
As
202020 play football & hockey ⇒ e+g=20e + g = 20e+g=20 -----(5)
As
161616 play hockey & cricket ⇒ f+g=16f + g = 16f+g=16 -----(6)
Adding these, d +++ e +++ f +++ 333g =50= 50=50
⇒ Q
+++ 333R =50= 50=50 -----(3)
Solving equations (1), (2) & (3), we get P = 62, Q = 32 & R = 6.
R
=g=6 = \bm{g = 6}=g=6
Substituting these in (4), (5) and (6), we get
d=8,e=14,f=10\bm{d = 8, e = 14, f = 10}d=8,e=14,f=10

Substituting these values,
a=40−d−f−g=16\bm{a} = 40 - d - f - g = \bm{16}a=40−d−f−g=16
b=44−d−e−g=16\bm{b} = 44 - d - e - g = \bm{16}b=44−d−e−g=16
c=60−e−f−g=30\bm{c} = 60 - e - f - g = \bm{30}c=60−e−f−g=30

Case I:
6%6 \%6% of the students play all games.
Case II: Number of students who play hockey but not football
=x=c+f=30+10=40= x = c + f = 30 + 10 = 40=x=c+f=30+10=40
Number of students in these who play cricket
=f=10= f = 10=f=10
Percent of Cricket players in
x=1040×100=25%x = \dfrac{10}{40} \times 100 = 25 \%x=4010​×100=25%
Case III: Students playing hockey or cricket but not both
=a+d+c+e=68%= a + d + c + e = 68 \%=a+d+c+e=68%

Answer: (I)
6%6 \%6%; (II) 25%25 \%25%; (III) 68%68 \%68%


Example 9

There are 200200200 students studying in Josh College of Management, which has 333 clubs - A, B and C. Students are allowed to be a member in more than 111 club. 20%20 \%20% of the students are a member in all clubs. 303030 students are not a member of Club A alone. Clubs B and C have 100100100 and 120120120 students as members respectively.

(I) How many students are not members in clubs B or C?

(1)
303030            (2) 404040            (3) 505050            (4) Cannot be determined           

(II) How many students are not in club A?

(1)
303030            (2) 404040            (3) 505050            (4) Cannot be determined           

(III) What percentage of the students who passed in C did not pass in B?

(1)
41.7%41.7 \%41.7%            (2) 58.3%58.3 \%58.3%            (3) 66.7%66.7 \%66.7%            (4) Cannot be determined           

(IV) At most how many students are in exactly one of A or C?

(1)
100100100            (2) 130130130            (3) 160160160            (4) Cannot be determined           

Solution

This case, despite requiring a
333-set venn diagram, has very little information.
Therefore, we shall start by drawing a venn diagram
with the
888 different regions marked with variables from aaa to hhh as follows.

Members in all
333 clubs
=== g\bm{g}g =20100×200== \dfrac{20}{100} \times 200 ==10020​×200= 40\bm{40}40

If
303030 members are not in Club A alone,
then they are definitely members in both Clubs B and C. Therefore,
e=\bm{e} =e= 30\bold{30}30



B === b+d+e+g=100b + d + e + g = 100b+d+e+g=100 ⇒ b+d+30+40=100 b + d + 30 + 40 = 100b+d+30+40=100
⇒
b+d\bm{b + d}b+d =30= \boldsymbol{30}=30

C
=c+f+e+g=120= c + f + e + g = 120=c+f+e+g=120 ⇒ c+f+30+40=120c + f + 30 + 40 = 120c+f+30+40=120
⇒
c+f\bm{c + f}c+f =50= \boldsymbol{50}=50

U
=a+b+c+d+e+f+g+h=200= a + b + c + d + e + f + g + h = 200=a+b+c+d+e+f+g+h=200
⇒ B
+a+f+c+h=200+ a + f + c + h = 200+a+f+c+h=200
⇒
100+a+50+h=200100 + a + 50 + h = 200100+a+50+h=200
⇒
a+h\bm{a + h}a+h =50= \boldsymbol{50}=50

(I) To select students who are not members in clubs B or C, we need to identify elements not coming under the circles B or C, i.e.
a+h\bm{a + h}a+h =50= \boldsymbol{50}=50

(II) Members who are not in club A
=b+e+c+h=b+c+h+30= b + e + c + h = b + c + h + 30=b+e+c+h=b+c+h+30
As we cannot solve for
b,cb, cb,c or hhh, this value cannot be determined.

(III)
120120120 students passed in C. Of this, number of students who did not pass in B =c+f=50= c + f = 50=c+f=50
∴ Of the students who passed in C, the
% \%% who did not pass in B =50120×100%= \dfrac{50}{120} \times 100 \%=12050​×100% =41.7%= \boldsymbol{41.7 \%}=41.7%

(IV) Students in exactly
111 of A or C === Students in A but not in C +++ Students in C but not in A =a+d+c+e=a+d+c+30= a + d + c + e = a + d + c + 30=a+d+c+e=a+d+c+30

From the
333 equations we formed earlier ⇒ b+d=30,c+f=50b + d = 30, c + f = 50b+d=30,c+f=50 and a+h=50a + h = 50a+h=50, maximum possible values of a,da, da,d and ccc are 50,3050, 3050,30 and 505050 respectively.

∴ Maximum number of students in exactly
111 of A or C =a+d+c+30=50+30+50+30== a + d + c + 30 = 50 + 30 + 50 + 30 ==a+d+c+30=50+30+50+30= 160\boldsymbol{160}160

Answer: (I) (3)
505050; (II) (4) Cannot be determined; (III) (1) 41.7%41.7 \%41.7%; (IV) (3) 160160160

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