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Set Theory

Set Theory

MODULES

Set Notations
Types of Sets
Set Operations & Venn Diagram
2 Set Venn Diagrams
3 Set Venn Diagrams
4 Set Venn Diagrams
Maximum and Minimum
Past Questions

CONCEPTS & CHEATSHEET

Concept Revision Video

SPEED CONCEPTS

Set Theory - 1
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PRACTICE

Set Theory : Level 1
Set Theory : Level 2
Set Theory : Level 3
ALL MODULES

CAT 2025 Lesson : Set Theory - 2 Set Venn Diagrams

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Example 3

In a survey of 808080 people, each of them responded with the languages they speak. It was noted that 474747 people speak Bengali while 555555 people speak Hindi. If all of them speak at least one of these two languages, then how many people speak both?

Solution

Let B and H be the set of people who speak Bengali and Hindi respectively. ∴ nnn(B) =47= 47=47 and nnn(H) =55= 55=55

As everybody speaks one of these two languages, nnn(B ∪\cup∪ H) =80= 80=80

nnn(B ∪\cup∪ H) === nnn(B) +++ nnn(H) −-− nnn(B ∩\cap∩ H)
⇒ 80=47+55−80 = 47 + 55 -80=47+55− nnn(B ∩\cap∩ H)
⇒ nnn(B ∩\cap∩ H) === 22\bold{22}22

Alternatively (Venn Diagram)

The question can be represented as a Venn diagram. Here, aaa and bbb are the number of people who speak only Bengali and only Hindi. ccc is the number of people who speak both.

Number of people who speak neither language is given to be 000.
∴ a+b+c=80a + b + c = 80a+b+c=80 -----(1)
Number of people who speak Bengali =a+c=47= a + c = 47=a+c=47 ----(2)


From (1) and (2), we get b=33b = 33b=33
Number of people who speak Hindi =b+c=55= b + c = 55=b+c=55
⇒ 33+c=5533 + c = 5533+c=55 ⇒ c=22\bm{c = 22}c=22

Answer: 222222


Example 4

50%50 \%50% of the students in a class failed in Maths and 60%60 \%60% failed in Physics, while 10%10 \%10% passed in both Maths and Physics.
(I) If there are 606060 students in total, then how many failed in both???
(II) If there are 606060 students who passed in Maths alone, then how many students are there in the class???

Solution

Let M and P be the sets comprising of students who passed Maths and Physics respectively.

As 50%50 \%50% failed in Maths, the remaining 50%50 \%50% passed in Maths.

Likewise, in Physics the remaining 40%40 \%40% must have passed.

Students who passed both Maths & Physics =10%= 10 \%=10%.
Students who passed only Maths =50%–10%== 50 \% – 10 \% ==50%–10%= 40%\bm{40 \%}40%

Students who passed only Physics =40%–10%== 40 \% – 10 \% ==40%–10%= 30%\bm{30 \%}30%

Students failed in both subjects =100%–40%–10%–30%= 100 \% – 40 \% – 10 \% – 30 \%=100%–40%–10%–30% =20%= \bm{20 \%}=20%

Case I: Number of students who failed in both =20%= 20 \%=20% of 60=60 =60= 12\bm{12}12

Case II: Let the total students in the class be xxx. Number of students who passed Maths alone =40100×x=60= \dfrac{40}{100} \times x = 60=10040​×x=60 ⇒ x=150x = \bm{150}x=150

Answer: (I) 121212 ; (II) 150150150


Example 5

A division has 100100100 employees. Number of people who are only engineers but not managers is the same as those that are not engineers, which is twice that of those who are both engineers and managers. How many are both engineers and managers???

Solution

Only Engineer but not Managers = Non-Engineers= Twice of those that are both Engineers & Managers.

⇒ a=b+d=2ca = b + d = 2ca=b+d=2c
Also given in the question is
a+b+c+d=100a + b + c + d = 100a+b+c+d=100

⇒ a+(b+d)+c=100a + (b + d) + c = 100a+(b+d)+c=100
⇒ a+a+a2=100a + a + \dfrac{a}{2} = 100a+a+2a​=100 ⇒ a=40a = 40a=40

Number of people who are both engineers and managers =c=a2=20= c = \dfrac{a}{2} = 20=c=2a​=20

Answer: 202020


Example 6

In a class of 200200200 students, 130130130 students play carrom, 150150150 students play chess and 202020 students play neither of these games.

(I) How many students play both the sports?
(II) How many students do not play carrom?
(III) How many students play exactly one of the games?
(IV) How many students do not play at least one of the 2 games?

Solution

As shown in the Venn diagram, number of people - playing Carrom =a+c=130= a + c = 130=a+c=130 - playing Chess =b+c=150= b + c = 150=b+c=150 - playing Carrom or Chess =a+b+c=180= a + b + c = 180=a+b+c=180

Solving the above equations, we get a=30a = 30a=30, b=50b = 50b=50 & c=100c = 100c=100

Case I: Students playing both sports =c=100= c = \bm{100}=c=100

Case II: Students who do not play carrom =b+d=70= b + d = \bm{70}=b+d=70

Case III: Students who play exactly one of the games =a+b=80= a + b = \bm{80}=a+b=80

Case IV: Students who do not play at least one of the 222 games =a+b+d=100= a + b + d = \bm{100}=a+b+d=100

Answer: (I) 100100100; (II) 707070; (III) 808080; (IV) 100100100


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