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Set Theory

Set Theory

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Set Operations & Venn Diagram
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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 - Maximum and Minimum

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6. Minimum and Maximum

6.1 Regions expressed as variable

In these questions, the number of elements of a region will be written as a variable, say xxx. Other regions will also be expressed in terms of xxx. We would then need to find the minimum and maximum values for xxx.

Example 12

In a club, 120120120 members order tea, 100100100 members order coffee and 909090 members order milk. 404040 members order tea & coffee, 505050 members order coffee & milk, and 555555 members order milk & tea.
(I) What is the minimum and maximum number of people who could have ordered all 333 drinks?
(II) What is the minimum number of people who did not order tea?
(III) What is the the maximum number of people who ordered at least one drink?

Solution

Let the number of people who had all 333 drinks be x\bm{x}x.

Number of people who had
- Only tea & coffee === 40 - x
- Only coffee & milk === 50 - x
- Only milk & tea === 55 - x

Number of people who had
- Only Tea === 120−(40−x+55−x+x)120 - (40 - x + 55 - x + x)120−(40−x+55−x+x)
= x + 25
- Only Coffee =100−(40−x+50−x+x)= 100 - (40 - x + 50 - x + x)=100−(40−x+50−x+x)
= x + 10
- Only Milk =90−(50−x+55−x+x)= 90 - (50 - x + 55 - x + x) =90−(50−x+55−x+x)
= x - 15
Case I: None of the 777 regions can be negative.
The numbers x−15x - 15x−15 and 40−x40 - x40−x
set the minimum and maximum values for xxx.
x−15>0x - 15 > 0x−15>0 ⇒ x>15 x > 15x>15
40−x>040 - x > 040−x>0 ⇒ x<40x < 40x<40
∴ Minimum and maximum number
of people who had all 333 drinks is 15 and 40 respectively.


Case II: Number of people who did not order tea =x+10+50−x+x−15== x + 10 + 50 - x + x - 15 ==x+10+50−x+x−15= x + 45

Substituting the minimum value of x=15x = 15x=15 that we derived in case I, we get the minimum number of people who did not order tea to be 15+45=15 + 45 =15+45= 60.

Case III: Number of people who ordered at least one drink === Sum of numbers in all 777 regions === x + 165
Substituting the maximum possible value of xxx ⇒ 40+165=40 + 165 =40+165= 205

Answer: (I) 151515 & 404040; (II) 606060; (III) 205205205


6.2 Minimum & Maximum number of elements

When number of elements in each set (i.e., nnn(A), nnn(B), ...) and that of universal set (i.e., nnn(U)) are given, then

1) Maximum number of elements in all sets === Minimum of (nnn(A), nnn(B), ...)
2) Minimum number of elements in all sets === U −-− (nnn(A‾\overline{A}A) +++ nnn(B‾\overline{B}B) +++ ...)
(Note: nnn(A‾\overline{A}A) === 100−n100 - n100−n(A), nnn(B‾\overline{B}B) =100−n= 100 - n=100−n(B), etc. Also for this method, there should be no other constraint like every element is in at least 111 set, etc.)

Example 13

In a school, if 90%90 \%90% of the students play football, 95%95 \%95% play hockey, 85%85 \%85% play cricket, 80%80 \%80% play basketball and 90%90 \%90% play kabaddi, then what is the minimum number of students who play all sports and what is the minimum and maximum percentage of students who play all sports?

Solution

Maximum students who play all sports === Smallest of the 555 sets === 80%\bm{80 \%}80%

Minimum students who play all sports
=100−[(100−90)+(100−95)+(100−85)+(100−80)+(100−90)]= 100 - [(100 - 90) + (100 - 95) + (100 - 85) + (100 - 80) + (100 - 90)]=100−[(100−90)+(100−95)+(100−85)+(100−80)+(100−90)]
=100−[10+5+15+20+10]= 100 - [10 + 5 + 15 + 20 + 10]=100−[10+5+15+20+10]
=100−60== 100 - 60 ==100−60= 40%\bm{40 \%}40%

Answer: 40%40 \%40% and 80%80 \%80%


Example 14

In a class of 120120120 students, 606060 passed in Maths while 404040 passed in Science.
(I) What is the maximum number of students who failed in both?
(II) What is the minimum number of students who failed in both?
(III) What is the maximum number of students who passed in both?
(IV) What is the minimum number of students who passed in both?
(V) What is the minimum number of students who passed in exactly 1 subject?
(VI) What is the maximum number of students who passed in exactly 1 subject?

Solution

Minimum number of students who passed in both exams =120−[(120−60)+(120−40)]=−20= 120 - [(120 - 60) + (120 - 40)] = -20=120−[(120−60)+(120−40)]=−20 As number of students cannot be negative, minimum number who passed in both === 0\bm{0}0

Maximum number of students who passed in both exams === Lowest (60,4060, 4060,40) === 40\bm{40}40

The following are 222 Venn diagrams for these two minimum and maximum cases. We answer the questions finding the minimum or maximum between these 222 Venn diagrams.


Case I: Maximum number of students who failed in both === d=60\bm{d = 60}d=60
Case II: Minimum number of students who failed in both === d=20\bm{d = 20}d=20
Case III: Maximum number of students who passed in both === c=40\bm{c = 40}c=40
Case IV: Minimum number of students who passed in both === c=0\bm{c = 0}c=0
Case V: Minimum number of students who passed in exactly 111 subject === a+b=20+0=20\bm{a + b = 20 + 0 = 20}a+b=20+0=20
Case VI: Maximum number of students who passed in exactly 111 subject === a+b=60+40=100\bm{a + b = 60 + 40 = 100}a+b=60+40=100

  Answer: (I) 606060; (II) 202020; (III) 404040; (IV) 000; (V) 202020; (VI) 100100100


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