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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 xx. Other regions will also be expressed in terms of xx. We would then need to find the minimum and maximum values for xx.

Example 12

In a club, 120120 members order tea, 100100 members order coffee and 9090 members order milk. 4040 members order tea & coffee, 5050 members order coffee & milk, and 5555 members order milk & tea.
(I) What is the minimum and maximum number of people who could have ordered all 33 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 33 drinks be x\bm{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)
= x + 25
- Only Coffee =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)
= x - 15
Case I: None of the 77 regions can be negative.
The numbers x−15x - 15 and 40−x40 - x
set the minimum and maximum values for xx.
x−15>0x - 15 > 0 ⇒ x>15 x > 15
40−x>040 - x > 0 ⇒ x<40x < 40
∴ Minimum and maximum number
of people who had all 33 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 + 45

Substituting the minimum value of x=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 = 60.

Case III: Number of people who ordered at least one drink == Sum of numbers in all 77 regions == x + 165
Substituting the maximum possible value of xx ⇒ 40+165=40 + 165 = 205

Answer: (I) 1515 & 4040; (II) 6060; (III) 205205


6.2 Minimum & Maximum number of elements

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

1) Maximum number of elements in all sets == Minimum of (nn(A), nn(B), ...)
2) Minimum number of elements in all sets == U −- (nn(A‾\overline{A}) ++ nn(B‾\overline{B}) ++ ...)
(Note: nn(A‾\overline{A}) == 100−n100 - n(A), nn(B‾\overline{B}) =100−n= 100 - n(B), etc. Also for this method, there should be no other constraint like every element is in at least 11 set, etc.)

Example 13

In a school, if 90%90 \% of the students play football, 95%95 \% play hockey, 85%85 \% play cricket, 80%80 \% play basketball and 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 55 sets == 80%\bm{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−[10+5+15+20+10]= 100 - [10 + 5 + 15 + 20 + 10]
=100−60== 100 - 60 = 40%\bm{40 \%}

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


Example 14

In a class of 120120 students, 6060 passed in Maths while 4040 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 As number of students cannot be negative, minimum number who passed in both == 0\bm{0}

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

The following are 22 Venn diagrams for these two minimum and maximum cases. We answer the questions finding the minimum or maximum between these 22 Venn diagrams.


Case I: Maximum number of students who failed in both == d=60\bm{d = 60}
Case II: Minimum number of students who failed in both == d=20\bm{d = 20}
Case III: Maximum number of students who passed in both == c=40\bm{c = 40}
Case IV: Minimum number of students who passed in both == c=0\bm{c = 0}
Case V: Minimum number of students who passed in exactly 11 subject == a+b=20+0=20\bm{a + b = 20 + 0 = 20}
Case VI: Maximum number of students who passed in exactly 11 subject == a+b=60+40=100\bm{a + b = 60 + 40 = 100}

  Answer: (I) 6060; (II) 2020; (III) 4040; (IV) 00; (V) 2020; (VI) 100100


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