# Set Theory - Maxima & Minima concepts - Kamal Lohia

• 1. According to a survey, at least 70% of people like apples, at least 75% like bananas and at least 80% like cherries. What is the minimum percentage of people who like all three ?

Answer: Let's first calculate the surplus:

percentage of people who like apples + percentage of people who like bananas + percentage of people who like cherries = 70% + 75% + 80% = 225% = a surplus of 125%.

Now this surplus can be accommodated by adding elements to either intersection of only two sets or to intersection of only three sets. As the intersection of only two sets can accommodate only a surplus of 100%, the surplus of 25% will still be left. This surplus of 25% can be accommodated by adding elements to intersection of three sets. For that we have to take 25% out of the intersection of only two sets and add it to intersection of three sets. Therefore, the minimum percentage of people who like all three = 25%

The question can be solved mathematically also. Let the elements added to intersection of only two sets and intersection of three sets be x and y, respectively. These elements will have to cover the surplus.
x + 2y = 125%, where x + y =100%. For minimum value of y, we need maximum value of x.
x = 75%, y = 25%.

2. In a college, where every student follows at least one of the three activities- drama, sports, or arts- 65% follow drama, 86% follow sports, and 57% follow arts. What can be the maximum and minimum percentage of students who follow
1 all three activities
2.exactly two activities

Answer: Let us again see the surplus: Percentage of students who follow drama + Percentage of students who follow sports + Percentage of students who follow arts = 65% + 86% + 57% = 208% =surplus of 108%.

This surplus can be accommodated through adding elements either to intersection of only two sets or to intersection of only three sets. As the intersection of only two sets can accommodate only a surplus of 100%, the
surplus of 8% will still be left. This surplus of 8% can be accommodated by adding elements to intersection of three sets. For that we have to take 8% out of the intersection of only two sets and add it to intersection of three sets. Therefore, the minimum percentage of people who like all three = 8%.

In this case the percentage of students who follow exactly two activities will be maximum = 92%.

The surplus of 108% can also be accommodated through adding elements to only intersection of three sets. As adding 1 element to intersection of three sets give a surplus of 2 sets, adding 54% to intersection of three sets will give a surplus of 108%. Therefore, the maximum value of students who follow all three activities is 54%. In this case the percentage of students who follow exactly two activities will be minimum = 0%.

We can also solve it mathematically , x + 2y = 108%, where x + y = 100%. The maximum value of x will give minimum value of y, whereas minimum value of x will give maximum value of y.

3. Out of 210 interviews of IIM- Ahmedabad, 105 CAT crackers were offered tea by the interview panel, 50 were offered biscuits, and 56 were offered toffees. 32 CAT crackers were offered tea and biscuits, 30 were offered biscuits and toffees, and 45 were offered toffees and tea. What is the

a. maximum and minimum number of CAT crackers who were offered all three snacks?
b. maximum and minimum number of CAT crackers who were offered at least one snack?

Answer: In this case we assign the variables to every area of the Venn diagram and form the conditions keeping two things in mind:
a. try to express the areas in the Venn diagram through least number of variables.
b. all the numbers will be zero or positive. No number can be negative.

Let’s make the Venn diagram for this question. Since we want to assume least number of variables, we can see that assuming a variable for the number of students who were offered all three snacks will help us express all the other areas. Let the number of students who were offered all three snacks = x.

In the above diagram, we have expressed all the areas in terms of x. To decide maximum value of x, we note that x can’t be more than 30, otherwise 30-x will be negative.. Therefore, the maximum value of x will be 30. (30 is the lowest among 30, 32 and 45). To decide minimum value of x, we note that x cannot be less than 19.

Therefore, maximum and minimum number of CAT crackers who were offered all three snacks = 30 and 19.

The number of CAT crackers who were offered at least one snack = Total number of CAT crackers in the Venn diagram = x + 28 + 32 - x + x + 45 - x + x - 19 + 30 - x + x - 12 = 104 + x.

As the maximum and minimum values of x are 30 and 19, respectively, the maximum and minimum value of 104 + x will be 134 and 123, respectively.

Maximum and minimum number of CAT crackers who were offered at least one snack = 134 and 123.

4) A survey was conducted and the details of CAs in the industry were collected :

Total: 2600.Percentage wise distribution of different Queries are:

 Name Age Address Experience Phone Email 100% 80% 50% 40% 50% 90%

For example, of the total number of CAs of only 50% viz, 1300 people gave their experiences.

1: For at most how many CAs exactly five of six features are available:

(1) 1300(2) 1950(3) 2000(4) 1820(5) None of the above.

Solution:Number of CAs with atleast 1 query + Number of CAs with atleast 2 query  + Number of CAs with atleast 3 query + Number of CAs with atleast 4 query + Number of CAs with atleast 5 query + Number of CAs with atleast 6 query = 100% + 80% + 50% + 40% + 50% + 90% = 410%

If we want to maximize the number of CAs with exactly five query, then number of CAs with exactly six query should be minimized to zero if possible.

Also note that number of CAs with atleast one, two, three or four query cannot be less than those with atleast five queries.

(410 - 100 - 90 - 80)/2 =  70%

So Maximum of 70% i.e. 1820 out of 2600 CAs have answered to exactly 5 queries.

Modus Operandi:

If we want to maximise exactly five then atleast six will be minimum(zero if possible) and atleast one, two, three and four equal to or almost equal to atleast five as these values can't go lower than atleast five.
Now in this case: If I am making exactly six to be zero then I need to divide 41% in five categories and for maximum value of exactly five I will distribute 410 equally i.e. 410/5 = 82% each.
But least value for 'atleast one' is 100%. So removing that one figure, using same logic as earlier, I will divide remaining 310% (410% - 100%) equally in four categories i.e. 310/4 = 77.5% each.
But again least value for 'atleast two' is 90%. So going by same reasoning, remaining 220% (410% - 100% - 90%) should be divided equally in remaing three categories i.e. 220/3 = 73.3% each.
Oh No! Once more the least value for 'atleast three' is 80%. Now removing this one, I need to divide the remaining 140% (410% - 100% - 90% - 80%) equally in remaining two categories i.e. 140/2 = 70% each.

Therefore the maximum number for exactly 5 is 70% of 2600 = 1820 :)

Note: This would be true,If the greatest value > 50% &If the sum of the two greatest value > 100%Otherwise, you have to workout the minimum value on ‘atleast 2’ and ‘atleast 3’.

5) The following table gives the number of students who secured more than 90% marks in each of the five subjects from class 6th to 10th at a school in year 2006

 class/subject english physics chemistry maths biology 6th 12 16 15 22 18 7th 15 22 22 21 15 8th 7 18 16 23 17 9th 10 19 15 22 18 10th 15 25 21 29 16

The number of students in the different classes in the year were as below

 Class No. of students 6th 30 7th 35 8th 28 9th 36 10th 40

Q1- In the class 7th the number of students who scored more than 90% in at least two of the five subjects is at least

a) 12
b) 15

c) 18
d) 20
e) none of these

For class 7th, number of students scoring more than 90% in atleast (one  + two + three + four + five) subjects = 15 + 22 + 22 + 21 + 15 = 95.
As I want to minimize the figure for atleast two, atleast one must be maximized i.e. 35 and the remaining 60 (95 - 35) must be divided equally in remaining four categories that means 60/4 = 15 each. Hence option (b) is correct.

Q2-The number of students in class 10th who scored more than 90% in exactly three subjects is at most

a) 32
b) 34
c) 35
d) 38
e) none of these

For class 10th, number of students scoring more than 90% in atleast (one  + two + three + four + five) subjects = 15 + 25 + 21 + 29 + 16 = 106.

As I want to maximize exactly three, I will try to minimize the figures for atleast 4 and 5. So distributing equal and maximum values atleast one , two and three i.e. [106/3] = 35 each. Now I am left with 1 which can be assigned to atleast 1 so getting the value for exactly three as 35 . Hence option (c) is correct.

In above solution, I am saying number of students scoring more than 90% marks in :

exactly no subject = 4
exactly one subject = 1
exactly two subject = 0
exatly three subject = 35
exactly four subject = 0
exactly five subject = 0

Q3-The no. of students who scored more than 90% in exactly four subjects in all the four classes together is at most

a) 28
b) 61
c) 96
d) 101
e) none of these

Atleast(1+2+3+4) in 6th,7th,8th,9th&10th follows in order,22+21+21+2024+24+24+2323+20+19+1922+21+21+2029+26+26+25

Therefore, total = 20+23+19+20+25 = 107

Q4- The number of students in class 6th who scored more than 90% in at most two subjects is at most-

a) 19
b) 21
c) 22
d) 26
e) 30

Ans: 22

1

1

14

1

1

1

1

28