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Maxima and Minima in Set Theory

Possibility of maxima and minima occurs in a set theory due to the fact that intersections of two or more than two values, in most of the cases, have a finite range. In other words, they can lie in between x and y, where x and y can be any number depending upon the given conditions.
 
Let us understand this with the following example:
 
Example-1
In a class of 50 students, a test for 2 subjects maths and physics was conducted. 30 students passed in maths and 40 passed in physics.
  • What is the maximum number of students who passed in both the subjects?
  • What is the minimum number of students who passed in both the subjects?
Solution
  • Since 30 students passed in maths, so we can assume that these are the same 30 students out of 40 students who passed physics also. So, the maximum number of students who passed in both the subjects = 30. (Remember that in this case there was no restriction regarding the number of students who passed any of the subject. Had it been given that all these 50 students are passing atleast one subject, then we would not have been in a position to do this.)
     
    The Venn-diagram representation of the same is given below.
     
    Description: 17-8.tif
  • To find out the minimum number of students who passed in both the subjects, intersection (math ∩ physics) should be minimum, i.e., individually math and physics each should be maximum.
Let us do a bit of hit and trial before we finally move on to do it with its proper solution.
If we take (math ∩ physics) = 0, then the situation would be like
Description: 17-9.tif
 
It can be seen that total number of students in this situation = 30 + 40 = 70, which is not true.
Similarly, if we take (math ∩ physics) = 10, then the situation would be like
 
Description: 17-10.tif
It can be seen that total number of students in this situation = 20 + 10 + 30 = 60, which is again not true.
Hence, now we can conclude that since the number of students = 50, so (math ∩ physics) = 20. And this is the minimum value of (math ∩ physics).
Description: 17-11.tif
 
Alternatively, question (ii) can be done by the following set theory formula also: n(maths U physics) = n(maths) + n(physics) – n(maths ∩ physics)
 
 

Remember In case of two sets, by using this formula of set theory we get the minimum value, but if we use the same formula in case of three sets, then we obtain the maximum value.

 
Example-2
There are 200 students in a school. Out of these, 100 students play cricket, 50 students play hockey and 60 students play basketball. 30 students play both cricket and hockey, 35 students play both hockey and basketball, and 45 students play both basketball and cricket.
  • What is the maximum number of students who play at least one game?
  • What is the maximum number of students who play all the 3 games?
  • What is the minimum number of students playing at least one game?
  • What is the minimum number of students playing all the 3 games?
Solution
Description: 17-12.tif
 
Consider the venn diagram given above:
 
At first we will convert all the values in terms of x, which can be seen above.
 
Since the number of students cannot be negative,
x – 15 ≥ 0
∴ x – 20 ≥ 0
 
So, iv. For the minimum number of students playing all three games, i.e., x = 20
 
For the maximum value of x, again none of the categories should have –ve number of students.
∴ 30 – x ≥ 0
x ≤ 30
 
If x is more than 30, 30−x would be –ve which is not possible.
Total number of students playing at least one game,
= 100 + x − 15 + 35 – x + x – 20
= 100 + x
 
So, the minimum number of students playing at least one game = 100 + 20 = 120
 
Hence, the maximum number of students playing at least one game = 100 + 30 = 130
 

Example-3
In a class of 50 students, 70% students pass in QA and 60% pass in RC. What is the minimum percentage of students who pass in both the papers?
Solution
We know that n(A∪B) = n(A) + n(B) – n(A ∩ B)
 
Assuming that n(Q) = number of students passing QA and n(R) = number of students passing RC.
 
Hence, n(Q∪R) = n(Q) + n(R) – n(Q∩R)
 
Or, n(Q∩R) = n(Q) + n(R) – n(Q∪R) = 70% + 60% − n(Q∪R)
 
Now the minimum value of n(Q∩R) will occur for the maximum value of n(Q∪R). The maximum possible value of n(Q∪R) = 100%
 
So, the minimum value of n(Q∩R) = 30%
 

Example-4
In a class of 50 students, 70% students pass in QA, 90% pass in EU and 60% pass in RC. What is the minimum percentage of students who pass in all the papers?
Solution
We can do this question by using either the formula for three sets, or we can simply keep on applying the formula for two sets required a number of times.
 
We know that n(A∪B) = n(A) + n(B) – n(A∩B)
 
Assuming that n(Q) = number of students passing QA, n(E) = number of students passing EU and n(R) = number of students passing RC.
 
Hence, n(Q∪R) = n(Q) + n(R) – n(Q∩R)
 
Or, n(Q∩R) = n(Q) + n(R) – n(Q∪R) = 70% + 60% − n(Q∪R)
 
Now the minimum value of n(Q∩R) will occur for the maximum value of n(Q∪R). The maximum possible value of n(Q∪R) = 100%
So, the minimum value of n(Q∩R) = 30%
 
Now, the minimum value of EU, QA and RC will be obtained by finding the minimum of EU and (QA∩RC).
 
The minimum value of EU and (QA∩RC) = 90% + 30% − 100% = 20%.
 
 
Example-5
In a class of 50 students, 70% students pass in QA, 90% pass in EU, 85% pass in DI and 60% pass in RC. What is the minimum percentage of students who pass in all the papers?
Solution
This question is just the extension of the earlier question.
 
The minimum value of EU and (QA∩RC) = 90% + 30% − 100% = 20%.
 
The minimum value of DI and EU and (QA∩RC) = 20% + 85% − 100% = 5%
 




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