Numerical Problems I: Number of Elements in a Set (Two Sets)

NP1: Sets A and B are such that A has 25 members, B has 20 members and A ∪ B has 35 members. Draw a Venn diagram to represent the above situation. Find the number of elements in the intersection of sets A and B, i.e., A ∩ B 

Solution:

Given:

n(A)=25 → number of elements in set A

n(B)=20 → number of elements in set B

n(A ∪ B) = 35 → number of elements in the union of A and B

To find:

• n(A ∩ B) ) → number of elements common to both A and B

The formula for union of two sets is:

n(A ∪ B) = n(A)+ n(B)− n(A ∩ B) 

35 = 25 + 20 − n(A ∩ B)

35 = 45 − n(A ∩ B)

n( A ∩ B) = 45 − 35 = 10

∴ The number of elements in the intersection of sets A and B is 10

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NP2: In a class of 25 students of Economics and Politics 12 students have taken Economics. Out of these 8 have taken Economics but not Politics. Find the number of students who have taken Economics and Politics and those who have taken Politics but not Economics.

Solution:

Let,

E → Economics

P → Politics

Given:

Total number of students, n(E ∪ P) = 25

Number of students who have taken Economics, n(E) = 12

Number of students who have taken Economics but not Politics, n(E) - n(E ∪ P) = 8

To find:

i). Number of students who have taken both Economics and Politics, n(E ∩ P) 

ii). Number of students who have taken Politics but not Economics, n(P) - n(E ∪ P)

∴ Number of students who have taken both Economics and Politics, n(E ∩ P) = 12 − 8 = 4 

We know that,

n(E ∪ P) = n(E) + n(P) − n(E ∩ P)

⇒ 25 = 12 + n(P) −4

⇒ n(P) = 25 − 8 = 17​ 

∴ Number of students who have taken Politics but not Economics = n(P) − n(E ∩ P) = 17 − 4 = 13

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NP3: In a class of 50 students, 30 students take Economics, 25 students take Mathematics and 10 take both. Find the number of students taking neither of the two subjects.

Solution:

Let,

E → Economics

M → Mathematics

Given:

Total students, n(Ω) = 50

Students taking Economics n(E) = 30  

Students taking Mathematics, n(M) = 25 

Students taking both n(E ∩ M) = 10 

To find:  Number of students taking neither of the two subjects,  n(Ω) − n(E ∪ M) 

Using the Union Formula,

n(E ∪ M) = n(E) + n(M) − n(E ∩ M)  = 45

n(E ∪ M) = 30 + 25 − 10

n(E ∪ M)  = 45

Next, n(Ω) − n(E ∪ M) = 50 − 45 = 5

∴ Number of students taking neither of the two subjects, n(Ω) − n(E ∪ M) =  5

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NP4. In a group of 60 students, 25 like Math (M), 30 like Econometrics (E) and 10 like both Find how many like only Math, only Economics, and neither.

Solution:

Given:
Total number of students,
$n$$($$\mathrm{\Omega }$$)$$=$$60$ 
Students who like Mathematics,
$n$$($$M$$)$$=$$25$ 
Students who like Econometrics,
$n$$($$E$$)$$=$$\mathrm{30 }$$\mathrm{<br>}$$\mathrm{<br>}$$\mathrm{<br>}$S
Students who like both,
$n$$($$E$$\cap$$M$$)$$=$$10$ 
Number of students who like only Mathematics
Number of students who like only Econometrics 
Number of students who like neither subject$$$$$$n(E\cup M)=n(E)+n(M)-n(E\cap M)$$$$$$$$n(E\cup M)=30+25-10=45$$
Next,
Number of students who like only Mathematics, n(M) − n(E∩M) = 25 − 10 = 15
Number of students who like only Econometrics, n(E) − n(E∩M) = 30 − 10 = 20
Number of students who like neither subject, n(Ω) − n(E∪M) = 60 − 45 = 15

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NP5: In a class, 40 students play football, 35 students play basketball, and 18 students play both games. Find the number of students who play either football or basketball or both.

Solution:
B → Set of students who play Basketball

Let,

F → Set of students who play Football

Given:

Total number of students, n$(\Omega )$ = Not given (so we’ll find only union)
Students who play Football, n$(F)=40$
Students who play Basketball, $n(B)=35$

Students who play both games, $n(F\cap B)=18$

To Find: Number of students who play either Football or Basketball or both, i.e., n(F∪B)

Using the Union Formula:

n(F∪B) = n(F) + n(B) − n(F∩B)
$$n(F\cup B)=40+35-18=57$$

n(F∪B) = 40 + 35 − 18 = 57
 
$$\text{Number of students who play either Football or Basketball or both} = \boxed{57}$$

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