For solving problems on intersection of two sets we have to consider the following rules :

1) n ( A ∪ B ) = n (A) + n(B) – n ( A ∩ B )

2) If n ( A ∩ B ) = 0 then sets A and B are disjoint sets, and

n ( A ∪ B) = n ( A) + n (B)

3) n ( A – B) = n ( A) – n ( A ∩ B )

4) n ( B – A ) = n ( B) – n ( A ∩ B )

5) n ( A ∪ B )’ = n ( U) – n ( A ∪ B)

1) In a school, all pupils play either Hockey or Football or both. 300 play Football, 250 play Hockey and 110 play both the games. Draw a Venn-diagram and find :

(i) The number of pupils who play Football only;

(ii) The number of pupils who play Hockey only;

(iii) The total number of pupils in the school.

H = Hockey and F = Football

n (H ) = 250 n (F)= 300

n ( H ∩ F) = 110

(i) The number of pupils who only play Football = n (F – H )

n (F – H ) = n(F) – n( F ∩ H )

= 300 – 110

= 190

(ii) The number of pupils who only play Hockey = n (H – F )

n (H– F ) = n(H) – n( F ∩ H )

= 250 – 110

= 140

(iii) The total number of pupils in school

= n(H) + n(F) – n (F ∩ H)

= 250 + 140 - 110

= 440

2) In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee?(problems on intersection of two sets)

Total number of students = U ; T = tea and C = coffee

n( U) = 600 n(T) = 150 n(C) = 225 and n ( T ∩ C ) = 100

Students who take only tea = n (T – C )

n (T – C ) = n(T ) – n( T ∩ C)

= 150 – 100

n (T – C ) = 50

Students who take only coffee = n (C – T )

n (C – T ) = n(C ) – n ( T ∩ C )

= 225 – 100

n (C – T ) = 125

Students were taking neither tea nor coffee is a complement of n ( T ∪ C ) = n ( T ∪ C )’

n ( T ∪ C )’ = n (U ) – [ n( T – C ) + n (C – T) + n ( T ∩ C)]

= 600 – ( 50 + 125 + 100)

= 600 – 275

n ( T ∪ C )’ = 325

3) In a school, there are 20 teachers who teach Mathematics or Physics. Of these, 12 teach Mathematics and 4 teach both Physics and Mathematics. How many teach Physics only?

M = Mathematics P = Physics

n(M) = 12 n( M ∩ P) = 4 and n (M ∪ P )= 20

n (M ∪ P ) = n(M) + n(P) – n(M ∩ P)

20 = 12 + n(P) – 4

20 = 8 + n(P)

⇒ n(P) = 20 – 8

⇒ n(P) = 12

Number of teachers teach Physics only = n( P – M)

n( P – M) = n(P) – n( M ∩ P )

= 12 – 4

n( P – M) = 8

Number of teachers teach physics only = 8.

These are the problems on intersection of two sets.If you have any queries on this then please

• Sets

• Representation of Set

• Cardinal Number

• Types of Sets

• Pairs of Sets

• Subset

• Complement of Set

• Union of the Sets

• Intersection of Sets

• Operations on Sets

• De Morgan's Law

• Venn Diagrams

• Venn-diagrams for sets

• Venn-diagrams for different situations

• Problems on Intersection of Two Sets

• Problems on Intersection of Three Sets

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