In a group of 30 students, the number of students studying French, English and Sanskrit were found to be as
follows:
French = 17, English = 13, Sanskrit = 15, French and English = 9, English and Sanskrit = 4, French and
Sanskrit = 5, English, French and Sanskrit = 3. Based on the above information answer the following-
A) French only.
I. 6
II. 3
III. 9
IV. 20
B) English only.
I. 6
II. 3
III. 9
IV. 20
C) Sanskrit only.
I. 6
II. 3
III. 9
IV. 20
D) None of the three languages.
I. 6
II. 3
III. 9
IV. 0
E) At least one of the three languages.
I. 30
II. 3
III. 9
IV. 20
Answers
Answer:
Let,
French = set A,
English = set B,
Sanskrit = set C
n(A)=17,n(B)=13,n(C)=15
n(A∩B)=9,n(B∩C)=4,n(A∩C)=5
n(A∩B∩C)=3
n(u)=50
n(A∩
B
ˉ
∩
C
ˉ
)=n(A)−n(A∩B)−n(A∩C)+n(A∩B∩C)
=17−9−5+3
French only =6
n(
A
ˉ
∩B∩
C
ˉ
)=n(B)−n(B∩C)−n(A∩B)+n(A∩B∩C)
=13−4−9+3
English only =3
n(
A
ˉ
∩
B
ˉ
∩C)=n(C)−n(A∩C)−n(B∩C)+n(A∩B∩C)
=15−4−5+3
Sanskrit only =9
n(
A
ˉ
∩B∩C)=n(B∩C)−n(A∩B∩C)
French, English and Sanskrit =4−3=1
n(A∩
B
ˉ
∩C)=n(A∩C)−n(A∩B∩C)=5−3=2
n(A∩B∩
C
ˉ
)=n(A∩B)−n(A∩B∩C)=9−3=6
n(name of A orB or C )=5a−(A∩B∩C)
=5a−{n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(A∩C)+n(A∩B∩C)}
=50−{17+13+15−9−4−5+3}
=20
∴n(at least of A or B or C)=50−20=30