1

Question

In a class of 42 students, 23 are studying Mathematics, 24 are studying Physics, 19 are studying Chemistry. If 12 are studying both Mathematics and Physics, 9 are studying both Mathematics and Chemistry, 7 are studying both Physics and Chemistry and 4 are studying all the three subjects, then the number of students studying exactly one subject, is

Open in App

Solution

The correct option is **C** 22

Let M,P,C denote the set of students studying Mathematics, Physics and Chemistry respectively.

n(M)=23, n(P)=24, n(C)=19n(M∩P)=12, n(M∩C)=9n(P∩C)=7, n(M∩P∩C)=4

Now, n(M∩P′∩C′)

n(M∩(P∪C)′)

n(M)−n(M∩(P∪C))

=n(M)−n[(M∩P)∪(M∩C)]

=n(M)−[n(M∩P)+n(M∩C)]+n(M∩P∩C)=23−[12+9]+4=6

Similarly, n(P∩M′∩C′)

=n(P)−[n(P∩M)+n(P∩C)]+n(M∩P∩C)=24−[12+7]+4=9

and n(C∩P′∩M′)

=n(C)−[n(C∩P)+n(C∩M)]+n(M∩P∩C)=19−[9+7]+4=7

Hence, number of students studying exactly one subject =6+9+7=22

Alternatively, using Venn diagram,

Given, a+b+c+d=23 ⋯(1)

b+c+e+f=24 ⋯(2)

d+c+f+g=19 ⋯(3)

b+c=12, d+c=9, c+f=7

and c=4

∴b=8, d=5, f=3

From (1), a=6

From (2), e=9

From (3), g=7

Required number =a+e+g=22

Let M,P,C denote the set of students studying Mathematics, Physics and Chemistry respectively.

n(M)=23, n(P)=24, n(C)=19n(M∩P)=12, n(M∩C)=9n(P∩C)=7, n(M∩P∩C)=4

Now, n(M∩P′∩C′)

n(M∩(P∪C)′)

n(M)−n(M∩(P∪C))

=n(M)−n[(M∩P)∪(M∩C)]

=n(M)−[n(M∩P)+n(M∩C)]+n(M∩P∩C)=23−[12+9]+4=6

Similarly, n(P∩M′∩C′)

=n(P)−[n(P∩M)+n(P∩C)]+n(M∩P∩C)=24−[12+7]+4=9

and n(C∩P′∩M′)

=n(C)−[n(C∩P)+n(C∩M)]+n(M∩P∩C)=19−[9+7]+4=7

Hence, number of students studying exactly one subject =6+9+7=22

Alternatively, using Venn diagram,

Given, a+b+c+d=23 ⋯(1)

b+c+e+f=24 ⋯(2)

d+c+f+g=19 ⋯(3)

b+c=12, d+c=9, c+f=7

and c=4

∴b=8, d=5, f=3

From (1), a=6

From (2), e=9

From (3), g=7

Required number =a+e+g=22

0

View More

Join BYJU'S Learning Program

Join BYJU'S Learning Program