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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

A
15
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B
30
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C
22
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D
27
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Solution

## The correct option is C 22Let 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

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