in a survey of 200 students, 120 study mathematics, 90 study physics, and 70 study chem. 40 study maths and physics, 30 study phyc and chem, 0 study chem and mathand 20 study none of the subjects.find:
(i)Those who study all the three subjects,
Answers
Answer:
Correct option is B)
Let M, P and C denote the students studying Mathematics, Physics and Chemistry
And U represents total students
So, n(U)=200,
n(M)=120,n(P)=90
n(C)=70,n(M∩P)=40,n(P∩C)=30
n(M∩C)=50,n(M∪P∪C)
′
=20
∴n(M∪P∪C)
′
=n(U)−n(M∪P∪C)
⇒20=200−n(M∪P∪C)
⇒n(M∪P∪C)=180
⇒n(M∪P∪C)=n(M)+n(P)+n(C)
−n(M∩P)−n(P∩C)−n(C∩M)+n(C∩M∩P)
∴180=120+90+70−40−30−50+n(C∩M∩P)
⇒180=280−120+n(C∩M∩P)
⇒n(P∩C∩M)=300−280=20
Hence, the number of students studying all three subjects is 20.
Answer:
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Explanation:
Let M, P and C denote the students studying Mathematics, Physics and Chemistry
And U represents total students
So, n(U)=200,
n(M)=120,n(P)=90
n(C)=70,n(M∩P)=40,n(P∩C)=30
n(M∩C)=50,n(M∪P∪C)′=20
∴n(M∪P∪C)′=n(U)−n(M∪P∪C)
⇒20=200−n(M∪P∪C)
⇒n(M∪P∪C)=180
⇒n(M∪P∪C)=n(M)+n(P)+n(C)
−n(M∩P)−n(P∩C)−n(C∩M)+n(C∩M∩P)
∴180=120+90+70−40−30−50+n(C∩M∩P)
⇒180=280−120+n(C∩M∩P)
⇒n(P∩C∩M)=300−280=20
Hence, the number of students studying all three subjects is 20.