Question

out of 300 s students donot study both of the subjecttudent in a class, 60% students study physics, 35% students study chemistry and 20% students do not study both of the subject. then how many students study both subjects
SET

Answer 1

Answer:

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 2

Step-by-step explanation:

Step-by-step explanation:Let M, P and C denote the students studying Mathematics, Physics and Chemistry

Step-by-step explanation:Let M, P and C denote the students studying Mathematics, Physics and ChemistryAnd U represents total students

Step-by-step explanation:Let M, P and C denote the students studying Mathematics, Physics and ChemistryAnd U represents total studentsSo, n(U)=200,

Step-by-step explanation:Let M, P and C denote the students studying Mathematics, Physics and ChemistryAnd U represents total studentsSo, n(U)=200,n(M)=120,n(P)=90

Step-by-step explanation:Let M, P and C denote the students studying Mathematics, Physics and ChemistryAnd U represents total studentsSo, n(U)=200,n(M)=120,n(P)=90n(C)=70,n(M∩P)=40,n(P∩C)=30

Step-by-step explanation:Let M, P and C denote the students studying Mathematics, Physics and ChemistryAnd U represents total studentsSo, n(U)=200,n(M)=120,n(P)=90n(C)=70,n(M∩P)=40,n(P∩C)=30n(M∩C)=50,n(M∪P∪C)

Step-by-step explanation:Let M, P and C denote the students studying Mathematics, Physics and ChemistryAnd U represents total studentsSo, n(U)=200,n(M)=120,n(P)=90n(C)=70,n(M∩P)=40,n(P∩C)=30n(M∩C)=50,n(M∪P∪C) ′

Step-by-step explanation:Let M, P and C denote the students studying Mathematics, Physics and ChemistryAnd U represents total studentsSo, n(U)=200,n(M)=120,n(P)=90n(C)=70,n(M∩P)=40,n(P∩C)=30n(M∩C)=50,n(M∪P∪C) ′ =20

Step-by-step explanation:Let M, P and C denote the students studying Mathematics, Physics and ChemistryAnd U represents total studentsSo, n(U)=200,n(M)=120,n(P)=90n(C)=70,n(M∩P)=40,n(P∩C)=30n(M∩C)=50,n(M∪P∪C) ′ =20∴n(M∪P∪C)

Step-by-step explanation:Let M, P and C denote the students studying Mathematics, Physics and ChemistryAnd U represents total studentsSo, n(U)=200,n(M)=120,n(P)=90n(C)=70,n(M∩P)=40,n(P∩C)=30n(M∩C)=50,n(M∪P∪C) ′ =20∴n(M∪P∪C) ′

Step-by-step explanation:Let M, P and C denote the students studying Mathematics, Physics and ChemistryAnd U represents total studentsSo, n(U)=200,n(M)=120,n(P)=90n(C)=70,n(M∩P)=40,n(P∩C)=30n(M∩C)=50,n(M∪P∪C) ′ =20∴n(M∪P∪C) ′ =n(U)−n(M∪P∪C)

Step-by-step explanation:Let M, P and C denote the students studying Mathematics, Physics and ChemistryAnd U represents total studentsSo, n(U)=200,n(M)=120,n(P)=90n(C)=70,n(M∩P)=40,n(P∩C)=30n(M∩C)=50,n(M∪P∪C) ′ =20∴n(M∪P∪C) ′ =n(U)−n(M∪P∪C)⇒20=200−n(M∪P∪C)

Step-by-step explanation:Let M, P and C denote the students studying Mathematics, Physics and ChemistryAnd U represents total studentsSo, n(U)=200,n(M)=120,n(P)=90n(C)=70,n(M∩P)=40,n(P∩C)=30n(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

Step-by-step explanation:Let M, P and C denote the students studying Mathematics, Physics and ChemistryAnd U represents total studentsSo, n(U)=200,n(M)=120,n(P)=90n(C)=70,n(M∩P)=40,n(P∩C)=30n(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)

Step-by-step explanation:Let M, P and C denote the students studying Mathematics, Physics and ChemistryAnd U represents total studentsSo, n(U)=200,n(M)=120,n(P)=90n(C)=70,n(M∩P)=40,n(P∩C)=30n(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)

Step-by-step explanation:Let M, P and C denote the students studying Mathematics, Physics and ChemistryAnd U represents total studentsSo, n(U)=200,n(M)=120,n(P)=90n(C)=70,n(M∩P)=40,n(P∩C)=30n(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)

Step-by-step explanation:Let M, P and C denote the students studying Mathematics, Physics and ChemistryAnd U represents total studentsSo, n(U)=200,n(M)=120,n(P)=90n(C)=70,n(M∩P)=40,n(P∩C)=30n(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)

Step-by-step explanation:Let M, P and C denote the students studying Mathematics, Physics and ChemistryAnd U represents total studentsSo, n(U)=200,n(M)=120,n(P)=90n(C)=70,n(M∩P)=40,n(P∩C)=30n(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

Step-by-step explanation:Let M, P and C denote the students studying Mathematics, Physics and ChemistryAnd U represents total studentsSo, n(U)=200,n(M)=120,n(P)=90n(C)=70,n(M∩P)=40,n(P∩C)=30n(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=20Hence, the number of students studying all three subjects is 20.