Question

From 50 students taking examinations in Mathematics, Physics and Chemistry, each of the students has passed in at least one of the subject. 37 passed Mathematics, 24 Physics and 43 Chemistry. At most 19 passed Mathematics and Physics, at most 29 Mathematics and Chemistry and at most 20 Physics and Chemistry. Then the largest possible number that could have passed all three examinations is​

Answer 1

Step-by-step explanation:

Now, let M, P, and C is the sets of students taking examinations in Mathematics, Physics, and Chemistry, respectively.

So, it is given that:

37 passed in Mathematics, therefore: n(M)=37 and 24 passed in Physics, therefore: n(P)=24, and 43 passed in Chemistry, therefore: n(C)=43

Now, it is given that the total students taking the examination are 50, therefore: n(M∪P∪C)=50 .

Now it is given that at most 19 passed in Mathematics and Physics, therefore: n(M∩P)≤19, it is also given that at most 29 passed in Mathematics and Chemistry, therefore: n(M∩C)≤29, and finally it is given that at most 20 passed in Physics and Chemistry, therefore: n(P∩C)≤20.

Now, we know that according to inclusion formula:

n(M∪P∪C)=n(M)+n(P)+n(C)−n(M∩P)−n(M∩C)−n(P∩C)+n(M∩P∩C)Now, we will all put the values in this equation:

⇒50=37+24+43−{n(M∩P)+n(M∩C)+n(P∩C)}+n(M∩P∩C) ⇒{n(M∩P)+n(M∩C)+n(P∩C)}=n(M∩P∩C)+104−50 ⇒{n(M∩P)+n(M∩C)+n(P∩C)}=n(M∩P∩C)+54 ..

Now, it is given that: n(M∩P)≤19, n(M∩C)≤29 and n(P∩C)≤20.

We will now add these, therefore: n(M∩P)+n(M∩C)+n(P∩C)≤19+29+20

Therefore, we will use equation 1 to replace the left hand side:

n(M∩P∩C)+54≤68⇒n(M∩P∩C)≤14.

Therefore, the largest possible number that could have passed all the three examinations is 14. Hence, the correct option is D.

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Answer 2

Hope it helps you........