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Question

From 50 students taking examinations in Mathematics, Physics and Chemistry, 37 passed Mathematics, 24 Physics and 43 Chemistry. At most 19 passed Mathematics and Physics, at most 29 passed Mathematics and Chemistry and at most 20 passed Physics and Chemistry. The largest possible number that could have passed all three examinations is

(a) 11
12
13
14

The correct answer is: 14

Let $M comma P$ and $C$ be the sets of students taking examinations in Mathematics, Physics and Chemistry respectively.
We have,
$n open parentheses M union P union C close parentheses equals 50 comma n open parentheses M close parentheses equals 37 comma n open parentheses P close parentheses equals 24 comma n open parentheses C close parentheses equals 43$
$n open parentheses M intersection P close parentheses less than 19 comma n left parenthesis M intersection C right parenthesis less or equal than 29 comma n left parenthesis P intersection C right parenthesis less or equal than 20$
Now,
$n open parentheses M union P union C close parentheses equals n open parentheses M close parentheses plus n open parentheses P close parentheses plus n open parentheses C close parentheses minus n left parenthesis M intersection P right parenthesis$
$negative n open parentheses M intersection C close parentheses minus n open parentheses P intersection C close parentheses plus n blank left parenthesis M intersection P intersection C right parenthesis$
$rightwards double arrow 50 equals 37 plus 24 plus 43 minus left curly bracket n open parentheses M intersection P close parentheses plus n open parentheses M intersection C close parentheses plus n left parenthesis P intersection C right parenthesis right curly bracket$
$plus n left parenthesis M intersection P intersection C right parenthesis$
$rightwards double arrow n open parentheses M intersection P intersection C close parentheses equals n open parentheses M intersection P close parentheses plus n open parentheses M intersection C close parentheses plus n open parentheses P intersection C close parentheses minus 54$
$rightwards double arrow n open parentheses M intersection P close parentheses plus n open parentheses M intersection C close parentheses plus n left parenthesis P intersection C right parenthesis$
$equals n open parentheses M intersection P intersection C close parentheses plus 54$ …(i)
Now,
$n left parenthesis M intersection P right parenthesis less or equal than 19 comma n left parenthesis M intersection C right parenthesis less or equal than 29 comma n left parenthesis P intersection C right parenthesis less or equal than 20$
$rightwards double arrow n open parentheses M intersection P close parentheses plus n open parentheses M intersection C close parentheses plus n open parentheses P intersection C close parentheses less or equal than 19 plus 29 plus 20$ [Using (i)]
$rightwards double arrow n open parentheses M intersection P intersection C close parentheses plus 54 less or equal than 68$
$rightwards double arrow n left parenthesis M intersection P intersection C right parenthesis less or equal than 14$

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