In a class of 100 students 70 passed in physics, 62 passed in mathemat

Bunuel
I think that you should add one more option 0f 5 because that is the answer according to me is 5. Here is my logic
please check if it is correct
chetan2u
help me understand this question

This was a new one for me (4 variables) so it took some time to figure out. Although the first answer is essentially correct, it needs elaboration to explain the assumptions made and the rationale behind them.
In order to determine the maximum number of students who failed we have to assume maximum overlap i.e. all those who passed, apart from the 37 who passed in all 4 subjects, passed in exactly 3 subjects. In other words, no student passed in exactly one or exactly two subjects. Let the number who passed in Math, Physics and Chemistry be denoted by 'mpc', Math Physics and English by 'mpe' and so on. Then:

Breakdown of students who passed in Physics: pmc + mpe + pec + 37 = 70
.................................................... Math: pmc + emc + mpe +37 = 62
.................................................... English: emc + pec + mpe + 37 = 84
.................................................... Chem: pmc + pec + emc + 37 = 82

3(pmc+mpe+pec+emc) + 148 = 298.....> (pmc+mpe+pec+emc)=50. Total number of students who passed is 50+37=87 so the number who failed in all subjects is 100-87=13.
ANS: C

P.S. From the above it can be calculated that: emc=17, pec=25, pmc=3 and mpe=5

Bunuel
What is the solution to this problem?

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First we add up the counts of all students that have passed a test, disregarding the subjects and any overlap.
Total Pass=70+62+84+82=298

Now, if 37 students passed all 4 tests, that means each category must include these same 37 students. Since we have 4 categories:
Total 4-Pass=37∗4=148

Now we subtract the numbers. The difference is the number of students that have passed at least a single test, but did not pass all N tests
Remaining=298−148=150

Now to solve the problem, we must maximize the overlap by assuming these kids all passed exactly 3 tests. It doesn't matter which tests exactly, and the numbers technically dont need to be evenly distributed. So we can just take the average
Avg 3-Pass = 150/3=50

At this point the original numbers and categories dont even matter anymore as we've slimmed the numbers down to include 100% of the kids that have passed, but we've now distributed them across only 3 tests instead of 4. So we can add these 2 numbers together
Total max passed =37+50=87


So out of 100 students, a maximum of 87 have all passed identical tests (100% overlap). And we can easily subtract these 2 to assume that all of the remaining kids have all failed all tests.
Max failed students=100−87=13

Bunuel what do you think of this analysis? maybe you have a clearer approach or explanation

HiBunuel - Please share OA for this.