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15 votes

In a class of $300$ students in an M.Tech programme, each student is required to take at least one subject from the following three:

- M600: Advanced Engineering Mathematics
- C600: Computational Methods for Engineers
- E600: Experimental Techniques for Engineers

The registration data for the M.Tech class shows that $100$ students have taken M600, $200$ students have taken C600, and $60$ students have taken E600. What is the maximum possible number of students in the class who have taken all the above three subjects?

- $20$
- $30$
- $40$
- $50$

16 votes

Best answer

@Deepak Poonia Sir @Lakshman Patel RJIT Sir why are we using this inequality sign “>?. 100-x + 200-x + 60-x = 300-x should always be equal .

0

Analyze the value: $(100 – x + 200 – x + 60 -x) $

The $(100-x)$ and $(200-x)$ may have something in common (Those students who have taken both $M600$ and $C600$ But Not taken $E600$).

The $(100-x)$ and $(60-x)$ may have something in common (Those students who have taken both $M600$ and $E600$ But Not taken $C600$).

The $(200-x)$ and $(60-x)$ may have something in common (Those students who have taken both $C600$ and $E600$ But Not taken $M600$).

That’s why, we can write that $(100 – x + 200 – x + 60 -x) \geq (300-x) $

The $(100-x)$ and $(200-x)$ may have something in common (Those students who have taken both $M600$ and $C600$ But Not taken $E600$).

The $(100-x)$ and $(60-x)$ may have something in common (Those students who have taken both $M600$ and $E600$ But Not taken $C600$).

The $(200-x)$ and $(60-x)$ may have something in common (Those students who have taken both $C600$ and $E600$ But Not taken $M600$).

That’s why, we can write that $(100 – x + 200 – x + 60 -x) \geq (300-x) $

2

4 votes

Let the number of students who have taken all $3$ subjects are $x$

Remaining students(Out of total $300$) = $300 - x$

Subtract $x$ from the number of students in $M600$ , $C600$, $E600$ to get the number of students who have taken either one subject or two subjects.

In $M600$ we have $100-x$ such students

In $C600$ we have $200-x$ such students

In $E600$ we have $60 - x$ such students

Now, $(100-x) + (200-x) + (60-x) \geq 300$

We get $ x \leq 30$ hence **Option B** is the correct answer

(**p.s** The above inequality will become an equality if number of students who have taken exactly $2$ subjects is $0$)

2 votes

Total Number of student is 300.

n(M-600) = 100, n(C-600) = 200, n(E-600) = 60 (Given data in question)

We need to find max((M-600)∩(C-600)∩(E-600)).............................?

n(M-600)+n(C-600)+n(E-600) = 100+200+60 = 360 (Every one has taken at-least one subject) then 360 - 300 = 60 (Remeber)

we have 60 people who has been enrolled for more than one subject in M.tech Program.

so **60 / 2 = 30 (To maximize the 3 subject student we can enroll 30 student max for all three subject).**

Please comment if it's easy to understand i will upload the pictorial representation with some small example.