in Quantitative Aptitude edited by
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15 votes
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?

  1. $20$
  2. $30$
  3. $40$
  4. $50$
in Quantitative Aptitude edited by
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2 Comments

(B) is correct.
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(B) is the correct option!

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3 Answers

17 votes
17 votes
Best answer
Let the no. of students who took all courses be $x$.

Since every one must take at least 1 course, we have

$200 - x + 100 - x + 60 - x \geq 300 -x \\\implies 360-2x \geq 300 \implies x \leq 30.$

Correct Answer: $B$
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7 Comments

why is this sufficient condition?
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@Arjun Sir can we do like this,

for maximum number students, we assume, there is no student who take exactly 2 courses everyone either  takes 1 course or take 3 courses so the equation will be 

300=100+200+60-x-x-x+x

x=30

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I think your approach is correct ..
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Nope. The sign will always be greater than equals to. As we are removing the number of students who have common three subjects and their sum can be equal or greater than total students minus the students with all three subjects
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200−x+100−x+60−x ≥ 300−x

 Can anybody explain why '>‘ sign in this line??

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@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 .

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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) $
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4 votes
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 $0$)

1 comment

Your inequality will give `$x \leq 20$`. You probably missed a minus x in the RHS.
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2 votes
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.

2 Comments

why 30, why not 20?
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See as we need to assign at least 1 Subject to 300 Student then 360 - 300 = 60.

Now we can max assign two subject to 60/2 = 30 Student

So, Max 30 student will have 3 subject and rest 270 will have 1 subject.

Total subject = 3 Subject + 1 Subject Student

360 = 30*3 + 270 #Answer
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Answer:

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