in Quantitative Aptitude edited by
2,537 views
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
2.5k views

2 Comments

(B) is correct.
4
4

(B) is the correct option!

3
3

3 Answers

16 votes
16 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: $B$
edited by
by

4 Comments

200−x+100−x+60−x ≥ 300−x

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

0
0

@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
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) $
2
2
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.
0
0
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?
0
0
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
0
0
Answer:

Related questions