How many students took AI?

		A	B	C	F	Total
CS	AI
	ML
Non CS	AI					2x
	ML					5x

Clue 1 says,

The numbers of non-CS students who took AI and ML were in the ratio 2 : 5.

Let it be, 2x and 5x

		A	B	C	F	Total
CS	AI					7x
	ML					7x
Non CS	AI					2x
	ML					5x

Clue 2 says,

The number of non-CS students who took either AI or ML was equal to the number of CS students.

Non-CS students who took either AI or ML = 2x + 5x = 7x

therefore, total number of CS students = 7x

And we know, All CS students took both the courses

=> In CS, AI = ML = 7x

		A	B	C	F	Total
CS	AI					7x
	ML			2y		7x
Non CS	AI				y	2x
	ML				y	5x

Clue 3,

The numbers of non-CS students who failed in the two courses were the same and their total is equal to the number of CS students who got a C grade in ML.

Let the number of non-CS students who failed in AI / ML = y

=> number of CS students who got a C grade in ML is also = y + y = 2y.

		A	B	C	F	Total
CS	AI				0	7x
	ML			2y		7x
Non CS	AI	0			y	2x
	ML				y	5x

Clue 5,

No CS student failed in AI, while no non-CS student got an A grade in AI.

=> $0$ CS students got F grade in AI

And, $0$ Non CS students got an A grade in ML.

		A	B	C	F	Total
CS	AI	3a	5a	2a	0	7x
	ML	4b	5b	2b		7x
Non CS	AI	0			b	2x
	ML				b	5x

Clue 6,

The numbers of CS students who got A, B and C grades respectively in AI were in the ratio 3 : 5 : 2, while in ML the ratio was 4 : 5 : 2.

Therefore, numbers of CS students who got A, B and C grades respectively in AI = 3a, 5a, and 2a

Also , in ML = 4b, 5b, 2b,

Now, we know CS students with C grade in ML = y

=> 2y = 2b

Also, For AI , 3a + 5a + 2a + 0 = 7x

=> 10a = 7x ... (1)

		A	B	C	F	Total
CS	AI	3a	5a	2a	0	7x
	ML	4b	5b	2b	2b/3	7x
Non CS	AI	0			b	2x
	ML				b	5x

Clue 7,

The ratio of the total number of non-CS students failing in one of the two courses to the number of CS students failing in one of the two courses was 3 : 1.

We know, total number of non-CS students failing in one of the two courses = b + b = 2b

=> number of CS students failing in one of the two courses will be = 2b/3

		A	B	C	F	Total
CS	AI	63	105	42	0	210
	ML	72	90	36	12	210
Non CS	AI	0			18	60
	ML				18	150

Clue 8,

30 students failed in ML.

we know, total students who failed in ML = $2b/3 + b$

=> $2b/3 + b = 30$

=> $b =18$

Substitute b's value in the table.

Which will give us x's value as ;

$36 + 45 + 18 + 24 = 7x$

=> $210 = 7x$

=> $x = 30$

=> since, $3a + 5a+ 2a = 210$

=> $10a = 210$

=> $a = 21$

		A	B	C	F	Total
CS	AI	63	105	42	0	210
	ML	72	90	36	12	210
Non CS	AI	0	21		18	60
	ML		75		18	150

Clue 4,

In both the courses, 50% of the students who passed got a B grade.

Now, total number of students who passed in ML ;

i) in CS = $210 - 12 = 198$

ii) in non CS = $150 - 18 = 132$

=> Half of $198 + 132 = 330$ will get B grade

=>$165$ got B grade in AI

out of which, 90 got it in CS

Therefore, 165 - 90 = 75 got B grade in non CS, ML.

Similairly for AI ;

total number of students who passed in AI ;

i) in CS = 210

ii) in non CS = 60 - 18 = 42

=> Half of 210 + 42 = 252 will get B grade

=> 126 got B grade in AI

out of which, 105 got it in CS

Therefore, 126 - 105 = 21 got B grade in non CS, AI.

		A	B	C	F	Total
CS	AI	63	105	42	0	210
	ML	72	90	36	12	210
Non CS	AI	0	21	21	18	60
	ML	27	75	30	18	150

Continuing with clue 4,

while the numbers of students who got A and C grades were the same for AI, they were in the ratio 3 : 2 for ML.

Since numbers of students who got A and C grades were the same for AI

=> $63 + 0 = 42 +$ non CS (AI)

=> non CS , AI = 21 students with C grade.

Now, the ratio is 3:2 for ML

Let the number of students with grade A in ML non CS = $m$

and, the number of students with grade c in ML non CS = $n$

=>$(72 + m) / (36 + n) = 3 / 2$

=> $3n - 2m = 36$...(1)

also, $m + 75 + n + 18 = 150$

=> $m + n = 57$...(2)

Solving eq (1) & (2) , we get,

$m = 27$

and $n = 30$

		A	B	C	F	Total
CS	AI	63	105	42	0	210
	ML	72	90	36	12	210
Non CS	AI	0	21	21	18	60
	ML	27	75	30	18	150

Total number of students taking AI = 210 + 60 = 270 ans.