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

From statement 3 we can say that the number of non- CS students who failed in AI and ML are b i
n each category. Now from statement 8, we can say that number of CS students who failed in MI i
s equal to 30-b. 

2 b 3

From statement 7, we can say that, 30− b = 1

or, 2b = 90-3b or, b= 18

CS students take both the AI and ML courses, therefore 10a= 10b +30 or, a= b+3 = 21

From statement 2, we can say that 10a = 7x or, x = 30 

Substituting the values of a, b and x in the above table. 

A total of 270 students took AI out of which 252 students passed and a total of 360 students took
ML out of which 330 students passed.

From statement 4, we can say that out of the 252 students who passed in AI, 126 of them got Gra
de B and 63 got Grade A and 63 got Grade C.

Similarly, We can say that out of the 330 students who passed in ML, 165 of them got Grade C and
99 got Grade A and 66 got Grade C. Therefore, the final table which we get is