Given, a+b+c=50 and a+b+c+x+y+z=190 => x+y+z=140.

Also, let E1=k => E2=2k and E3=3k

Also, E1 = (x+a+0+b) , E2 = (y+b+0+c) , E3 = (z+a+0+c) ----------------2

From 1, E1+E2+E3 = 6k and from 2, E1+E2+E3 = (x+y+z)+2(a+b+c) = 140 + 2*50 = 240

=> 6k = 240

=> k = 40

Thus our venn diagram will look like this -

Option A,  the number of students choosing E1 is 40.

Option B, number of students choosing either E1 or E2 or both, but not E3 = total - E3 = 190-120 = 70. Or, it is equal to (40-a-b) + b + (80-b-c) = 120 - a - b - c = 120 - (a+b+c), and (a+b+c) = 50. Thus, the number of students is = 120 - 50 = 70.

Option C, number of students choosing both E1 and E2 => this can not be obtained, because it is equal to b and the value of b is not known.

Option D, number of students choosing E3 = 3k = 120.

Option E, number of students choosing exactly one elective = Out of 190, 50 are choosing two electives, hence 190-50 = 140 are choosing exactly one elective. Or, it is equal to (40-a-b) + (80-b-c) + (120-a-c) = 240 - 2(a+b+c) = 240 - 2(50) = 140.