Exactly 2 = 75.

Not posting the solution so others can try

I am getting 5

Math = 25 + 15 (all-3)
German = 15 + 15 (all-3)
English = 20 + 15 (all-3)

i.e. we have 15 students taking all-3 + 60 students
not taking all-3 for a total of 75.

Since there are 70 students in all, 5 of them should be
taking 2 courses at the same time.

Is the OA 5 ?

let :
M : students in maths, G : all students in geman ; E : all students in English
DM : are students either with nath and english or with maths and german; so are DG and DE
T : students in 3 class

( M - DM -T) +( G - DG- T) +( E-DE - T) = 70


summimg the following expessions : M+G+E - ( DM+DE+DG) - 3T= 70

Given M+E+G = 70
T = 15
BUT, (carefully) DM+DE+DG =2 sum of students with all double subjects , because DM has double components

LEt the sum of all students with double subjects be x;

70 =105 - 2X- 45

Here I can not go farther, because x would be negative,

Am I wrong.
Any comment,

Oooops, I was wrong, but I can not give up
in fact, the equation USING Venn Diagrams is, (SEE ATTACHED FIGURE)

LEt the sum of all students with double subjects be x;

M+G+E - x - 2T= 70

Given M+E+G = 40+35+30 = 105
T = 15

then x = 5

I hope it's clear

here's the oa:

MuEuG = M + E + G - MnE - MnG - EnG - 2(MnEnG)
MnE + MnG + EnG = M + E + G - 2(MnEnG) - MuEuG
MnE + MnG+ EnG = 40 + 30 + 35 - 2(15) - 70 = 105 - 30 - 70 = 5

Whenever an intersection occurs between 2 sets, (MnEnG) is counted twice, therefore you deduct one of it. If the intersection occurs between 3 sets, it is counted thrice; therefore you deduct two of it. And so forth.
If there are four sets, then the formula is A + B + C + D -(two) -(three)*2 -(four)*3 = total