First, draw a big square or rectangle to show the universal set.
In a corner of this big square, write ε=120 to denote the universal set.
1. Draw in the centre of the rectangle, three circles, one for each of Physics, Math and Chemistry.
2. Starting from the intersection of the three, write "x" to show we don't know how many students study the three subjects.
3. Where Physics and Chemistry intersect, write "8"; for Math and Physics write "12" and for Chemistry and Math write "7".
NOTE: We're writing the numbers 'as is' because it is signified "ONLY".
Now write in the space for Maths "60-(12+7+x)"; write "40-(12+8+x)" for Physics, and write "55-(8+7+x)" for Chemistry.
PS: What "60-(12+7+x)" means is;
A total of 60 students offer Maths (either alone or combined with other subjects).
Now subtract the number of people that offer Math and one other subject, and those that offer all three, and you'd get the number of students who offer Math ONLY. (Apply this knowledge to others).
Now to solve for x, which represents the number of students offering all three subjects, we write:
[60-(19+x)] + [40-(20+x)] + [55-(15+x)] + 27 + x + 10 = 120
60-19+40-20+55-15+27+10-x-x-x+x=120 (collect like terms)
-2x=120-138 (collect like terms)
Therefore, 9 students offer all three subjects.
You can substitute 9 for x in each category and check.
9 students offer all three subjects
7 offer only Math and Chemistry
8 offer Physics and Chemistry only
12 offer Physics and Math only
32 offer only Math
11 offer Physics only
31 offer Chemistry only and, finally,
10 students offer none of the three.
9+7+8+12+32+11+31+10 = 120