Let the no. of schools with three of the facilities was the same irrespective of which three were be x.
Number of schools without facilities be 'n' from 1, n=80.
Number of schools where only F1 and F2 be 'b'
Number of schools where only F1 and F3 be 'c'
Number of schools where only F1 and F4 be 'd'
From the the question we will get the following Venn diagram.
From 5, b+141+3x=313
⇒ b+3x=172....(i)
From 8, b+x+40+x=162
⇒ b+2x=122....(ii)
equation (ii)-(i) gives x=50 and b=22
From 9, 237+3x+c+d=279+3x=d ⇒ c=42
Total number of schools =600 ⇒ 313+25+c+x+d+26+24+20+80=600
⇒ d=20.
Now the table looks will be :