SOLUTION: In a survey of 100 out-patients who reported at a hospital one day, it was found that 70 complained of fever,50 has stomach ache and 30 were injured. All 100 patients had at least

Let F denotes the set of those who complain fever (and the number of elements in this set, at the same time).

Let S denotes the set of those who complain stomach ache (and the number of elements in this set, at the same time).

Let I denotes the set of those who complain injuries (and the number of elements in this set, at the same time).


Let FS denotes the intersection of the sets F and S (and the number of elements in this set, at the same time).

Let FI denotes the intersection of the sets F and I (and the number of elements in this set, at the same time).

Let SI denotes the intersection of the sets S and I (and the number of elements in this set, at the same time).


Finally, let FSI be the intersetion of the three sets F, S, and I (and the number of elements in this set, at the same time).


Then the number of those who has exactly 2 complaints is

    44 = FS + FI + SI - 3*FSI.     (1)


We have also the second equation from the condition


    100 = F + S + I - (FS + FI + SI) + FSI.     (2)


Now substitute the given values F= 70, S= 50  and  I = 30 into equation (2)


    100 = 70 + 50 + 30 - (FS + FI + SI) + FSI


and simplify it to get

    50 = (FS + FI + SI) - FSI.     (3)


Last step is to subtract eq(1) from eq(3). You will get


     6 = 2*FSI,


which implies  FSI = 6/2 = 3.


Answer.  The number of patients who have all the three complaints is 3.