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

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Question 1096428: In a survey of the 100 out-patients who reported at a hospital one day, it was found out that 70 complained of fever, and 50 complained of stomach ache and 30 were injured. All 100 patients had at least one of the complaints and 44 had exactly two of the complaints. How many patients had all three complaints?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

The given numbers are impossible.

By the inclusion-exclusion principle, if the number of patients having all three complaints is x, then we must have





not possible!


Note: I disagree with tutor ikleyn's solution to the problem. In her solution, she took the number 44 to be the total number of patients who had AT LEAST 2 of the 3 complaints. But the problem says the 44 is the total number who had EXACTLY 2 of the 3 complaints.
Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
.
Let F be the set and the number of those who complained of fever        (F = 70);
    S be the set and the number of those who complained of stomach ache (S = 50);
    I be the set and the number of those who wee injured                (I = 30).


Do not be worried that I denoted by the same symbol the set and the number: I made it for simplicity, 
and you always can distinct from the context what I am talking about.


Let FS be the intersection of the sets F and S and the number of elements in this set at the same time.

Let FI be the intersection of the sets F and I and the number of elements in this set at the same time.

Let SI be the intersection of the sets S and I and the number of elements in this set at the same time.



    Let FSI = x be the intersection of the sets F, S and I and the number of elements in this intersection at the same time.

    I called the last quantity as "x", since it is our major unknown in this problem.



From the elementary theory of finite sets, we have this equation 

100 = F + S + I - FS - FI - SI + x.       (1)


From the given part of the condition, we have this equation

(FS-x) + (FI-x) + (SI-x) = 44.            (2)


Now, we can re-write equation (1) in this form

100 = 70 + 50 + 30 - (FS-x) - (FI-x) - (SI-x) -3x + x = 

    = 150 - [(FS-x) + (FI-x) + (SI-x)] - 2x = 150 - 44 - 2x = 106 - 2x,

which gives us 

2x = 106 - 100 = 6.


Hence,  x = 3.


Answer.  The number of patients who had all three complaints was 3.

For the key equation (1) see my lessons
    - Counting elements in sub-sets of a given finite set
and especially
    - Advanced problems on counting elements in sub-sets of a given finite set
in this site.


================

Regarding the note by  @greenestamps  in his post,  my comment is   THIS : 

    He has his right to have his own opinion, but in the given case it is  W R O N G.


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