In a survey of the 100 out-patients who reported at a hospital one day it was found out that 70 complained of fever 50 complained of stomach ache and 30 were injured all 100 patients had at least one the complaint and 44 had exactly 2 of the complaint how many patients had all the three complaints
To find the number of patients who had all three complaints, we can use the principle of inclusion-exclusion.
Let's define:
A = Number of patients with fever
B = Number of patients with stomach ache
C = Number of patients who were injured
From the provided information, we know that:
A = 70 (patients with fever)
B = 50 (patients with stomach ache)
C = 30 (patients who were injured)
A ∩ B = 44 (patients with both fever and stomach ache)
To find the number of patients with all three complaints, we need to calculate A ∩ B ∩ C.
Using the principle of inclusion-exclusion:
A ∪ B ∪ C = A + B + C - (A ∩ B) - (A ∩ C) - (B ∩ C) + (A ∩ B ∩ C)
From the given values, we can fill in what we know and calculate:
100 = 70 + 50 + 30 - 44 - (A ∩ C) - (B ∩ C) + (A ∩ B ∩ C)
Simplifying the equation:
100 = 106 - (A ∩ C) - (B ∩ C) + (A ∩ B ∩ C)
We can rearrange the equation to solve for (A ∩ B ∩ C):
(A ∩ B ∩ C) = 106 - 100 + (A ∩ C) + (B ∩ C)
Let's calculate (A ∩ B ∩ C):
(A ∩ B ∩ C) = 106 - 100 + (A ∩ C) + (B ∩ C)
(A ∩ B ∩ C) = 6 + (A ∩ C) + (B ∩ C)
Since we don't have the exact values for (A ∩ C) and (B ∩ C), we cannot determine the exact number of patients who had all three complaints. However, we can conclude that it is at least 6.
To find out how many patients had all three complaints, we need to analyze the information provided.
We know that there were a total of 100 out-patients who reported at the hospital and that all of them had at least one of the complaints.
Let's break down the information and make a diagram:
- Number of patients who complained of fever (F): 70
- Number of patients who complained of stomach ache (S): 50
- Number of patients who were injured (I): 30
We can represent this information as a Venn diagram. Assume that the circles represent fever (F), stomach ache (S), and injury (I), respectively.
First, let's place the number of patients who complained of all three complaints in the center intersection of the three circles: X
Next, let's fill in the number of patients who have exactly two complaints. We know that 44 patients fall into this category. However, we cannot determine the exact count for each combination (e.g., FS, FI, or SI) based on the given information. So, we'll represent the number of patients in these regions as Y for now.
Finally, let's calculate the remaining patients who have only one complaint:
- Number of patients with only fever (F): 70 - X - Y
- Number of patients with only stomach ache (S): 50 - X - Y
- Number of patients with only injury (I): 30 - X - Y
Since all the patients accounted for in the diagram add up to 100, we can set up the following equation:
(X) + (Y) + (70 - X - Y) + (50 - X - Y) + (30 - X - Y) = 100
Simplifying the equation:
X + Y + 70 - X - Y + 50 - X - Y + 30 - X - Y = 100
Combine like terms:
150 - 4X - 4Y = 100
Rearrange the equation:
4X + 4Y = 50
Now, we can solve for X (the number of patients with all three complaints):
X = (50 - 4Y) / 4
To find the value of X, we can substitute different values for Y, starting from 0. We need to find a value for Y that satisfies the equation and gives us a whole number value for X.
By substitution, we find:
- Y = 6, X = (50 - 4(6)) / 4 = 8
Therefore, according to the given information, there are 8 patients who had all three complaints.
70+50+30 - (44+3x) + x = 100
x = 3