Question

In a survey of 220 students of a higher secondary school, it was found that at least 125 and at most 130 students studied Mathematics; at least 85 and at most 95 studied Physics; at least 75 and at most 90 studied Chemistry; 30 studied both Physics and Chemistry; 50 studied both Chemistry and Mathematics; 40 studied both Mathematics and Physics and 10 studied none of these subjects. Let m and n respectively be the least and the most number of students who studied all the three subjects. Then m + n is equal to _____

Answer 1

Correct Answer: 45

Input Data:
Let:
\(M\) = count of students studying Mathematics.
\(P\) = count of students studying Physics.
\(C\) = count of students studying Chemistry.

Constraints:
\(125 \leq M \leq 130\)
\(85 \leq P \leq 95\)
\(75 \leq C \leq 90\)

Students studying two subjects:
\(|P \cap C| = 30\)
\(|C \cap M| = 50\)
\(|M \cap P| = 40\)

Students studying no subjects = 10.
This implies the total number of students in at least one subject is \(|M \cup P \cup C| = 210\), calculated as Total Students - Students studying none.

Set Union Formula Application:
The formula for the union of three sets is:
\(|M \cup P \cup C| = M + P + C - |M \cap P| - |P \cap C| - |C \cap M| + |M \cap P \cap C|\)

Substituting known values:
\(210 = M + P + C - 40 - 30 - 50 + x\)
where \(x\) represents the count of students studying all three subjects.

Simplified equation:
\(M + P + C + x = 330\)

Determining the Range for \(x\):
Using the given ranges for \(M\), \(P\), and \(C\):
\(125 \leq M \leq 130\)
\(85 \leq P \leq 95\)
\(75 \leq C \leq 90\)

The calculated range for \(x\) is:
\(15 \leq x \leq 30\)

Calculation of \(m + n\):
Let \(m\) be the minimum value of \(x\) and \(n\) be the maximum value of \(x\).
\(m = 15\)
\(n = 30\)

Therefore, \(m + n = 15 + 30 = 45\).