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 _____
Given data:
Let:
\(M =\) number of students who studied Mathematics,
\(P =\) number of students who studied Physics,
\(C =\) number of students who studied Chemistry.
Given conditions:
\[ 125 \leq M \leq 130, \quad 85 \leq P \leq 95, \quad 75 \leq C \leq 90. \]
Number of students studying two subjects:
\[ |P \cap C| = 30, \quad |C \cap M| = 50, \quad |M \cap P| = 40. \]
Number of students studying none:
\[ |U| - |M \cup P \cup C| = 10 \implies |M \cup P \cup C| = 210. \]
Using the formula for the union of three sets:
\[ |M \cup P \cup C| = M + P + C - |M \cap P| - |P \cap C| - |C \cap M| + |M \cap P \cap C|. \]
Substituting the values: \[ 210 = M + P + C - 40 - 30 - 50 + x, \]
where \(x\) is the number of students who studied all three subjects.
Simplifying: \[ M + P + C + x = 330. \]
Finding the range for \(x\):
From the given bounds: \[ 125 \leq M \leq 130, \quad 85 \leq P \leq 95, \quad 75 \leq C \leq 90. \]
Therefore: \[ 15 \leq x \leq 30. \]
Calculating \(m + n\):
\[ m = 15, \quad n = 30. \] \[ m + n = 15 + 30 = 45. \]
If \[ f(x) = \int \frac{1}{x^{1/4} (1 + x^{1/4})} \, dx, \quad f(0) = -6 \], then f(1) is equal to:
If the system of equations \[ (\lambda - 1)x + (\lambda - 4)y + \lambda z = 5 \] \[ \lambda x + (\lambda - 1)y + (\lambda - 4)z = 7 \] \[ (\lambda + 1)x + (\lambda + 2)y - (\lambda + 2)z = 9 \] has infinitely many solutions, then \( \lambda^2 + \lambda \) is equal to: