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

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 \(m + n\): 
\[ m = 15, \quad n = 30. \] \[ m + n = 15 + 30 = 45. \]

Answer 2

Step 1: Given data.
Total students = 220
10 students studied none of the subjects.
Therefore, number of students who studied at least one subject = 220 - 10 = 210.

Let:
M = number of students who studied Mathematics
P = number of students who studied Physics
C = number of students who studied Chemistry
Let x = number of students who studied all three subjects.

Also given:
At least 125 and at most 130 studied Mathematics → \( 125 \leq M \leq 130 \)
At least 85 and at most 95 studied Physics → \( 85 \leq P \leq 95 \)
At least 75 and at most 90 studied Chemistry → \( 75 \leq C \leq 90 \)
\( n(P \cap C) = 30, \; n(C \cap M) = 50, \; n(M \cap P) = 40. \)

Step 2: Apply the formula for union of three sets.
\[ n(M \cup P \cup C) = M + P + C - [n(M \cap P) + n(P \cap C) + n(C \cap M)] + n(M \cap P \cap C) \] Substitute the given values: \[ 210 = M + P + C - (40 + 30 + 50) + x \] \[ 210 = M + P + C - 120 + x \] \[ x = 330 - (M + P + C) \]

Step 3: Find the least and the greatest possible values of x.
To find the least value of x (m), the expression \( 330 - (M + P + C) \) must be minimum.
This happens when \( M + P + C \) is maximum.

Maximum values are \( M = 130, P = 95, C = 90 \).
\[ M + P + C = 130 + 95 + 90 = 315. \] Thus, \[ m = 330 - 315 = 15. \]
To find the greatest value of x (n), \( 330 - (M + P + C) \) must be maximum.
This happens when \( M + P + C \) is minimum.

Minimum values are \( M = 125, P = 85, C = 75 \).
\[ M + P + C = 125 + 85 + 75 = 285. \] Thus, \[ n = 330 - 285 = 45. \]

Step 4: Final calculation.
\[ m + n = 15 + 30 = 45. \]

Final Answer:
\[ \boxed{45} \]