Out of a group of students taking examinations in Mathematics, Physics, and Chemistry, students passed Mathematics, passed Physics, and passed Chemistry. Additionally, no more than students passed both Mathematics and Physics, no more than passed both Mathematics and Chemistry, and no more than passed both Physics and Chemistry. What is the maximum number of students who could have passed all three examinations?

14 Explanation: Let the sets of students who passed Mathematics, Physics, and Chemistry be represented by , , and respectively. From the given data: Total students, We need to find the maximum possible value of students who passed all three subjects, which is . Let this value be . Step 1: Apply the Principle of Inclusion-Exclusion The total number of students who passed at least one exam satisfies the inequality: Using the set theory formula for three sets: Substituting the known values into the inequality: Step 2: Find upper bounds using individual set constraints Every student who passes all three exams ( ) is inherently part of the groups passing two exams. Therefore, the following conditions must hold true so that the number of students passing only two subjects remains non-negative: Additionally, the

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Explanation

Let the sets of students who passed Mathematics, Physics, and Chemistry be represented by $M$ , $P$ , and $C$ respectively.

From the given data:

Total students, $n (U) = 50$
$n (M) = 37$
$n (P) = 24$
$n (C) = 43$
$n (M \cap P) \leq 19$
$n (M \cap C) \leq 29$
$n (P \cap C) \leq 20$

We need to find the maximum possible value of students who passed all three subjects, which is $n (M \cap P \cap C)$ . Let this value be $x$ .

Step 1: Apply the Principle of Inclusion-Exclusion

The total number of students who passed at least one exam satisfies the inequality:

$n (M \cup P \cup C) \leq n (U)$

Using the set theory formula for three sets:

$n (M \cup P \cup C) = n (M) + n (P) + n (C) - n (M \cap P) - n (M \cap C) - n (P \cap C) + n (M \cap P \cap C)$

Substituting the known values into the inequality:

$37 + 24 + 43 - n (M \cap P) - n (M \cap C) - n (P \cap C) + x \leq 50$

$104 - [n (M \cap P) + n (M \cap C) + n (P \cap C)] + x \leq 50$

$x \leq [n (M \cap P) + n (M \cap C) + n (P \cap C)] - 54 — (Equation 1)$

Step 2: Find upper bounds using individual set constraints

Every student who passes all three exams ( $x$ ) is inherently part of the groups passing two exams. Therefore, the following conditions must hold true so that the number of students passing only two subjects remains non-negative:

$n (M \cap P) \geq x$
$n (M \cap C) \geq x$
$n (P \cap C) \geq x$

Additionally, the total number of elements belonging to any single set (like Physics) cannot exceed its total given count. Let's look at the constraint for the smallest set, Physics ( $P = 24$ ):

$n (P) \geq n (M \cap P) + n (P \cap C) - n (M \cap P \cap C)$

$24 \geq n (M \cap P) + n (P \cap C) - x$

$n (M \cap P) + n (P \cap C) \leq 24 + x — (Equation 2)$

Step 3: Maximize the intersection sum

To find the absolute upper limit for $x$ , we substitute Equation 2 and the maximum given limit for $n (M \cap C) \leq 29$ back into Equation 1:

$x \leq [n (M \cap P) + n (P \cap C)] + n (M \cap C) - 54$

$x \leq (24 + x) + 29 - 54$

$x \leq x + 53 - 54$

$x \leq x - 1$

Since this point defines a boundary on the total universe capacity, let's look at the maximum limits directly. To maximize $x$ , the two-set intersections must be as large as possible. Let's test the maximum values given in the question:

$n (M \cap P) = 19$

$n (M \cap C) = 29$

$n (P \cap C) = 20$

Substitute these maximum values directly into Equation 1:

$x \leq (19 + 29 + 20) - 54$

$x \leq 68 - 54$

$x \leq 14$

Thus, the maximum number of students who could have passed all three examinations is $14$ .

Correct Answer: A) 14

Explanation

Let the sets of students who passed Mathematics, Physics, and Chemistry be represented by $M$ , $P$ , and $C$ respectively.

From the given data:

Total students, $n (U) = 50$
$n (M) = 37$
$n (P) = 24$
$n (C) = 43$
$n (M \cap P) \leq 19$
$n (M \cap C) \leq 29$
$n (P \cap C) \leq 20$

We need to find the maximum possible value of students who passed all three subjects, which is $n (M \cap P \cap C)$ . Let this value be $x$ .

Step 1: Apply the Principle of Inclusion-Exclusion

The total number of students who passed at least one exam satisfies the inequality:

$n (M \cup P \cup C) \leq n (U)$

Using the set theory formula for three sets:

$n (M \cup P \cup C) = n (M) + n (P) + n (C) - n (M \cap P) - n (M \cap C) - n (P \cap C) + n (M \cap P \cap C)$

Substituting the known values into the inequality:

$37 + 24 + 43 - n (M \cap P) - n (M \cap C) - n (P \cap C) + x \leq 50$

$104 - [n (M \cap P) + n (M \cap C) + n (P \cap C)] + x \leq 50$

$x \leq [n (M \cap P) + n (M \cap C) + n (P \cap C)] - 54 — (Equation 1)$

Step 2: Find upper bounds using individual set constraints

$n (M \cap P) \geq x$
$n (M \cap C) \geq x$
$n (P \cap C) \geq x$

Additionally, the total number of elements belonging to any single set (like Physics) cannot exceed its total given count. Let's look at the constraint for the smallest set, Physics ( $P = 24$ ):

$n (P) \geq n (M \cap P) + n (P \cap C) - n (M \cap P \cap C)$

$24 \geq n (M \cap P) + n (P \cap C) - x$

$n (M \cap P) + n (P \cap C) \leq 24 + x — (Equation 2)$

Step 3: Maximize the intersection sum

To find the absolute upper limit for $x$ , we substitute Equation 2 and the maximum given limit for $n (M \cap C) \leq 29$ back into Equation 1:

$x \leq [n (M \cap P) + n (P \cap C)] + n (M \cap C) - 54$

$x \leq (24 + x) + 29 - 54$

$x \leq x + 53 - 54$

$x \leq x - 1$

$n (M \cap P) = 19$

$n (M \cap C) = 29$

$n (P \cap C) = 20$

Substitute these maximum values directly into Equation 1:

$x \leq (19 + 29 + 20) - 54$

$x \leq 68 - 54$

$x \leq 14$

Thus, the maximum number of students who could have passed all three examinations is $14$ .

Correct Answer: A) 14

NIMCET 2024 — Mathematics PYQ

Choose the correct answer:

Explanation

Step 1: Apply the Principle of Inclusion-Exclusion

Step 2: Find upper bounds using individual set constraints

Step 3: Maximize the intersection sum

Explanation

Step 1: Apply the Principle of Inclusion-Exclusion

Step 2: Find upper bounds using individual set constraints

Step 3: Maximize the intersection sum