Class JEE Mathematics Sets, Relations, and Functions Q #1030
COMPETENCY BASED
APPLY
4 Marks 2023 JEE Main 2023 (Online) 11th April Morning Shift MCQ SINGLE
An organization awarded $48$ medals in event 'A', $25$ in event 'B' and $18$ in event 'C'. If these medals went to total $60$ men and only five men got medals in all the three events, then, how many received medals in exactly two of three events?
(A) $10$
(B) $15$
(C) $21$
(D) $9$
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Correct Answer: C
Explanation
Let $|A|$ be the number of medals in event A, $|B|$ be the number of medals in event B, and $|C|$ be the number of medals in event C. We are given $|A| = 48$, $|B| = 25$, and $|C| = 18$. We are also given that the total number of people who received medals is $60$, so $|A \cup B \cup C| = 60$, and the number of people who received medals in all three events is $5$, so $|A \cap B \cap C| = 5$.

We want to find the number of people who received medals in exactly two of the three events. Let $x$ be the number of people who received medals in exactly two events. Using the Principle of Inclusion-Exclusion, we have:
$|A \cup B \cup C| = |A| + |B| + |C| - (|A \cap B| + |A \cap C| + |B \cap C|) + |A \cap B \cap C|$
$60 = 48 + 25 + 18 - (|A \cap B| + |A \cap C| + |B \cap C|) + 5$
$60 = 91 - (|A \cap B| + |A \cap C| + |B \cap C|) + 5$
$|A \cap B| + |A \cap C| + |B \cap C| = 91 + 5 - 60 = 36$

Now, let $N_{2}$ be the number of people who received medals in exactly two events. Then:
$|A \cap B| + |A \cap C| + |B \cap C| = N_{2} + 3|A \cap B \cap C|$
$36 = N_{2} + 3(5)$
$N_{2} = 36 - 15 = 21$
Therefore, the number of people who received medals in exactly two of the three events is $21$.

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Step-by-Step Solution

Let $n(A)$, $n(B)$, and $n(C)$ be the number of medals awarded in events A, B, and C, respectively.

Given: $n(A) = 48$, $n(B) = 25$, $n(C) = 18$.

Total number of men who received medals, $n(A \cup B \cup C) = 60$.

Number of men who got medals in all three events, $n(A \cap B \cap C) = 5$.

We know the formula for the union of three sets:

$n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(A \cap C) + n(A \cap B \cap C)$

Plugging in the given values:

$60 = 48 + 25 + 18 - n(A \cap B) - n(B \cap C) - n(A \cap C) + 5$

$60 = 96 - n(A \cap B) - n(B \cap C) - n(A \cap C)$

$n(A \cap B) + n(B \cap C) + n(A \cap C) = 96 - 60 = 36$

Let $x$ be the number of men who received medals in exactly two of the three events.

Let $n(A \cap B)$ be the number of men who received medals in both A and B, $n(B \cap C)$ be the number of men who received medals in both B and C, and $n(A \cap C)$ be the number of men who received medals in both A and C.

Then, $n(A \cap B) = $ (number of men who received medals in exactly A and B) + (number of men who received medals in A, B, and C)

Similarly for $n(B \cap C)$ and $n(A \cap C)$.

So, $n(A \cap B) + n(B \cap C) + n(A \cap C) = $ (number of men who received medals in exactly two events) + $3 \times$ (number of men who received medals in all three events)

$36 = x + 3 \times 5$

$36 = x + 15$

$x = 36 - 15 = 21$

Correct Answer: 21

AI Suggestion: Option C

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because it requires the student to apply the principle of inclusion and exclusion to solve a word problem.
Knowledge Dimension: CONCEPTUAL
Justification: The question requires understanding of set theory concepts like union and intersection, and applying the principle of inclusion-exclusion.
Syllabus Audit: In the context of JEE, this is classified as COMPETENCY. It is not a direct textbook question, but rather an application of set theory principles to solve a problem.

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