Question

Let awards in event A is 48 and awards in event B is 25 and awards in event C is 18 and also n(A ∪ B ∪ C) = 60, n(A ⋂ B ⋂ C) = 5, then how many got exactly two awards is?

• A. 21
• B. 25
• C. 24
• D. 23

Answer 1

The Correct Option is A

Let the total number of men be represented by the set \( A \cup B \cup C = 60 \) where: - \( |A| = 48 \) (men who received medals in event A) - \( |B| = 25 \) (men who received medals in event B) - \( |C| = 18 \) (men who received medals in event C) - \( |A \cup B \cup C| = 60 \) (total number of men) The number of men who received medals in all three events is given by: \[ |A \cap B \cap C| = 5 \] 

We need to find how many men received medals in exactly two events, which is calculated by: \[ |A \cap B| + |B \cap C| + |C \cap A| - 3|A \cap B \cap C| \] Using the inclusion-exclusion principle, we get: \[ |A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |B \cap C| - |C \cap A| + |A \cap B \cap C| \] Substituting the values we know: \[ 60 = 48 + 25 + 18 - |A \cap B| - |B \cap C| - |C \cap A| + 5 \] \[ 60 = 91 - |A \cap B| - |B \cap C| - |C \cap A| + 5 \] \[ |A \cap B| + |B \cap C| + |C \cap A| = 36 \] Now, to find the number of men who received exactly two medals, we use the formula: \[ \text{No. of men who received exactly 2 medals} = |A \cap B| + |B \cap C| + |C \cap A| - 3|A \cap B \cap C| \] Substituting the values: \[ \text{No. of men who received exactly 2 medals} = 36 - 15 = 21 \] Thus, the number of men who received exactly two medals is 21. 

Answer 2

Let the total number of men be represented by the set \( A \cup B \cup C = 60 \) where: - \( |A| = 48 \) (men who received medals in event A) - \( |B| = 25 \) (men who received medals in event B) - \( |C| = 18 \) (men who received medals in event C) - \( |A \cup B \cup C| = 60 \) (total number of men) The number of men who received medals in all three events is given by: \[ |A \cap B \cap C| = 5 \] \begin{figure} \centering \includegraphics[width=0.5\linewidth]{image4.png} \caption{Enter Caption} \label{fig:enter-label} \end{figure} We need to find how many men received medals in exactly two events, which is calculated by: \[ |A \cap B| + |B \cap C| + |C \cap A| - 3|A \cap B \cap C| \] Using the inclusion-exclusion principle, we get: \[ |A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |B \cap C| - |C \cap A| + |A \cap B \cap C| \] Substituting the values we know: \[ 60 = 48 + 25 + 18 - |A \cap B| - |B \cap C| - |C \cap A| + 5 \] \[ 60 = 91 - |A \cap B| - |B \cap C| - |C \cap A| + 5 \] \[ |A \cap B| + |B \cap C| + |C \cap A| = 36 \] Now, to find the number of men who received exactly two medals, we use the formula: \[ \text{No. of men who received exactly 2 medals} = |A \cap B| + |B \cap C| + |C \cap A| - 3|A \cap B \cap C| \] Substituting the values: \[ \text{No. of men who received exactly 2 medals} = 36 - 15 = 21 \] Thus, the number of men who received exactly two medals is 21.