Question:
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?
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?
Show Hint
To solve inclusion-exclusion problems, always remember to account for the overlap of sets, subtracting the number of elements in the intersection of all sets when required.
Updated On: Mar 25, 2026
- 15
- 9
- 21
- 10
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
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.
- \( |A| = 48 \) (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.
Was this answer helpful?
0
1
Top JEE Main Mathematics Questions
- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
- If , then the general value of is:
- The general solution of
- If the constant term in the binomial expansion of $\left(\frac{x^{\frac{3}{2}}}{2}-\frac{4}{x^l}\right)^9$ is $-84$ and the coefficient of $x^{-3 l}$ is $2^\alpha \beta$, where $\beta<0$ is an odd number, then $|\alpha l-\beta|$ is equal to_____
View More Questions
Top JEE Main Sets and Relations Questions
- If the set $R = {(a, b) : a + 5b = 42, a, b \in \mathbb{N}}$ has $m$ elements and $\sum_{n=1}^m (1 + i^n) = x + iy$, where $i = \sqrt{-1}$, then the value of $m + x + y$ is:
Let \( A = \{-3, -2, -1, 0, 1, 2, 3\} \). A relation \( R \) is defined such that \( xRy \) if \( y = \max(x, 1) \). The number of elements required to make it reflexive is \( l \), the number of elements required to make it symmetric is \( m \), and the number of elements in the relation \( R \) is \( n \). Then the value of \( l + m + n \) is equal to:
- Let \( A = \{2,3,4,5,6\} \) be a set. Consider \( R \) be a relation on \( A \times A \) such that \( (x,y) \, R \, (a,b) \) implies \( x \) divides \( a \) and \( y \leq b \), then total number of elements in \( R \) is:
- Let $A = \{1, 2, 3, 4, 5\}$ and $B = \{a, b, c\}$ then total number of functions from $A$ to $B$ which are not onto are
- Let \[ S=\{x\in[-\pi,\pi]:\sin x(\sin x+\cos x)=a,\; a\in\mathbb{Z}\}. \] Then \(n(S)\) is equal to:
View More Questions
Top JEE Main Questions
- In a hydrogen like atom, when an electron jumps from the $M$ - shell to the $L$ - shell, the wavelength of emitted radiation is $\lambda$. If an electron jumps from $N$-shell to the $L$-shell, the wavelength of emitted radiation will be :
- Two masses $m$ and $\frac{m}{2}$ are connected at the two ends of a massless rigid rod of length $l$. The rod is suspended by a thin wire of torsional constant $k$ at the centre of mass of the rod-mass system(see figure). Because of torsional constant $k$, the restoring torque is $\tau =k\theta$ for angular displacement $\theta$. If the rod is rota ted by $\theta_0$ and released, the tension in it when it passes through its mean position will be:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)- The value of is
- The number of terms of an is even. The sum of the odd terms is and of the even terms is . The last term exceeds the first by . Then the number of terms in the series is ______.
View More Questions