Question

In a group of 200 Boys, 80 play Cricket, 60 play Chess and 35 play both Cricket and Chess. Number of boys play neither Cricket nor Chess is

• A. 95
• B. 90
• C. 105
• D. 85

Hint:

Using the quick formula:
\[ \text{Neither} = \text{Total} - (\text{Set A} + \text{Set B} - \text{Both}) \] \[ \text{Neither} = 200 - (80 + 60 - 35) = 200 - 105 = 95 \] Keep this simple subtraction path in mind for set theory questions.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This question is a practical application of set theory and Venn diagrams, requiring us to find the number of elements outside the union of two given sets.

Step 2: Key Formulas and approach:
Let the universal set of boys be \(U\), where \(n(U) = 200\).
Let the set of boys playing Cricket be \(C\) and those playing Chess be \(H\).
We are given:
\[ n(C) = 80 \] \[ n(H) = 60 \] The number of boys playing both games is the intersection:
\[ n(C \cap H) = 35 \] We use the Set Union Formula to find the number of boys playing at least one game:
\[ n(C \cup H) = n(C) + n(H) - n(C \cap H) \] The number of boys playing neither game is given by:
\[ \text{Neither} = n(U) - n(C \cup H) \]

Step 3: Detailed Explanation:

• Compute the number of boys playing at least one of the two sports:
\[ n(C \cup H) = 80 + 60 - 35 \] \[ n(C \cup H) = 140 - 35 = 105 \]

• Now, subtract this union set size from the total population of boys to find those who play neither sport:
\[ \text{Neither} = 200 - 105 = 95 \]

• Alternatively, we can calculate using individual exclusive regions in a Venn diagram:
Only Cricket \( = 80 - 35 = 45 \)
Only Chess \( = 60 - 35 = 25 \)
Both Cricket and Chess \( = 35 \)
Sum of active players \( = 45 + 25 + 35 = 105 \)
Inactive players \( = 200 - 105 = 95 \)


Step 4: Final Answer:
The number of boys who play neither sport is \(95\), making Option (A) the correct choice.