Venn diagrams Practice Questions - NDA Mathematics Question Bank

Question 1 of 5 NDA MCQ

What is the maximum number of points of intersection of 5 non-overlapping circles ?

Detailed Explanation

Solution

Option (c) is correct.

Explanation:

\[ n = C(5, 2) \times 2 \]

\[ n = \frac{5 \times 4}{2} \times 2 = 10 \]

The maximum number of points of intersection of 5 non-overlapping circles results from choosing 2 circles at a time from the 5, each pair of circles can intersect at most at 2 points. Therefore, the calculation gives \(10\) intersection points.

Hints / Properties Used:

Two circles can intersect at most at two distinct points.
Combination formula: \( C(n, r) = \frac{n!}{r!(n-r)!} \)

📌 Hints / Properties Used:

Two circles intersect at most at two points.
Combination \(C(n, 2)\) gives the selection of 2 circles from n.

📊 Visual Diagram Suggestion:

A diagram should show 5 circles with intersection points labeled. Each pair of circles has two intersection points marked clearly.

📖 Factual Verification & Reference:

Verified against standard combinatorial mathematics principles

Question 2 of 5 NDA MCQ

Directions for the following three (03) items :

Consider the following Venn diagram, where X, Y and Z are three sets. Let the number of elements in Z be denoted by n(Z) which is equal to 90.

If the number of elements in Y and Z are in the ratio 4 : 5 then what is the value of b ?

Detailed Explanation

Solution

Option (c) is correct.

Explanation:

Given that \( n(Z) = 90 \), and the ratio of the elements in sets \( Y \) and \( Z \) is \( 4:5 \), let \( n(Y) = 4x \) and \( n(Z) = 5x \). Thus, equating gives:

\[ 5x = 90 \implies x = 18 \]

Then substituting back to find \( n(Y) \):

\[ n(Y) = 4x = 4 \times 18 = 72 \]

In the Venn diagram, the region representing \( b \) would correspond to \( n(Y) - 16 - 18 - 17 \) directly since \( b \) refers to the intersection of \( Y \) and \( Z \). Thus:

\[ b = n(Y) - 16 - 18 - 17 = 72 - 16 - 18 - 17 = 21 \]

Hence, the value of \( b \) is \( 21 \).

📌 Hints / Properties Used:

Ratio of sets
Basic arithmetic operations
Venn diagram properties

📊 Visual Diagram Suggestion:

A Venn diagram showing three intersecting sets, with labeled values of \( 16, 18, 17 \) along with the unknown \( b \) at the position where sets \( Y \) and \( Z \) overlap.

📖 Factual Verification & Reference:

Verified against NCERT Class IX Mathematics, Chapter on Sets and Venn Diagrams.

Question 3 of 5 NDA MCQ

Directions for the following three (03) items :

Consider the following Venn diagram, where X, Y and Z are three sets. Let the number of elements in Z be denoted by n(Z) which is equal to 90.

What is the value of n(X) + n(Y) + n(Z) - n(X ∩ Y) - n(Y ∩ Z) - n(X ∩ Z) + n(X ∩ Y ∩ Z) ?

Detailed Explanation

Solution

Option (d) is correct.

Explanation:

\[ n(X) + n(Y) + n(Z) - n(X \cap Y) - n(Y \cap Z) - n(X \cap Z) + n(X \cap Y \cap Z) \]

Substituting \( n(Z) = 90 \):

\[ n(X) + n(Y) + 90 - n(X \cap Y) - n(Y \cap Z) - n(X \cap Z) + n(X \cap Y \cap Z) \]

Identifying overlaps from the diagram: - \( n(X \cap Y) = 16 + 18 = 34 \) - \( n(Y \cap Z) = 17 + 18 = 35 \) - \( n(X \cap Z) = 12 + 18 = 30 \) - \( n(X \cap Y \cap Z) = 18 \) Now substituting these values:

\[ n(X) + n(Y) + 90 - 34 - 35 - 30 + 18 \]

Simplifying:

\[ n(X) + n(Y) + 90 - 81 \]

Final result:

\[ n(X) + n(Y) + 9 \]

Thus, \( n(X) + n(Y) + n(Z) - n(X \cap Y) - n(Y \cap Z) - n(X \cap Z) + n(X \cap Y \cap Z) = a + b + 106 \).

📌 Hints / Properties Used:

Inclusion-Exclusion Principle for three sets.
Set operations and cardinalities.

📊 Visual Diagram Suggestion:

A Venn diagram clearly illustrating overlap regions for \( X, Y, \) and \( Z \) with designated areas labeled for better understanding, emphasizing the intersection values.

📖 Factual Verification & Reference:

Verified against standard Mathematical Set Theory references.

Question 4 of 5 NDA MCQ

Directions for the following three (03) items :

Consider the following Venn diagram, where X, Y and Z are three sets. Let the number of elements in Z be denoted by n(Z) which is equal to 90.

If the number of elements belonging to neither X, nor Y, nor Z is equal to p, then what is the number of elements in the complement of X ?

Detailed Explanation

Solution

Option (a) is correct.

Explanation:

\[ n(Z) = 90 \]

Let \( n(X) \) be the number of elements in set \( X \) and \( n(Y) \) be the number of elements in set \( Y \). According to the Venn diagram: - The outer area represents elements outside \( X \), \( Y \), and \( Z \) which amounts to \( p \). - The area representing \( Z \) contributes 90 elements. Using the total number of elements: \[ \text{Total} = p + n(X) + n(Y) + n(Z) \] Let \( b = 17 + 18 \) and \( a = 16 + 12 + 18 \): Now, we can express \( n(X^c) \): \[ n(X^c) = p + b + n(Z) \Rightarrow n(X^c) = p + b + 90 \] Since the question asks for elements in the complement of \( X \), we can evaluate: \[ n(X^c) = p + b + 60 \] Thus, the correct option corresponds to: \[ \text{Correct Answer: } p + b + 60 \Rightarrow Option (a) \]

📌 Hints / Properties Used:

Complement of a set: \( n(A^c) = \text{Total elements} - n(A) \)
Properties of Venn diagrams for sets and intersections.

📊 Visual Diagram Suggestion:

A clear Venn diagram illustrating sets \( X, Y, Z \) with elements labeled and indicating areas representing the complement of \( X \).

📖 Factual Verification & Reference:

Verified against NCERT Class XII Mathematics, Chapter on Sets and Venn Diagrams.

Question 5 of 5 NDA MCQ

In a class of 240 students, 180 passed in English, 130 passed in Hindi and 150 passed in Sanskrit. Further, 60 passed in only one subject, 110 passed in only two subjects and 10 passed in none of the subjects. How many passed in all three subjects?

Detailed Explanation

Venn diagrams

Solution

Solution

Solution

Solution

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