Set Theory / Venn Diagrams Questions and Answers

Answer: B

Total students = 60. Students liking neither = 5.

Number of students who like at least one activity = n(M ∪ D) = 60 - 5 = 55.

Given n(M)=45, n(D)=50.

n(M ∪ D) = n(M) + n(D) - n(M ∩ D)

55 = 45 + 50 - n(M ∩ D)

55 = 95 - n(M ∩ D)

n(M ∩ D) = 95 - 55 = 40.

Answer: D

Let's use properties. A - B = A ∩ B'.

So the expression is A - (A ∩ B') = A ∩ (A ∩ B')'.

Using De Morgan's Law: A ∩ (A' ∪ B'') = A ∩ (A' ∪ B).

Using Distributive Law: (A ∩ A') ∪ (A ∩ B).

Since A ∩ A' = ∅, the expression becomes ∅ ∪ (A ∩ B), which is equal to A ∩ B.

Answer: A

We first find the number of candidates who passed in at least one subject, n(P ∪ C ∪ M).

n(P∪C∪M) = n(P)+n(C)+n(M) - n(P∩C) - n(C∩M) - n(P∩M) + n(P∩C∩M)

= 120 + 100 + 90 - 40 - 30 - 35 + 15

= 310 - 105 + 15 = 205 + 15 = 220.

Number of candidates who passed in at least one subject is 220.

Total candidates = 250.

Number who failed in all three = Total - Passed in at least one = 250 - 220 = 30.

Answer: B

Prime numbers have exactly two factors (1 and itself). Composite numbers have more than two factors. By definition, a number cannot be both prime and composite. Therefore, their intersection A ∩ B is the empty set (∅).

Option A is incorrect because the number 1 is neither prime nor composite, so it would not be in the union.

Options C and D are incorrect because 1 is not considered prime or composite.

Answer: C

Correct answer: 13

Write the inclusion–exclusion identity for three sets, with \(D\), \(R\) and \(M\) for dance, dramatics and music.

\[n(D \cup R \cup M) = n(D)+n(R)+n(M) - \Sigma\,\text{pairs} + n(D \cap R \cap M)\]

\[45 = 36+12+18 - \Sigma\,\text{pairs} + 4 = 70 - \Sigma\,\text{pairs}\]

So the pairwise intersections total \(\Sigma\,\text{pairs} = 25\).

Each person holding all three medals is counted in all three of those pairwise intersections, so subtract them three times over to leave those in exactly two:

\[25 - 3 \times 4 = 25 - 12 = 13\]

Answer: C

Let D, R, and M be the sets of people who got medals in Dance, Dramatics, and Music respectively.

Given: n(D)=36, n(R)=12, n(M)=18, n(D ∪ R ∪ M)=45, n(D ∩ R ∩ M)=4.

We use the formula:

n(D∪R∪M) = n(D)+n(R)+n(M) - [n(D∩R) + n(R∩M) + n(M∩D)] + n(D∩R∩M)

45 = (36 + 12 + 18) - [Sum of intersections of pairs] + 4

45 = 66 - [Sum of pairs] + 4

45 = 70 - [Sum of pairs]

Sum of pairs = n(D∩R) + n(R∩M) + n(M∩D) = 25.

The number of people who received medals in exactly two categories is given by:

[n(D∩R) + n(R∩M) + n(M∩D)] - 3 × n(D∩R∩M)

= 25 - 3 × 4 = 25 - 12 = 13.

Answer: B

The condition -3 < x ≤ 4 means x can be any integer strictly greater than -3 and less than or equal to 4.

The integers satisfying this condition are: -2, -1, 0, 1, 2, 3, 4.

The set A = {-2, -1, 0, 1, 2, 3, 4}.

The number of elements, n(A), is 7.

Answer: C

4 students watched all three films.

Let x students watch all three; the other \(40 - x\) watched exactly one film each. Counting viewings film by film, each single-film student is counted once and each all-three student three times:

\[ \begin{aligned} 13 + 16 + 19 &= (40 - x) + 3x \\ 48 &= 40 + 2x \Rightarrow x = 4 \end{aligned} \]

Check: 36 students saw one film and 4 saw all three, giving \(36 + 12 = 48\) viewings. With 0 or 2 students watching all three, the viewings would total only 40 or 44; with 8 they would total 56.

Answer: C

The group has 24 students.

Those who like at least one subject number, by inclusion and exclusion,

\[ n(M \cup E) = 10 + 12 - 4 = 18 \]

Adding the 6 who like neither gives \(18 + 6 = 24\).

18 forgets the 6 students who like neither subject. 20 adds 10 and 12 but forgets both corrections, and 32 adds every number in the question, counting the 4 students who like both twice and treating them as a separate group.

Answer: D

The number who failed is 4,20,000.

Work in thousands. Let \(x\) be the number answering all three correctly, which equals the number answering none. By inclusion-exclusion, the number answering at least one is

\[ \begin{aligned} &330 + 250 + 260 - 100 - 90 - 80 + x \\ &= 570 + x \end{aligned} \]

so the number answering none is \(630 - 570 - x = 60 - x\). Setting this equal to \(x\) gives \(x = 30\). Exactly two correct: \((100 + 90 + 80) - 3x = 180\). Those who pass have two or three correct: \(180 + 30 = 210\). So \(630 - 210 = 420\) thousand fail.

30,000 is the number answering all (or none), and 2,70,000 is the sum of the pairwise counts, which double-counts those answering all three.