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Question

In a beauty contest, half the number of experts voted for Mr. A and two third voted for Mr. B. 10 voted for both and 6 did not for either. How many experts were there in all?

The correct answer is

24

Experts Voting Calculation: Beauty Contest Problem

This problem involves figuring out the total number of experts who voted in a beauty contest, given information about how many voted for Mr. A, Mr. B, both, or neither. We can solve this using principles of sets and basic algebra.

Understanding the Voting Data

Let's denote the total number of experts by $T$. We are given the following information:

  • Experts who voted for Mr. A: $\frac{1}{2} T$
  • Experts who voted for Mr. B: $\frac{2}{3} T$
  • Experts who voted for both Mr. A and Mr. B: 10
  • Experts who did not vote for either Mr. A or Mr. B: 6

Applying Set Theory Principles

Let $A$ be the set of experts who voted for Mr. A, and $B$ be the set of experts who voted for Mr. B.

  • The number of experts who voted for Mr. A is $|A| = \frac{1}{2} T$.
  • The number of experts who voted for Mr. B is $|B| = \frac{2}{3} T$.
  • The number of experts who voted for both is the intersection of the two sets: $|A \cap B| = 10$.
  • The number of experts who voted for at least one of them is the union of the two sets: $|A \cup B|$.
  • The number of experts who voted for neither is $6$. This means these experts are outside the union $A \cup B$. Therefore, the total number of experts $T$ can be expressed as the sum of those who voted for at least one candidate and those who voted for neither: $T = |A \cup B| + 6$.

From this, we can write the size of the union as: $|A \cup B| = T - 6$.

Using the Inclusion-Exclusion Principle

The principle of inclusion-exclusion for two sets states that:

$$ |A \cup B| = |A| + |B| - |A \cap B| $$

Now, substitute the given values and expressions into this formula:

$$ T - 6 = \left( \frac{1}{2} T \right) + \left( \frac{2}{3} T \right) - 10 $$

Solving for the Total Number of Experts

Let's solve the equation for $T$:

  1. Combine the terms involving $T$ on one side of the equation. First, find a common denominator for $\frac{1}{2}$ and $\frac{2}{3}$, which is 6: $$ \frac{1}{2} T + \frac{2}{3} T = \frac{3}{6} T + \frac{4}{6} T = \frac{7}{6} T $$
  2. Substitute this back into the equation: $$ T - 6 = \frac{7}{6} T - 10 $$
  3. Rearrange the equation to isolate $T$. Move the $T$ terms to one side and the constant terms to the other: $$ 10 - 6 = \frac{7}{6} T - T $$
  4. Simplify both sides: $$ 4 = \left( \frac{7}{6} - 1 \right) T $$ $$ 4 = \left( \frac{7}{6} - \frac{6}{6} \right) T $$ $$ 4 = \frac{1}{6} T $$
  5. Solve for $T$ by multiplying both sides by 6: $$ T = 4 \times 6 $$ $$ T = 24 $$

So, there were 24 experts in total.

Verifying the Solution

Let's check if this total number works with the given information:

  • Total experts: $T = 24$
  • Voted for Mr. A: $\frac{1}{2} \times 24 = 12$
  • Voted for Mr. B: $\frac{2}{3} \times 24 = 16$
  • Voted for both: 10
  • Voted for A only: $12 - 10 = 2$
  • Voted for B only: $16 - 10 = 6$
  • Total who voted (union): $12 + 16 - 10 = 18$
  • Did not vote for either: $24 - 18 = 6$. This matches the given information.

The calculations confirm that the total number of experts is 24.

Final Answer

The total number of experts who participated in the voting is 24.

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Important Questions from Operations on Sets

  1. Match List I with List II

    Let R 1= {(1, 1), (2, 2), (3, 3)} and R 2 = {(1, 1), (1, 2), (1, 3), (1, 4)}

    List I

    List II

    (A) R 1∪ R 2

    (I) {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (3, 3)}

    (B) R 1- R 2

    (II) {1, 1}

    (C) R 1∩ R 2

    (III) {(1, 2), (1, 3), (1, 4)}

    (D) R 2- R 1

    (IV) {(2, 2), (3, 3)}

    Choose the correct answer from the options given below:

  2. Let R be a relation on a set A such that R = R-1, then R is

  3. Which of the following is an open set?

  4. If C = { 2, 4, 6, 8, 10, 12, 14, 16 }, and D = {5, 10, 15, 20}, then the number of elements in the set D - C is:

  5. In a survey where 100 students reported which subjects they like, 32 students in total liked Mathematics, 38 students liked Business and 30 students liked Literature. Moreover 7 students liked both Mathematics and Literature, 10 students liked both Mathematics and Business, 8 students liked both Business and Literature, 5 students liked all three subjects.

    Then the number of people who liked exactly one subject is

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