
In a society, 25% of the residents won both Cricket and Hockey games. 37.5% of the residents lost in both the games, while 60% of the residents lost in Hockey. The difference between the residents who lost only in Cricket and those who lost only in hockey is 30. Find the total number of residents.

Let's break down the information provided and use set theory and Venn diagrams to solve this problem.

### Step 1: Define the sets.
Let:
- \( A \) = the set of residents who won the **Cricket** game.
- \( B \) = the set of residents who won the **Hockey** game.
- \( |A^c| \) = the set of residents who **lost** the Cricket game.
- \( |B^c| \) = the set of residents who **lost** the Hockey game.
- \( n(U) \) = total number of residents.

### Step 2: Translate the problem into equations.

- \( 25\% \) of the residents won both **Cricket** and **Hockey**.
  \[
  P(A \cap B) = 0.25n(U)
  \]
- \( 37.5\% \) of the residents lost both **Cricket** and **Hockey**.
  \[
  P(A^c \cap B^c) = 0.375n(U)
  \]
- \( 60\% \) of the residents lost **Hockey**.
  \[
  P(B^c) = 0.60n(U)
  \]
- The difference between the number of residents who lost only in **Cricket** and those who lost only in **Hockey** is 30. This gives:
  \[
  |A^c \cap B| - |A \cap B^c| = 30
  \]

### Step 3: Express relationships using set theory.

- \( P(B^c) = P(A^c \cap B^c) + P(A \cap B^c) \), because people who lost **Hockey** either lost both games or won **Cricket**.
  \[
  0.60n(U) = 0.375n(U) + P(A \cap B^c)
  \]
  Solving for \( P(A \cap B^c) \):
  \[
  P(A \cap B^c) = 0.60n(U) - 0.375n(U) = 0.225n(U)
  \]

### Step 4: Find the residents who lost only **Cricket**.

- Similarly, \( P(A^c) = P(A^c \cap B^c) + P(A^c \cap B) \), because people who lost **Cricket** either lost both games or won **Hockey**.
- But we are not given \( P(A^c) \), so we need to use the difference condition:
  \[
  |A^c \cap B| - |A \cap B^c| = 30
  \]
  Substitute \( P(A \cap B^c) = 0.225n(U) \):
  \[
  |A^c \cap B| = 0.225n(U) + 30
  \]
  This is the number of people who lost only **Cricket**.

### Step 5: Total population equation.

Now we know:
- \( P(A \cap B) = 0.25n(U) \) (won both),
- \( P(A^c \cap B^c) = 0.375n(U) \) (lost both),
- \( P(A \cap B^c) = 0.225n(U) \) (won **Cricket**, lost **Hockey**),
- \( P(A^c \cap B) = 0.225n(U) + 30 \) (won **Hockey**, lost **Cricket**).

The total population is the sum of all these disjoint sets:
\[
n(U) = P(A \cap B) + P(A^c \cap B^c) + P(A \cap B^c) + P(A^c \cap B)
\]
Substituting the values:
\[
n(U) = 0.25n(U) + 0.375n(U) + 0.225n(U) + (0.225n(U) + 30)
\]
Simplifying:
\[
n(U) = 1.075n(U) + 30
\]
\[
n(U) - 1.075n(U) = 30
\]
\[
-0.075n(U) = 30
\]
\[
n(U) = \frac{30}{0.075} = 400
\]

### Final Answer:
The total number of residents is **400**.

Would you like further explanation on any step?

Here are five related questions to consider:
1. How would the problem change if the percentage of people who lost both games increased?
2. What happens if the number of residents winning both games is decreased?
3. How can Venn diagrams be used to visualize this problem?
4. What if 70% of the residents won at least one game, how would that affect the solution?
5. How could this problem be extended to include more games?

**Tip**: When solving word problems, breaking the information into parts and creating equations often simplifies the process.

In a society, 25% of the residents won both Cricket and Hockey games. 37.5% of the residents lost in both the games, while 60% of the residents lost in Hockey. The difference between the residents who lost only in Cricket and those who lost only in hockey is 30. Find the total number of residents.

Solve a real-world problem using set theory and Venn diagrams. Find the total number of residents where 25% won both Cricket and Hockey, 37.5% lost both, and 60% lost Hockey. Includes step-by-step algebraic solution and percentage calculations. Suitable for students in Grades 9-12.

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