A school awarded 42 medals in hockey, 18 in basketball and 23 in cricket. If these medals were bagged by a total of 65 students and only 4 students got medals in all the three sports, how many students received medals in exactly two of the three sports ?

Explanation

To solve the problem, we need to find how many students received medals in exactly two of the three sports: hockey, basketball, and cricket. We are given the following information:

- Total medals in hockey (H) = 42
- Total medals in basketball (B) = 18
- Total medals in cricket (C) = 23
- Total students = 65
- Students who received medals in all three sports = 4

Let's denote:
- \( |H| \) = number of students who received medals in hockey
- \( |B| \) = number of students who received medals in basketball
- \( |C| \) = number of students who received medals in cricket
- \( |H \cap B| \) = number of students who received medals in both hockey and basketball
- \( |B \cap C| \) = number of students who received medals in both basketball and cricket
- \( |C \cap H| \) = number of students who received medals in both cricket and hockey
- \( |H \cap B \cap C| \) = number of students who received medals in all three sports

### Step 1: Set up the equation using the principle of inclusion-exclusion
We can use the principle of inclusion-exclusion to express the total number of students who received medals in at least one sport:

\[ |H \cup B \cup C| = |H| + |B| + |C| - |H \cap B| - |B \cap C| - |C \cap H| + |H \cap B \cap C| \]

### Step 2: Substitute known values into the equation
We know:
- \( |H| = 42 \)
- \( |B| = 18 \)
- \( |C| = 23 \)
- \( |H \cap B \cap C| = 4 \)
- \( |H \cup B \cup C| = 65 \)

Substituting these values into the equation gives:

\[ 65 = 42 + 18 + 23 - |H \cap B| - |B \cap C| - |C \cap H| + 4 \]

### Step 3: Simplify the equation
Now, simplifying the equation:

\[ 65 = 42 + 18 + 23 + 4 - (|H \cap B| + |B \cap C| + |C \cap H|) \]

Calculating the left side:

\[ 65 = 87 - (|H \cap B| + |B \cap C| + |C \cap H|) \]

### Step 4: Solve for the sum of intersections
Rearranging the equation gives:

\[ |H \cap B| + |B \cap C| + |C \cap H| = 87 - 65 \]

\[ |H \cap B| + |B \cap C| + |C \cap H| = 22 \]

### Step 5: Find the number of students who received medals in exactly two sports
Let:
- \( x = |H \cap B| - |H \cap B \cap C| \)
- \( y = |B \cap C| - |H \cap B \cap C| \)
- \( z = |C \cap H| - |H \cap B \cap C| \)

Thus, we can express the number of students who received medals in exactly two sports as:

\[ x + y + z = (|H \cap B| - 4) + (|B \cap C| - 4) + (|C \cap H| - 4) \]

Substituting \( |H \cap B| + |B \cap C| + |C \cap H| = 22 \):

\[ x + y + z = 22 - 3 \times 4 \]

\[ x + y + z = 22 - 12 = 10 \]

### Conclusion
Therefore, the number of students who received medals in exactly two of the three sports is **10**.