In a group of children, 35 play football out of which 20 play football only, 22 play hockey, 25 play cricket out of which 11 play cricket only. Out of these 7 play cricket and football but not hockey, 3 play football and hockey but not cricket and 12 play football and cricket both.
How many play all the three games ? How many play cricket and hockey but not football, how many play hockey only? What is the total number of children in the group?

Explanation

To solve the problem step by step, we will use a Venn diagram to represent the sets of children playing football, hockey, and cricket. Let's denote:

- F: Set of children playing football
- H: Set of children playing hockey
- C: Set of children playing cricket

### Step 1: Define the Variables
We will define the following variables based on the information given:

- \( a \): Number of children playing only cricket
- \( b \): Number of children playing only hockey
- \( c \): Number of children playing hockey and cricket but not football
- \( d \): Number of children playing football and hockey but not cricket
- \( e \): Number of children playing only football
- \( f \): Number of children playing cricket and football but not hockey
- \( g \): Number of children playing all three games (football, hockey, cricket)

### Step 2: Write Down the Given Information
From the problem statement, we have the following information:

1. Total children playing football: \( e + d + f + g = 35 \)
2. Children playing only football: \( e = 20 \)
3. Total children playing hockey: \( b + d + c + g = 22 \)
4. Total children playing cricket: \( a + f + g + c = 25 \)
5. Children playing cricket and football but not hockey: \( f = 7 \)
6. Children playing football and hockey but not cricket: \( d = 3 \)
7. Total children playing football and cricket: \( f + g = 12 \)

### Step 3: Solve for Each Variable
From the equations, we can substitute and solve for the unknowns.

1. From \( f + g = 12 \) and \( f = 7 \):
\[ 7 + g = 12 \implies g = 12 - 7 = 5 \]

2. Now substitute \( g = 5 \) into \( d \):
\[ d = 3 \quad (\text{already known}) \]

3. Substitute \( f = 7 \) and \( g = 5 \) into the equation for cricket:
\[ a + 7 + 5 + c = 25 \implies a + c + 12 = 25 \implies a + c = 13 \]

4. Substitute \( g = 5 \) into the hockey equation:
\[ b + 3 + c + 5 = 22 \implies b + c + 8 = 22 \implies b + c = 14 \]

### Step 4: Solve for \( b \) and \( c \)
Now we have two equations:
1. \( a + c = 13 \)
2. \( b + c = 14 \)

From \( a + c = 13 \), we can express \( a \):
\[ a = 13 - c \]

Substituting \( a \) into the total number of children playing football:
\[ e + b + d + g + f + a + c = 35 \]

### Step 5: Calculate Each Value
Now we can find the values of \( a \), \( b \), and \( c \) using the equations derived:

1. From \( b + c = 14 \) and \( a + c = 13 \):
- Subtract the two equations:
\[ (b + c) - (a + c) = 14 - 13 \implies b - a = 1 \implies b = a + 1 \]

2. Substitute \( b = a + 1 \) into \( b + c = 14 \):
\[ (a + 1) + c = 14 \implies a + c = 13 \quad (\text{consistent}) \]

3. Now, substituting \( a = 11 \) (from \( a + c = 13 \)) gives:
\[ c = 13 - 11 = 2 \]
\[ b = 14 - c = 14 - 2 = 12 \]

### Step 6: Final Values
Now we have:
- \( a = 11 \) (only cricket)
- \( b = 12 \) (only hockey)
- \( c = 2 \) (hockey and cricket but not football)
- \( d = 3 \) (football and hockey but not cricket)
- \( e = 20 \) (only football)
- \( f = 7 \) (cricket and football but not hockey)
- \( g = 5 \) (all three games)

### Step 7: Total Number of Children
Total number of children:
\[ a + b + c + d + e + f + g = 11 + 12 + 2 + 3 + 20 + 7 + 5 = 60 \]

### Summary of Results
1. Number of children playing all three games: \( g = 5 \)
2. Number of children playing cricket and hockey but not football: \( c = 2 \)
3. Number of children playing hockey only: \( b = 12 \)
4. Total number of children in the group: \( 60 \)