Out of 40 children , 30 can swim,27 can play chess and 5 can do neither. How many children can swim only?

Options

A. 30
B. 22
C. 12
D. 8

Explanation

To solve the problem step by step, we can follow this approach:

### Step 1: Understand the given information
We have:
- Total number of children = 40
- Number of children who can swim (N(A)) = 30
- Number of children who can play chess (N(B)) = 27
- Number of children who can do neither = 5

### Step 2: Calculate the number of children who can do at least one activity
To find the number of children who can do at least one activity (swim or play chess), we subtract the number of children who can do neither from the total number of children:
\[ \text{Children who can do at least one activity} = \text{Total children} - \text{Children who can do neither} \]
\[ = 40 - 5 = 35 \]

### Step 3: Use the formula for union of two sets
We can use the formula for the union of two sets to find the number of children who can swim and play chess:
\[ N(A \cup B) = N(A) + N(B) - N(A \cap B) \]
Where:
- \(N(A \cup B)\) = Number of children who can swim or play chess
- \(N(A)\) = Number of children who can swim
- \(N(B)\) = Number of children who can play chess
- \(N(A \cap B)\) = Number of children who can swim and play chess

### Step 4: Substitute the known values into the formula
We know:
- \(N(A \cup B) = 35\)
- \(N(A) = 30\)
- \(N(B) = 27\)

Substituting these values into the formula gives:
\[ 35 = 30 + 27 - N(A \cap B) \]

### Step 5: Solve for \(N(A \cap B)\)
Rearranging the equation to find \(N(A \cap B)\):
\[ N(A \cap B) = 30 + 27 - 35 \]
\[ N(A \cap B) = 57 - 35 = 22 \]

### Step 6: Find the number of children who can swim only
To find the number of children who can swim only, we subtract the number of children who can swim and play chess from the total number of children who can swim:
\[ \text{Children who can swim only} = N(A) - N(A \cap B) \]
\[ = 30 - 22 = 8 \]

### Final Answer
Thus, the number of children who can swim only is **8**.

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