If `n(mu)=100,n(A)=50,n(B)=20` and `n(A cap B)`=10, then `n[(AcupB)]=`

Options

A. 60
B. 30
C. 40
D. 20

Explanation

To solve the problem step by step, we need to find \( n(A \cup B') \), where \( B' \) is the complement of set B.

Given:
- \( n(\mu) = 100 \) (the total number of elements in the universal set)
- \( n(A) = 50 \)
- \( n(B) = 20 \)
- \( n(A \cap B) = 10 \)

### Step 1: Calculate \( n(A \cup B) \)

Using the formula for the union of two sets:
\[ n(A \cup B) = n(A) + n(B) - n(A \cap B) \]

Substituting the values:
\[ n(A \cup B) = 50 + 20 - 10 \]
\[ n(A \cup B) = 60 \]

### Step 2: Calculate \( n(A \cup B') \)

To find \( n(A \cup B') \), we first need to find \( n(B') \). The complement of set B includes all elements in the universal set that are not in B.

Using the formula for the complement:
\[ n(B') = n(\mu) - n(B) \]

Substituting the values:
\[ n(B') = 100 - 20 \]
\[ n(B') = 80 \]

### Step 3: Calculate \( n(A \cup B') \)

Now, we can find \( n(A \cup B') \) using the formula:
\[ n(A \cup B') = n(A) + n(B') - n(A \cap B') \]

To find \( n(A \cap B') \), we can use the relationship:
\[ n(A) = n(A \cap B) + n(A \cap B') \]
Thus,
\[ n(A \cap B') = n(A) - n(A \cap B) = 50 - 10 = 40 \]

Now substituting back into the union formula:
\[ n(A \cup B') = n(A) + n(B') - n(A \cap B') \]
\[ n(A \cup B') = 50 + 80 - 40 \]
\[ n(A \cup B') = 90 \]

### Final Answer:
\[ n(A \cup B') = 90 \]