If A and B are two sets such that ` n( A-B) =24, n(B-A) =19 and n(Acap B) =11`, find :
`(i) n(A)`
(ii) `n(B)`
(iii) ` n(A cupB)`

Explanation

To solve the problem, we need to find the number of elements in sets A and B, as well as the number of elements in their union. We are given the following information:

- \( n(A - B) = 24 \) (elements in A but not in B)
- \( n(B - A) = 19 \) (elements in B but not in A)
- \( n(A \cap B) = 11 \) (elements common to both A and B)

### Step-by-step Solution:

**(i) Finding \( n(A) \)**

1. The number of elements in set A can be expressed as:
\[ n(A) = n(A - B) + n(A \cap B) \]
Here, \( n(A - B) \) is the number of elements in A but not in B, and \( n(A \cap B) \) is the number of elements common to both sets.

2. Substitute the known values:
\[ n(A) = 24 + 11 \]

3. Calculate:
\[ n(A) = 35 \]

**(ii) Finding \( n(B) \)**

1. The number of elements in set B can be expressed as:
\[ n(B) = n(B - A) + n(A \cap B) \]
Here, \( n(B - A) \) is the number of elements in B but not in A.

2. Substitute the known values:
\[ n(B) = 19 + 11 \]

3. Calculate:
\[ n(B) = 30 \]

**(iii) Finding \( n(A \cup B) \)**

1. The number of elements in the union of sets A and B can be expressed using the formula:
\[ n(A \cup B) = n(A) + n(B) - n(A \cap B) \]

2. Substitute the values we found:
\[ n(A \cup B) = 35 + 30 - 11 \]

3. Calculate:
\[ n(A \cup B) = 54 \]

### Final Answers:
- \( n(A) = 35 \)
- \( n(B) = 30 \)
- \( n(A \cup B) = 54 \)