To solve the problem, we are given specific set operations and their cardinalities. Let's denote the sets and their operations as follows:
1. [tex]\( n(A - B) = 25 \)[/tex]
2. [tex]\( n(B - A) = 15 \)[/tex]
3. [tex]\( n(A \cup B) = 60 \)[/tex]
Our goal is to find:
i) [tex]\( n(A) \)[/tex]
ii) [tex]\( n(B) \)[/tex]
iii) [tex]\( n(A \cap B) \)[/tex]
We'll use the principle of set operations to derive these values.
### Step-by-Step Solution
#### 1. Understand the given cardinalities:
- [tex]\( n(A - B) \)[/tex]: number of elements in A but not in B.
- [tex]\( n(B - A) \)[/tex]: number of elements in B but not in A.
- [tex]\( n(A \cup B) \)[/tex]: number of elements either in A or B or in both.
#### 2. Formula for the union of two sets:
[tex]\[ n(A \cup B) = n(A) + n(B) - n(A \cap B) \][/tex]
Let's denote [tex]\( x = n(A \cap B) \)[/tex], the number of elements in both sets A and B.
#### 3. Express total number of elements:
[tex]\[ n(A \cup B) = n((A - B) \cup (A \cap B) \cup (B - A)) \][/tex]
#### 4. Combine known values:
[tex]\[ n(A) = n(A - B) + n(A \cap B) \][/tex]
[tex]\[ n(B) = n(B - A) + n(A \cap B) \][/tex]
#### 5. Substitute values and solve for [tex]\( x \)[/tex]:
[tex]\[ n(A \cup B) = n(A - B) + n(A \cap B) + n(B - A) + n(A \cap B) \][/tex]
[tex]\[ 60 = 25 + x + 15 + x \][/tex]
[tex]\[ 60 = 40 + 2x \][/tex]
[tex]\[ 60 - 40 = 2x \][/tex]
[tex]\[ 20 = 2x \][/tex]
[tex]\[ x = 10 \][/tex]
So, [tex]\( n(A \cap B) = 10 \)[/tex].
#### 6. Calculate [tex]\( n(A) \)[/tex] and [tex]\( n(B) \)[/tex]:
[tex]\[ n(A) = n(A - B) + n(A \cap B) \][/tex]
[tex]\[ n(A) = 25 + 10 \][/tex]
[tex]\[ n(A) = 35 \][/tex]
[tex]\[ n(B) = n(B - A) + n(A \cap B) \][/tex]
[tex]\[ n(B) = 15 + 10 \][/tex]
[tex]\[ n(B) = 25 \][/tex]
### Summary of Results:
i) [tex]\( n(A) = 35 \)[/tex]
ii) [tex]\( n(B) = 25 \)[/tex]
iii) [tex]\( n(A \cap B) = 10 \)[/tex]
These are the correct values as derived through our step-by-step calculations.
