Given sets A = 1, 3, 5 and E = 1, 2, 3, 4, find the symmetric difference A Δ E.

$A - E = \{x \mid x \in A \text{ and } x \notin E\}$
$A - E = \{1, 3, 5\} - \{1, 2, 3, 4\}$
The elements in A that are not in E are just 5 (since 1 and 3 are in E).
$A - E = \{5\}$

$E - A = \{x \mid x \in E \text{ and } x \notin A\}$
$E - A = \{1, 2, 3, 4\} - \{1, 3, 5\}$
The elements in E that are not in A are 2 and 4 (since 1 and 3 are in A).
$E - A = \{2, 4\}$

$A \Delta E = (A - E) \cup (E - A)$
$A \Delta E = \{5\} \cup \{2, 4\}$
$A \Delta E = \{2, 4, 5\}$

Alternatively, we can use the formula:
$A \Delta E = (A \cup E) - (A \cap E)$

$A \cup E = \{1, 3, 5\} \cup \{1, 2, 3, 4\}$
$A \cup E = \{1, 2, 3, 4, 5\}$

$A \cap E = \{1, 3, 5\} \cap \{1, 2, 3, 4\}$
$A \cap E = \{1, 3\}$

$A \Delta E = \{1, 2, 3, 4, 5\} - \{1, 3\}$
$A \Delta E = \{2, 4, 5\}$

Both methods yield the same result.

Comparing this with the given options:
A. $\{1, 2\}$
B. $\{2, 4, 5\}$
C. $\{1, 2, 3, 4, 5\}$
D. $\{1, 3\}$

The correct option is B.

The final answer is $\boxed{\text{B. } \{2,4,5\}}$.