Let U = \ 1, 2, 3, , 60\ be a universal set. Let P be the subset of...

Let $U = \{1, 2, 3, \dots, 60\}$.

The set $P$ contains multiples of $4$ up to $60$:
$P = \{4, 8, 12, \dots, 60\}$
Number of terms in $P$, $n(P) = \dfrac{60}{4} = 15$.
Sum of elements in $P = \dfrac{15}{2}(4 + 60) = 15 \times 32 = 480$.

The set $Q$ contains multiples of $6$ up to $60$:
$Q = \{6, 12, 18, \dots, 60\}$
Number of terms in $Q$, $n(Q) = \dfrac{60}{6} = 10$.
Sum of elements in $Q = \dfrac{10}{2}(6 + 60) = 5 \times 66 = 330$.

The intersection $P \cap Q$ contains multiples of $\text{LCM}(4, 6) = 12$ up to $60$:
$P \cap Q = \{12, 24, 36, 48, 60\}$
Number of terms in $P \cap Q$, $n(P \cap Q) = \dfrac{60}{12} = 5$.
Sum of elements in $P \cap Q = \dfrac{5}{2}(12 + 60) = 5 \times 36 = 180$.

The elements belonging to exactly one of the sets $P$ or $Q$ form the symmetric difference $P \Delta Q$.
Sum of elements in $P \Delta Q = \text{Sum}(P) + \text{Sum}(Q) - 2 \times \text{Sum}(P \cap Q)$
$= 480 + 330 - 2(180)$
$= 810 - 360 = 450$.

Answer: $450$