5. Reversing the digits of 1746 gives 6471, the new number is 4725 more than the original number. Then, among all four-digit numbers, the number of numbers that satisfy the condition that the new number is 4725 more than the original number is ( ) .
(A) 16
(B) 20
(C) 24
(D) 28
5. D.
Let the new number be $\overline{a b c d}$, then the original number is $\overline{d c b a}$. From the given condition, we have $\overline{a b c d}-\overline{d c b a}=999(a-d)+90(b-c)=4725$, which simplifies to $111(a-d)+10(b-c)=525$.
Clearly, 5 divides $(a-d)$.
Noting that $0<a-d<9$, it can only take $a-d=5$, yielding $b-c=-3$.
Therefore, when $a=6,7,8,9$, $d$ correspondingly takes $1,2,3,4$; when $b=0,1,2,3,4,5,6$, $c$ correspondingly takes $3,4,5,6,7,8,9$.
Thus, there are 28 numbers that satisfy the condition.