1972 AMC 12 Problem 33

Let the hundreds, tens, and units digits be $H,$ $T,$ and $U,$ and consider which position should contain the largest digit

At a minimum $U=9$; then maximize $T$ and minimize $H$ subject to distinct nonzero digits

Let $Q$ denote the quotient. Then $$Q=1+\frac{99H+9T}{H+T+U}.$$ Interchanging $U$ with a larger digit in either earlier position decreases the quotient, so the units digit must be the largest. Increasing that units digit lowers a quotient greater than $1,$ so $U=9.$ Then $$\frac{T+11H}{T+H+9} =1+\frac{10H-9}{T+H+9}.$$ This is minimized by taking the largest available $T=8$ and then the smallest $H=1.$ The number is $189,$ and $$\frac{189}{1+8+9}=\frac{189}{18}=10.5.$$

Therefore, the correct answer is C.