First of all, we find the values of $x$ and $y$.  Since the average is $157$, we have  $(100+115+135+140+ 180+197+203+230+x+y = 157 \times 10 = 1570$  $1300+x+y =1570$  $x+y = 270$.  There is a unique mode, $135$, then $135$ must appear at least twice. Therefore, one of $x, y$ is $135$. It is easy to deduce that both $x$ and $y$ are $135$.  List the weights in increasing order: $100,115, 135, 135, 135, 140, 180, 197, 203, 230$.  $np=10(0.56)=5.6 \uparrow 6$. The $56$-th percentile is $140$. 
The following are the weights (in pounds) of ten people: $100,115, 135, 140, 180, 197, 203, 230, x, y$ (not necessarily in increasing order). It is also given that the average weight of these ten people is $157$ pounds, and there is a unique mode of $135$. Find the $56$-th percentile. 
Options:
(1) $$100$$ 
(2) $$115$$ 
(3) $$125$$ 
(4) $$135$$ 