From $f(1)=0$, we know that $a+b+c=0$.  From the first inequality, we get $20\textless9 a+3 b+c\textless24$.  Subtracting $a+b+c=0$ from this gives us $20\textless8 a+2 b\textless24$, and thus $10\textless4 a+b\textless12$. Since $4a+b$ must be an integer, it follows that $4 a+b=11$.  Similarly, from the second inequality, we get $36\textless16a+4 b+c\textless40$. Again subtracting $a+b+c=0$ from this gives us $36\textless15 a+3 b\textless40$, or $12\textless5 a+b\textless\frac{40}{3}$.  It follows from this that $5 a+b=13$. We now have a system of three equations: $a+b+c=0,4a+b=11$, and $5a+b=13$. Solving gives us $(a, b, c)=(2,3,-5)$ and from this we find that $f(10)=2(10)^{2}+3(10)-5=225$. We find that $k=22 \rightarrow(\mathbf{B})22$. 
Let $f(x)=a x^{2}+b x+c$, where $a$, $b$, and $c$ are integers. Suppose that $f(1)=0,20\textless f(3)\textless24,36\textless f(4)\textless40$, $10 k\textless f(10)\textless10(k+1)$ for some integer $k$. What is $k$? (Adapted From2011 AMC 12A Problems, Question 20) 
(A) $$11$$ 
(B) $$22$$ 
(C) $$23$$ 
(D) $$14$$ 
(E) $$25$$ 