cost of providing school dinners 
The graph of the cost of providing school dinners versus the number of children is a straight line not passing through the origin and increases as the number of pupils increases. For $200$ students the cost is Rs. $9$ each, for $300$ students the cost is Rs. $8$ each. The cost per student can never be below
a) $8$
b) $7$
c) $6$
d) $10$

The correct answer is option c. Can anyone tell me how to reach the answer?
The solution scheme is given as below in my book:

We have $C + 200x = 200 \times 9$, where $C$ is the fixed cost and $x$ is the variable cost.
$C + 300x = 300 \times 8$; $C = 600$
Solving the equations, we get
Total cost $= 600 + 6N$.
Thus, cost per student $= 600/N + 6$ assuming the minimum cost/student;
= Rs. $6$ for $600$ students.

i have understood the first step of this solution but need explanation for rest of the steps.
 A: The solution is very poorly explained.
From the data for $200$ students we know that $C+200x=9\cdot200=1800$, and from the data for $300$ students we know that $C+300x=8\cdot300=2400$. We therefore have a system of two equations in two unknowns,
$$\left\{\begin{align*}
&C+200x=1800\\
&C+300x=2400\;.
\end{align*}\right.$$
Solve this system by any method that you know. I’d simply subtract the first equation from the second to get $100x=600$ and therefore $x=6$, so that $C+1800-200\cdot6=600$. This means that if we have $N$ students, the total cost is $$C+Nx=600+6N\;,$$ and the cost per student is therefore $$\frac{600+6N}N=\frac{600}N+6\;.$$
$\frac{600}N>0$ no matter how many students we have, so the cost per student is always greater than $6$. On the other hand, we can make $\frac{600}N$ as small as we like by taking $N$ large enough, so we can make the cost per student as close to $6$ as we want. For instance, if $N>600$, then $\frac{600}N<1$, and the cost per student is $\frac{600}N+6<7$.
