Use discrete fine rates to choose the feasible solution

The question already defines $x,y$. Option A gives $4kx+5(10-k)x+5(3)y=1455$, or $50x-kx+15y=1455$. Option B gives $5(10)x+5(3)y=1500$, or $50x+15y=1500$. Thus $y=100-\tfrac{10}{3}x$.

Subtract the Option A equation from Option B: $kx=45$. Since $x$ is a positive multiple of 5 cents and $k$ is an integer number of books with $1\le k\le10$, the candidates are $k=1,3,9$. If $k=1$, $x=45$ and $y=-50$, rejected because a fine is positive. If $k=3$, $x=15$ and $y=50$. If $k=9$, $x=5$ and $y=250/3=83\tfrac13$, rejected because $y$ is not a multiple of 5.

Therefore the daily fines are 15 cents per book and 50 cents per audio-visual item, with $k=3$. Option A checks as $4(3)(15)+5(7)(15)+5(3)(50)=1455$ cents; Option B as $5(10)(15)+5(3)(50)=1500$ cents. The lecturer’s suggested assumption is constant fines: the model charges the same per-item daily rate throughout these overdue days. The source’s final rejection annotation names $x$ as the multiple-of-5 condition, but $x=5$ already satisfies it; the non-multiple is $y=83\tfrac13$. The raw line is retained, and this working explicitly corrects that label.