Question

Can someone help me with this question? You don’t really have to tell me the answer but just how to solve it.

Answer 1

part A

if the paperbacks are 40% less, that means 100% - 40% = 60%, they actually are 60% of the price of the hardcover ones.

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{60\% of 24.99}}{\left( \cfrac{60}{100} \right)249.99}\implies 14.994$

part B

so, if they return it, they get 1/3 credit, if they try to sell it, is half off, which is more than 1/3.  22 got damaged.

$\bf \stackrel{\textit{damaged books at half-price}}{(22\cdot 14.994)\cfrac{1}{2}\implies 164.934}~\hfill \stackrel{\textit{damaged books at one-third price}}{(22\cdot 14.994)\cfrac{1}{3}\implies 109.956} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill 164.934 - 109.956 = 54.978~\hfill$

part C

$\bf \textit{First Day}~\hfill \stackrel{hardcover}{3(24.99)}+\stackrel{paperback}{8(14.994)}+\stackrel{damaged}{3\left( 14.994\cdot \frac{1}{2} \right)} \implies 217.413 \\\\\\ \textit{Second Day}~\hfill \stackrel{hardcover}{4(24.99)}+\stackrel{paperback}{5(14.994)}+\stackrel{damaged}{8\left( 14.994\cdot \frac{1}{2} \right)}\implies 234.906 \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \boxed{17.493}$

I don't think she is incorrect, $17.5 is close to $20, she didn't say it was exactly $20, just thereabouts, and 17.5 is around that much.