Question

A radio which a dealer bought for $600.00 and marked to give a profit of 30% was reduced in sales by 10%. Find the sales price and the percentage profit.

Answer 1

well, the radio was marked up by 30%, so the marked up price is 600 plus 30% of that

`\begin{array} \cline{1-1} \textit{\textit{\LARGE a}\% of \textit{\LARGE b}}\\ \cline{1-1} \\ \left( \cfrac{\textit{\LARGE a}}{100} \right)\cdot \textit{\LARGE b} \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{30\% of 600}}{\left( \cfrac{30}{100} \right)600}\implies 180~\hfill \stackrel{\textit{marked up price}}{780}`

but he had a sale going on, so it reduced it by 10%

`\begin{array}c \cline{1-1} \textit{\textit{\LARGE a}\% of \textit{\LARGE b}}\\ \cline{1-1} \\ \left( \cfrac{\textit{\LARGE a}}{100} \right)\cdot \textit{\LARGE b} \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{10\% of 780}}{\left( \cfrac{10}{100} \right)780}\implies 78~\hfill \underset{\textit{sales price}}{\stackrel{\textit{reduced price}}{702}}`

so, what's the profit?

well, hell he bought it for 600 bucks, and sold it for 702 bucks, so that's a profit of 102 bucks, if we take 600(origin amount) to be the 100%, what's 102 off of it in percentage anyway?

`\begin{array}{ccll} amount&\%\\ \cline{1-2} 600 & 100\\ 102& p \end{array} \implies \cfrac{600}{102}~~=~~\cfrac{100}{p} \\\\\\ \cfrac{ 100 }{ 17 } ~~=~~ \cfrac{ 100 }{ p }\implies 100p=1700\implies p=\cfrac{1700}{100}\implies \stackrel{\textit{17\% in profit}}{p=17}`