Answer:
![\mathbf{r = (400p )/( 500 -p)}](https://img.qammunity.org/2021/formulas/mathematics/high-school/z2bc7cl0jag6ogp1hc1gkfxna3cbh0o7s2.png)
Explanation:
Given that:
Last Sunday a certain store sold copies of Newspaper A for $1.00 each
A = 1
and copies of Newspaper B for $1.25 each,
B = 1.25
and the store sold no other newspapers that day. If r percent of the store's revenue from newspaper sales was from Newspaper A and if p percent of the newspapers that the store sold were copies of Newspaper B.
Then ;
--- (1)
--- (2)
From (2)
![A+B = (A)/(p) * 100](https://img.qammunity.org/2021/formulas/mathematics/high-school/5bej3yagbk026ktc70rx6mt7j0p6yo4val.png)
![B = (A)/(p) * 100-A](https://img.qammunity.org/2021/formulas/mathematics/high-school/i0udoom8vubo0zpe43079fw4uosyi5l8r0.png)
![B = (A( 100-p))/(p)](https://img.qammunity.org/2021/formulas/mathematics/high-school/jhdeo04vl3kwfemv2fbnzvc5xthyjbimy2.png)
Substituting the value of B into equation (1), we have:
![r = \frac{A}{A+1.25 * { (A( 100-p))/(p)}} * 100](https://img.qammunity.org/2021/formulas/mathematics/high-school/p2s6k7jfpjde2poehxhlul4wub8ebh0trt.png)
![r = \frac{A}{A+1.25 * { A( 100-p)}} * 100p](https://img.qammunity.org/2021/formulas/mathematics/high-school/e2edmyuu9l32t2x461hpzdohc1k41gdq5s.png)
![r = (100p)/(p+125 -1.25 p)](https://img.qammunity.org/2021/formulas/mathematics/high-school/bn88hzbigeg0mwvmzq4ub4e1sr7n5pnygs.png)
![r = (100p)/(125 -0.25 p)](https://img.qammunity.org/2021/formulas/mathematics/high-school/ibbflsxe1y0vqkhhvde2bgiv4ycextj5nr.png)
multiplying both numerator and denominator by 4; we have
![r = (4(100p) )/( 4(125 -0.25) p)](https://img.qammunity.org/2021/formulas/mathematics/high-school/c3nb0o0yfcxqpboerkui8xiz9dxbqloju3.png)
![\mathbf{r = (400p )/( 500 -p)}](https://img.qammunity.org/2021/formulas/mathematics/high-school/z2bc7cl0jag6ogp1hc1gkfxna3cbh0o7s2.png)