Question

Last Sunday a certain store sold copies of Newspaper A for $1.00eachandcopiesofNewspaperBfor$1.25 each, 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 A, which of the following expresses r in terms of p ?

• A. $\frac{100p}{125-p}$
• B. $\frac{150p}{250-p}$
• C. $\frac{300p}{375-p}$
• D. $\frac{400p}{500-p}$
• E. $\frac{500p}{625-p}$

Answer 1

The correct option is D $\frac{400p}{500-p}$

$\Rightarrow$ Let $x$ be the number of copies of both newspaper sold.

$\Rightarrow$ Copies of newspaper A sold = $\left(\frac{p}{100}\right)x$

$\Rightarrow$ Copies of newspaper B sold = $(1-\frac{p}{100})x$

$\Rightarrow$ Revenue from newspaper A = $\$1\times \left(\frac{p}{100}\right)x=p\left(\frac{x}{100}\right)$

$\Rightarrow$ Revenue from newspaper B = $\$1.25\times (1-\frac{p}{100})x=1.25(100-p)\left(\frac{x}{100}\right)$

$\Rightarrow$ We have given that $r\%$ of the revenue came from newspaper A. We can translate this into an expression.

$\Rightarrow$ $\frac{r}{100}=\frac{RevenuefromnewspaperA}{RevenuefromnewspaperA+RevenuefromnewspaperB}$

$\Rightarrow$ $\frac{r}{100}=\frac{p\left(\frac{x}{100}\right)}{p\left(\frac{x}{100}\right)+1.25(100-p)\left(\frac{x}{100}\right)}$

$\Rightarrow$ $\frac{r}{100}=\frac{p}{p+1.25(100-p)}$

$\Rightarrow$ $\frac{r}{100}=\frac{4p}{4p+5(100-p)}$ [Multiplying right had side numerator and denominator by 4]

$\Rightarrow$ $\frac{r}{100}=\frac{4p}{4p+500-5p}$

$\Rightarrow$ $\frac{r}{100}=\frac{4p}{500-p}$

$\therefore$ $r=\frac{400p}{500-p}$