Question

Last Sunday a certain store sold copies of 256 Newspaper A for $1.00 each and copies of Newspaper B for $1.25 each, and the store sold no other newspapers that day. If r percent of the store's revenue from newspaper sales was from Newspaper A and if p percent of the newspapers that the store sold were copies of Newspaper A, which of the following expresses r in terms of p ?$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

Answer:

D. $\frac{400p}{500-p}$

Step-by-step explanation:

let's call A the number of copies sold of newspaper A and B the number of copies sold of newspaper B.

So, we can formulate the following equations from the sentence: a certain store sold copies of 256 Newspaper A for $1.00 each and copies of Newspaper B for $1.25 each as:

A + B = 256                               (1)

1A + 1.25 B = Total Revenue    (2)

Then, r and p are equal to:

$r=\frac{A}{Total Revenue}*100=\frac{100A}{A+1.25B}$     (3)

$p=\frac{A}{256} *100=\frac{100A}{256}$                            (4)

Isolating A from (4) and B from (1), we get:

$A=\frac{256p}{100} =2.56p$                (5)

$B=256-A=256-2.56p$                         (6)

Finally, replacing (5) and (6) in (3), we get:

$r=\frac{100A}{A+1.25B}=\frac{100(2.56p)}{2.56p+1.25(256-2.56p)}\\\\r=\frac{256p}{2.56p+320-3.2p}=\frac{256p}{320-0.64p}\\\\r=\frac{256p/0.64}{(320-0.64p)/0.64}=\frac{400p}{500-p}$