ashimasood wrote:
fluke wrote:
DeeptiM wrote:
A store selld newspaper A at price of $1 and B at price of $1.25. In a certain day, if r percent of the total income is from A, and p percent of the total copies sold are A, in terms of r, p=?
DeeptiM:
Please take care while posting question. This question looks as if you have formatted it in your own words; the options are also missing. Please mention the source of the question.
A's revenue: p*1=p
B's revenue: (100-p)*1.25=125-1.25p
Total revenue=p+125-1.25p=125-0.25p
We know, A's revenue is r% of the total revenue
\(\frac{p}{125-0.25p}=\frac{r}{100}\)
OR
\(r=\frac{100p}{125-0.25p}\)
Likewise, please solve to get p in terms of r. This looks time consuming to me.
Hi Fluke
Could you please explain it in detail.. Please elaborate your answer..
Which part you didn't understand?
p percent of total copies sold were "A". If p=20%, then B must have sold 80% i.e. (100-p)%
If A sold p percent; B must have sold (100-p)%.
Now, let's suppose the number of copies sold be 100.
A sold = p
B sold = 100-p
Money collected from A's sale=p*($1)=p
Money collected from B's sale=(100-p)*($1.25)=125-1.25p
Total income=p+125-1.25p=125-0.25p
Now,
In a certain day, if r percent of the total income is from A;
r% of total income= (r/100)*(125-0.25p)
A's income= p
\(p=(\frac{r}{100})*(125-0.25p)\)
\(100p=125r-0.25pr\)
\(100p+0.25pr=125r\)
\(10000p+25pr=12500r\)
\(400p+pr=500r\)
\(p(400+r)=500r\)
\(p=\frac{500r}{(400+r)}\)