Problem Solving

dkumar.83 wrote:

my way is too long, need optimal way
total number of sold=x
sold A-a
sold B-(x-a)
total income=a+(x-a)*5/4=1/4(5x-a)

a/((1/4)*(5x-a))=r/100
4a/(5x-a)=r/100

and a/x=p/100

and now simple algebra

100/r=(5x-a)/4a

100/r=5/4*(x/a)-1
and x/a=100/p

100/r=5/4(100/p)-1
100/r=(500/4p)-1
100/r=(500-4p)/4p
r/100=4*p*100/(500-4p)=100p/(125-p)
so A

The answer should be (D), try plugging in values...

Lets assume, total sold = 100 (because easy to calculate)
Last Sunday, News Paper - A sales = x then News-paper-B sales = (100-x)
they have given each price of NP-A & NP-B, then
Revenue from NP-A = x (1$) = $x
Revenue from NP-B = (100-x)($1.25) = $125-1.25x
Total Revenue = Store's Revenue = NP-A revenue + NP-B revenue = x + 125 - 1.25x = $(125 - 0.25x)

Given p% of total papers sold was NP-A. We can write as,
p% of total papers(100) = No.of NP-A sold (take values from the above statements)
(p/100) * (100) = x
then p = x

Given, r% of Store's Revenue was only from NP-A. we can write as,
r% of Store's Revenue(Total Revenue) = Revenue from NP-A
(r/100) * (125 - 0.25x) = x(1$)
r = (100x)/ (125-0.25x)

multiply & divide by 100/100

r = (100x)(100)/(12500 - 25x)
Cancel with 25 on both numerator & denominator on the Right-hand side.
r = (400x)/(500-x)

From above statement, substitute x=p;

r = (400p)/(500-p). So, CORRECT ANSWER is D.

HTH, GOOD LUCK,

Thanks,
Rajesh,
Loves GMAT....!!!