SP Newspaper A is $1.00 each
Let 'a' be the number of copies sold of Newspaper A
SP Newspaper B is $1.25 each ( Recall that 1.25 is \(\frac{5}{4}\)
Let 'b' be the number of copies sold of Newspaper B
Revenue earned by selling Newspaper A = number of copies sold * price per copy = a*1
Revenue earned by selling Newspaper B = number of copies sold * price per copy = b*\(\frac{5}{4}\)
% Revenue earned by selling copies of Newspaper A will be
\(\frac{(a*1)}{(a*1+ b*\frac{5}{4})}\) *100 = r
r= \(\frac{(4a)}{(4a+ 5b)}\) *100 --------------(I)
% copies sold of Newspaper A will be
\(\frac{(a)}{(a+ b)}\) *100 = p -----------------(II)
Look at equation I. Should we want to simplify that we need to express all the variable terms ( a ,b) in such a way that we have all the terms in a or b .
So from equation (II) we can get the value of b and substitute it in equation (I) . All the terms in equation (I) ( Numerator and denominator) will cancel out and also we will have a relation between r and p
So from Equation (II) we have b =\(\frac{a}{p}(100-1)\) -------(III)
Substituting (III) in (I) we have
r= \(\frac{(4a)}{(4a+ 5(\frac{a}{p}(100-1)))}\) *100
r= \(\frac{(4p)}{(4p+ 5(100-p)}\) *100
which is Answer D
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