RenanVivot wrote:
Last Friday, a grocery store sold bottles of Wine A and Wine B for 10 and 15 respectively and no other products were sold that day. If m percent of the grocery's revenues from wine sales was from Wine A and if n percent of the wine bottles that the grocery sold were from Wine A as well, what alternative describes m in terms of n?
A) 100n/(150 - 1.5n)
B) 150n/(150 - 7.5n)
C) 200n/(150 - 1.0n)
D) 300n/(750 - 1.0n)
E) 400n/(600 - 2n)
KushchokhaniNo timer is there as the question has not been posted in correct forum. I will do the needful.
This can have two immediate solutions.
A)
Substitute some values for n Let there be total 100 bottles out of which 20 are A and 80 are B. Thus, n is 20.
Total revenue = 20*10+80*15 = 1400
% from n bottles = \(100*\frac{200}{1400}=\frac{100}{7}\)
ONLY E fits in. => \(\frac{400n}{(600 - 2n)}=\frac{400*20}{600-2*20}=\frac{8000}{560}=\frac{100}{7}\)
B)
AlgebraicLet there be 100 bottles, so A is n bottles and B is 100-n bottles
Total Revenue = 10*n+(100-n)*15 = 1500-5n
10n as % of 1500-5n = \(100*\frac{10n}{1500-5n}=\frac{200n}{300-n}\), which is same as E.
E
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