Last Sunday a certain store sold copies of 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 revenues 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. 100p / (125 - p)
B. 150p / (250 - p)
C. 300p / (375 - p)
D. 400p / (500 - p)
E. 500p / (625 - p)
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Newspaper A
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shankar.ashwin
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We could substitute numbers here.
Let A sold = 10 and B sold =10
% Revenue from A (r) = 10/(10+12.5) = 44%
% of newspaper A sold (p) = 50%
Start with C; 300(50)/325 = 46% (Need lower hence try D)
D: 400(50)/450 = 8/9 *50 = 44% (Success)
Let A sold = 10 and B sold =10
% Revenue from A (r) = 10/(10+12.5) = 44%
% of newspaper A sold (p) = 50%
Start with C; 300(50)/325 = 46% (Need lower hence try D)
D: 400(50)/450 = 8/9 *50 = 44% (Success)
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copies of news paper A = p
copies of news paper B = 100 - P
Total revenue = p + (100 - P)1.25 = 125 - .25 p
as per given conditions
p/(125-.25p)=r/100
=>
r = 100 p/(125 -.25p)
= 400 p /500 -p multiplying 4 in both numerator and denominator
so IMO D
copies of news paper B = 100 - P
Total revenue = p + (100 - P)1.25 = 125 - .25 p
as per given conditions
p/(125-.25p)=r/100
=>
r = 100 p/(125 -.25p)
= 400 p /500 -p multiplying 4 in both numerator and denominator
so IMO D
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parul9
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%revenue of A = (revenue of A/Total revenue) * 100
p% of A is sold.
so, percentage of B sold = 100 - p
r = (1*p/(1*p + 1.25(100-p)))*100
= 100p/125-0.25p
= 100p/(125-p/4)
= 400p/(500-p)
So, Answer : D
p% of A is sold.
so, percentage of B sold = 100 - p
r = (1*p/(1*p + 1.25(100-p)))*100
= 100p/125-0.25p
= 100p/(125-p/4)
= 400p/(500-p)
So, Answer : D
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