SET 6 Q 28

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Q28:
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)

[kajcha]

I got D.. What's the OA? Long explanation. So don't want to type if this is not the OA

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Yes,The OA is D
Is it possible for you to explain

[camitava]

saurabh,
Refer the explanation below -
Say,
News A News B Total Rev Total No of News
--------- --------- ----------- -------------------
x y z k

Now x = pk/100
y = k - pk/100

RA (Revenue for A) = pk/100
RB (Revenue for B) = 125/100 * (k - pk/100)

RA + RB = z = 125k/100 - pk/100

Therefor, r = RA/(RA + RB) * 100
= 400p/(500 - p)

So the correct answer is - D. Now got it saurabh?

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Thanks Dude

[camitava]

Thank u sir! At your service any time ...

[kajcha]

This is how I did it

Suppose he sold x nos of Paper A and y nos of paper B

Revenues from papers = x*1+y*1.25

No if papers = x+y

Now the questions says
(x/(x+1.25y))*100=r => y/x= (100/r -1)*(4/5)----- (1)

(x/(x+y))*100=p => y/x=100/p - 1 -----(2)

Equate eq 1 and 2 and solve for r

you get 400P/(500-p)

PS: the trick is to write 1.25 as 5/4. This makes your calculation easy