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# Last Sunday a certain store sold copies of Newspaper A for

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Senior Manager
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Last Sunday a certain store sold copies of Newspaper A for [#permalink]

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23 Aug 2006, 06:06
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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 storeâ€™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 â€“ p)
B. 150p / (250 â€“ p)
C. 300p / (375 â€“ p)
D. 400p / (500 â€“ p)
E. 500p / (625 â€“ p)

Ans D had to plug in values to get the answer. Couldnt arrive at the answer algebraically within the 2 minute mark
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23 Aug 2006, 09:24
total volume = (p/100) of A + (1 - p/100) of B
total revenue = (p/100) * cost of 1 newspaperA + (1 - p/100) * cost of 1 newspaper b
= (p/100) + (1 - p/100)*1.25 --- (i)
%age of revenue contributed by newspaper A = r%
actual revenue of A = p/100 (from i above)
hence,
(r/100)*total revenue = p/100
solve for r in terms of p
ans. 400p/(500-p).
_________________

Cheers!
Prashrash.

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23 Aug 2006, 09:56
Value of each Newspaper A is $1 So, total Revenue(A) = Number of NewsPapers(A) Say total Revenue is R and Total number of News Papers is N So since price of one Newspaper A is 1, rR/100 = pN/100 i.e. r = pN/R R =pN/100 x 1 + [1-p/100] x 1.25N = (1.25-0.25p/100)N Thus, r = 100p/(1.25-0.25p) Simplify it and we get 400p/(500-p) Senior Manager Joined: 14 Jul 2006 Posts: 281 Followers: 1 Kudos [?]: 1 [0], given: 0 ### Show Tags 23 Aug 2006, 10:00 Thanks guys appreciate it Intern Joined: 22 Aug 2006 Posts: 46 Followers: 0 Kudos [?]: 0 [0], given: 0 ### Show Tags 23 Aug 2006, 12:17 I got the same question on GMAT yesterday. I did get it correct though. Senior Manager Joined: 14 Aug 2006 Posts: 363 Followers: 1 Kudos [?]: 45 [0], given: 0 ### Show Tags 23 Aug 2006, 19:54 gmatcrook: just a small correction: r = 100p/125-.25p = 400p/500-p Senior Manager Joined: 15 Aug 2004 Posts: 329 Followers: 1 Kudos [?]: 8 [0], given: 0 ### Show Tags 24 Aug 2006, 04:29 apollo168 wrote: Thanks guys appreciate it After this... Ruby wrote: I got the same question on GMAT yesterday. I did get it correct though. I guess where did you get this question from ? Senior Manager Joined: 15 Aug 2004 Posts: 329 Followers: 1 Kudos [?]: 8 [0], given: 0 ### Show Tags 24 Aug 2006, 04:57 prashrash wrote: total volume = (p/100) of A + (1 - p/100) of B total revenue = (p/100) * cost of 1 newspaperA + (1 - p/100) * cost of 1 newspaper b = (p/100) + (1 - p/100)*1.25 --- (i) %age of revenue contributed by newspaper A = r% actual revenue of A = p/100 (from i above) hence, (r/100)*total revenue = p/100 solve for r in terms of p ans. 400p/(500-p). I tried to plug in values.... But D does not come up properly....what can be used,, Senior Manager Joined: 14 Jul 2006 Posts: 281 Followers: 1 Kudos [?]: 1 [0], given: 0 ### Show Tags 24 Aug 2006, 06:59 Hi what I did was to i used 100 as the total revenue And I wanted A to comprise 40% of the value or 40 so to get B--> (5/4)B=60 B=48 newspapers which means A=40 newspapers p=40/88--> 45.4 percent and r=40% Intern Joined: 17 Apr 2006 Posts: 42 Followers: 0 Kudos [?]: 0 [0], given: 0 Re: Problem solving newspaper [#permalink] ### Show Tags 26 Aug 2006, 09:45 apollo168 wrote: 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 storeâ€™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 â€“ p)
B. 150p / (250 â€“ p)
C. 300p / (375 â€“ p)
D. 400p / (500 â€“ p)
E. 500p / (625 â€“ p)

Ans D had to plug in values to get the answer. Couldnt arrive at the answer algebraically within the 2 minute mark

This is fairly straightforward:

total revenue = a(1.00) + b(1.25)

where a and b are the number of newspapers of type A and B that were sold

now r% = a/(a+1.25b)
simplifying:

b = a/1.25 *(100 - r)/r --- (eq 1)

p% = a/(a+b)

Simplying and collecting terms of b:

b = a/p (100 - p ) --- (eq 2)

from 1 and 2,

1/p (100-p) = 0.8(100-r)/r

Simplify and collect terms of p on right side:

1000r - 10rp = 800p - 8rp

r(1000-2p) = 800p

therefore: r = 400p/(500-p) Ans: D
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26 Aug 2006, 12:53
let total papers =100

No. of papers of type A sold = p
No. of papers of type A sold = 100-p

No. of papers of type B sold = p.1
No. of papers of type B sold = (100-p).(1.25)

Percentage of sales from A

r = p.1/[p.1 + (100-p).(1.25)]
=400p/(500-p)
26 Aug 2006, 12:53
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