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Tough problem from tests

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Tough problem from tests [#permalink] New post 16 Dec 2009, 04:08
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

29% (00:00) correct 71% (02:41) wrong based on 8 sessions
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)

What is the simplest way to solve this??
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Re: Tough problem from tests [#permalink] New post 16 Dec 2009, 14:26
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Assume 1 copy of newspaper A was sold and 0 copies of newspaper B was sold
P = Percentage of Copies A sold = 100
R = Percentage of Revenue from A = 100
By substituting the value of P we easily know that D is the only answer choice to equal 100.

Hence D.

I feel bad about disrespecting such a fabulous question so I will provide the complete solution as well, if it helps anyone -

number of copies of Newspaper A sold = a.
number of copies of Newspaper B sold = b.

P/100=a/(a+b)
100/P = (a+b)/a
100/P = 1+b/a
b/a = 100/P -1

R/100=a/(a+1.25b)
100/R = (a+1.25b)/a
100/R = 1+5b/4a
substituting value of b/a
100/R = 1+125/P - 5/4
100/R = 125/P-1/4
100/R = (500-P)/4P
R=400P/(500-P)
Answer is D.
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Re: Tough problem from tests [#permalink] New post 16 Dec 2009, 05:09
zaarathelab 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 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)

What is the simplest way to solve this??


by plugging in numbers....will go with D. 400p / (500 – p)
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Re: Tough problem from tests [#permalink] New post 29 Dec 2009, 09:31
atish wrote:
Assume 1 copy of newspaper A was sold and 0 copies of newspaper B was sold
P = Percentage of Copies A sold = 100
R = Percentage of Revenue from A = 100
By substituting the value of P we easily know that D is the only answer choice to equal 100.

Hence D.

I feel bad about disrespecting such a fabulous question so I will provide the complete solution as well, if it helps anyone -

number of copies of Newspaper A sold = a.
number of copies of Newspaper B sold = b.

P/100=a/(a+b)
100/P = (a+b)/a
100/P = 1+b/a
b/a = 100/P -1

R/100=a/(a+1.25b)
100/R = (a+1.25b)/a
100/R = 1+5b/4a
substituting value of b/a
100/R = 1+125/P - 5/4
100/R = 125/P-1/4
100/R = (500-P)/4P
R=400P/(500-P)
Answer is D.








I think that a more intuitive way to solve the problem will be:

consider n=total copies sold
p=percentage of copy sold that were A
1-p=percentage of copy sold that were B
therefore
r= 1*p*n/(1*p*n+1.25*(1-p)*n)=p/(p+1.25-1.25p)=p/(1.25-0.25p)
multiply the solution for 400 and obtain
r=400p/(500-100p)
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Re: Tough problem from tests [#permalink] New post 05 Jan 2010, 22:25
I solved the problem by assuming that the store sold 100 newspapers. 50 NewspaperA and 50 NewspaperB. This means that "P"= 50%.

Next I calculated the Revenue.

Newspaper A 50x$1.00 = $50
Newspaper B 50x$1.25 = $62.50
Total Revenue $112.50

After that I found "r"

r = $50/$112.50 = .44444 x 100 = 44.4444%

Once I found "p" and "r", I just substituted "p" until one of the answers gave me the correct "r".

If you try plugging the answer into "A" first you will notice that 500/75 is too small to equal "r". So you can skip down to "E" or "D". Be careful with "D" because "D" says that "r" would equal 43.478. When you do the division you should get exactly 44.444, therefore the answer is "D".
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Re: Tough problem from tests [#permalink] New post 06 Jan 2010, 02:08
It's a straight forward question which I think should be dealt straight to arrive at the solution without any complexity.

Because the question deals with percent and only involves newspaper A, assuming the total number of papers sold i.e., A+B = 100 will make it very easy to understand as well as solve.

So, let total number of papers be 100.
Hence, no. of newspaper A sold = p% of 100 = p
and no. of newspaper B sold = 100 - p.

Now r% of total revenue from papers comes from selling A.
Here total revenue from papers = p * 1 + (100 - p) * 1.25 = 125 - 0.25p
[revenue from A = p * 1 and from B = (100 - p) * 1.25]

Now %age of total revenue coming from A = revenue from A /total revenue * 100
= p/(125 - 0.25p) * 100
But it's already given that p/ (125-0.25p) *100 = r
Since all the answer options have p in the denominator, we'll have to remove 0.25 from the denominator. Simple observation leads us that multiplying by 400 will remove 0.25.
Hence, r = 400p/(500 - p). Ans (d)

P.S. I have elaborated the solving method here else this could be solved in 4 lines. Also, if you are expert in maths then as Atish has mentioned in his shortcut solution assuming 0 papers for B will give you the answer in seconds.
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Re: Tough problem from tests [#permalink] New post 21 Jan 2010, 14:05
atish wrote:
Assume 1 copy of newspaper A was sold and 0 copies of newspaper B was sold
P = Percentage of Copies A sold = 100
R = Percentage of Revenue from A = 100
By substituting the value of P we easily know that D is the only answer choice to equal 100.

Hence D.

I feel bad about disrespecting such a fabulous question so I will provide the complete solution as well, if it helps anyone -

number of copies of Newspaper A sold = a.
number of copies of Newspaper B sold = b.

P/100=a/(a+b)
100/P = (a+b)/a
100/P = 1+b/a
b/a = 100/P -1

R/100=a/(a+1.25b)
100/R = (a+1.25b)/a
100/R = 1+5b/4a
substituting value of b/a
100/R = 1+125/P - 5/4
100/R = 125/P-1/4
100/R = (500-P)/4P
R=400P/(500-P)
Answer is D.




If the first approach is used (B) would also lead to the r=100...
Re: Tough problem from tests   [#permalink] 21 Jan 2010, 14:05
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