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The juice stall at the circus stocked just 2 brands of orang

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The juice stall at the circus stocked just 2 brands of orang  [#permalink]

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New post 04 Sep 2013, 10:25
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Difficulty:

  65% (hard)

Question Stats:

65% (02:56) correct 35% (03:04) wrong based on 139 sessions

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The juice stall at the circus stocked just 2 brands of orange juice tetra packs. Brand A costs $1 per pack and brand B costs $1.5 per pack. Last week , brand A contributed to m% of stall’s revenue and accounted for n% of sales of juice tetra packs. Which of the following expresses m in terms of n?

(A) 100n/(150 – n)
(B) 200n/(250-n)
(C) 200n/(300-n)
(D) 250n/(400-n)
(E) 300n/(500-n)

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Re: The juice stall at the circus stocked just 2 brands of orang  [#permalink]

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New post 04 Sep 2013, 10:31
suyash23n wrote:
The juice stall at the circus stocked just 2 brands of orange juice tetra packs. Brand A costs $1 per pack and brand B costs $1.5 per pack. Last week , brand A contributed to m% of stall’s revenue and accounted for n% of sales of juice tetra packs. Which of the following expresses m in terms of n?

(A) 100n/(150 – n)
(B) 200n/(250-n)
(C) 200n/(300-n)
(D) 250n/(400-n)
(E) 300n/(500-n)


"Stolen" OG question:
Quote:
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)


Discussed here: last-sunday-a-certain-store-sold-copies-of-newspaper-a-for-101739.html
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Re: The juice stall at the circus stocked just 2 brands of orang  [#permalink]

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New post 04 Sep 2013, 20:08
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suyash23n wrote:
The juice stall at the circus stocked just 2 brands of orange juice tetra packs. Brand A costs $1 per pack and brand B costs $1.5 per pack. Last week , brand A contributed to m% of stall’s revenue and accounted for n% of sales of juice tetra packs. Which of the following expresses m in terms of n?

(A) 100n/(150 – n)
(B) 200n/(250-n)
(C) 200n/(300-n)
(D) 250n/(400-n)
(E) 300n/(500-n)


1. Let the number of units sold be 100. Then n units of A are sold and (100-n) units of B are sold.

Total revenue= Revenue due to A + Revenue due to B

= (No.of units of A sold * Cost/unit of A) + (No.of units of B sold * cost/unit of B)

= n *1 + (100-n) *1.5
= n + (150-1.5n) --- (1)

2. But we know, revenue due to A = m%. of the total revenue --- (2)
3. To express m in terms of n we need to relate (1) and (2)
4. m% = n / ( n + (150-1.5n))
or m= 100n / 150-0.5n = 200n/(300-n)
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Re: The juice stall at the circus stocked just 2 brands of orang  [#permalink]

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New post 25 Jan 2014, 09:18
I like to use a combination of Smart Numbers and Backsolving to solve these questions. As Bunuel pointed out very similar to one found in the OG

So if I say total units = 100
and then n = 50%

So a =50%

I will get that 50/50+75=2/5

So m= 2/5=40%

So my target is 40 and my input variable is 50

I always like starting with C

200*50 / 5*250 = 40

Jackpot!

C it is

Hope it helps
Cheers!
J :)
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Re: The juice stall at the circus stocked just 2 brands of orang  [#permalink]

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New post 09 Oct 2018, 09:43
suyash23n wrote:
The juice stall at the circus stocked just 2 brands of orange juice tetra packs. Brand A costs $1 per pack and brand B costs $1.5 per pack. Last week , brand A contributed to m% of stall’s revenue and accounted for n% of sales of juice tetra packs. Which of the following expresses m in terms of n?

(A) 100n/(150 – n)
(B) 200n/(250-n)
(C) 200n/(300-n)
(D) 250n/(400-n)
(E) 300n/(500-n)


Let’s let p = the total number of juice tetra packs. Since brand A accounts for n% (or n/100) of the sales of the juice tetra packs, brand B accounts for (100-n)% (or (100 - n)/100) of the the sales of the juice tetra packs. Therefore, we have:

Sales of Brand A juice / total sales = fraction of stall revenue from Brand A sales

[1 x n/100 x p] / [1 x n/100 x p + 1.5 x (100-n)/100 x p] = m/100

[1 x n/100] / [1 x n/100 + 1.5 x (100-n)/100] = m/100

(n/100)/[n/100 + 1.5(100-n)/100] = m/100

n/[n + 1.5(100 - n)] = m/100

100n/[n + 150 - 1.5n] = m

100n/[150 - 0.5n] = m

200n/[300 - n] = m

Answer: C
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Re: The juice stall at the circus stocked just 2 brands of orang  [#permalink]

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New post 09 Oct 2018, 11:29
suyash23n wrote:
The juice stall at the circus stocked just 2 brands of orange juice tetra packs. Brand A costs $1 per pack and brand B costs $1.5 per pack. Last week , brand A contributed to m% of stall’s revenue and accounted for n% of sales of juice tetra packs. Which of the following expresses m in terms of n?

(A) 100n/(150 – n)
(B) 200n/(250-n)
(C) 200n/(300-n)
(D) 250n/(400-n)
(E) 300n/(500-n)

Let´s explore an AGGRESSIVE PARTICULAR CASE:

n = 100 : in this case, only brand A was sold, hence all revenue came from brand A and our FOCUS will be (the TARGET) m =100 (of course)!

\(\left. \begin{gathered}
\left( A \right)\,\,\,\frac{{{{10}^4}}}{{50}} \ne 100 \hfill \\
\left( B \right)\,\,\frac{{2 \cdot {{10}^4}}}{{150}} \ne 100 \hfill \\
\left( C \right)\,\,\frac{{2 \cdot {{10}^4}}}{{200}} = 100\,\,\,\,\, \hfill \\
\left( D \right)\,\,\frac{{25 \cdot {{10}^3}}}{{300}} \ne 100 \hfill \\
\left( E \right)\,\,\frac{{3 \cdot {{10}^4}}}{{400}} \ne 100 \hfill \\
\end{gathered} \right\}\,\,\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{only}}\,\,{\text{survivor}}\,{\text{!}}} \,\,\,\,\,\,\,\left( C \right)\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: The juice stall at the circus stocked just 2 brands of orang   [#permalink] 09 Oct 2018, 11:29
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