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The juice stall at the circus stocked just 2 brands of

by BTGmoderatorLU » Thu Oct 04, 2018 3:03 pm

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Source: Princeton Review

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?
$$A.\ \frac{100n}{150-n}$$
$$B.\ \frac{200n}{250-n}$$
$$C.\ \frac{200n}{300-n}$$
$$D.\ \frac{250n}{400-n}$$
$$E.\ \frac{300n}{500-n}$$
The OA is C.
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by [email protected] » Thu Oct 04, 2018 3:53 pm
Hi All,

You might recognize this as a 'lift' of an Official Guide question. You can find that question - and a discussion of how to solve it - here:

https://www.beatthegmat.com/what-mistak ... 76293.html

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
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by Jay@ManhattanReview » Thu Oct 04, 2018 9:50 pm
BTGmoderatorLU wrote:Source: Princeton Review

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?

$$A.\ \frac{100n}{150-n}$$
$$B.\ \frac{200n}{250-n}$$
$$C.\ \frac{200n}{300-n}$$
$$D.\ \frac{250n}{400-n}$$
$$E.\ \frac{300n}{500-n}$$
The OA is C.
Say the total number of packs sold = x

Thus,
the total number of brand A packs sold = nx/100; and
the total number of brand B packs sold = x - nx/100 = (100x - nx)/100 = x(100 - n)/100

Sales revenue from the sale of brand A = nx/100 * 1 = nx/100
Sales revenue from the sale of brand B = x(100 - n)/100 * 1.5 = 3x(100 - n)/200

=> Total sales revenue from the sale of both the brands = [nx/100 + 3x(100 - n)/200]

Since it is given that brand A contributed to m% of stall's revenue, we have

[nx/100] = m% of [nx/100 + 3x(100 - n)/200]

[n/100] = m[n/100 + 3(100 - n)/200]/100; x gets cancelled
n = m[n/100 + 3(100 - n)/200]; 100 gets cancelled

200n = m(2n + 300 - 3n)

m = (200 n)/(300 - n)

The correct answer: C

Hope this helps!

-Jay
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by Scott@TargetTestPrep » Tue Oct 09, 2018 9:40 am
BTGmoderatorLU wrote:Source: Princeton Review

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?
$$A.\ \frac{100n}{150-n}$$
$$B.\ \frac{200n}{250-n}$$
$$C.\ \frac{200n}{300-n}$$
$$D.\ \frac{250n}{400-n}$$
$$E.\ \frac{300n}{500-n}$$
The OA is C.
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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by fskilnik@GMATH » Tue Oct 09, 2018 11:23 am
BTGmoderatorLU wrote:Source: Princeton Review

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?
$$A.\ \frac{100n}{150-n} \,\,\,\,\,\,\, B.\ \frac{200n}{250-n} \,\,\,\,\,\,\, C.\ \frac{200n}{300-n} \,\,\,\,\,\,\, D.\ \frac{250n}{400-n} \,\,\,\,\,\,\, E.\ \frac{300n}{500-n}$$
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.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
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