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01 Oct 2014, 10:54
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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A vendor sells only Product A, for $6, and Product B, for $21. If Q% of the products sold are Product B, and if T% of the total revenue comes from sales of Product B, find Q in terms of T.
(A) \(1501T - 15T^2\)
(B) \(\frac{15T}{615 - 6T}\)
(C) \(\frac{21T}{621 - 6T}\)
(D) \(\frac{40T}{140 - T}\)
(E) \(\frac{60T}{160 - T}\)
For the OE on this problem, and more practice Word Problems, see:
https://magoosh.com/gmat/2014/gmat-prac ... -problems/
Mike
(A) \(1501T - 15T^2\)
(B) \(\frac{15T}{615 - 6T}\)
(C) \(\frac{21T}{621 - 6T}\)
(D) \(\frac{40T}{140 - T}\)
(E) \(\frac{60T}{160 - T}\)
For the OE on this problem, and more practice Word Problems, see:
https://magoosh.com/gmat/2014/gmat-prac ... -problems/
Mike
ShashankDave
Joined: 03 Apr 2013
Last visit: 26 Jan 2020
Posts: 215
Given Kudos: 872
Location: India
Concentration: Marketing, Finance
Schools: McDonough '21 (A) Kelley '21 ISB '21 (I)
GMAT 1: 740 Q50 V41
GPA: 3
12 Jul 2016, 21:53
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mikemcgarry
Quick way to do this one..
Let the total number of products sold be 100..
We will work with ratio of the individual revenues for Product A and Product B
\(\frac{21Q}{6(100-Q)} = \frac{T}{100-T}\)
Answer (D)
01 Oct 2014, 19:41
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....................... Product A .................... Product B ...................
Per Piece cost...... 6 .................................. 21 ....
Sales ................. \(x - \frac{qx}{100}\) ...................... \(\frac{qx}{100}\) ....................... x (Assume-Total sales)
Total Revenue ........\(r - \frac{tr}{100}\) ..................... \(\frac{tr}{100}\) ......................... r (Assume - Total revenue)
Two equations would be setup:
\(21 * \frac{qx}{100} = \frac{tr}{100}\)
\(\frac{r}{x} = \frac{21q}{t}\) .................... (1)
\(6 * (x - \frac{qx}{100}) = r - \frac{tr}{100}\) ......... (2)
Divide the complete equation (2) by x
\(6 * (1 - \frac{q}{100}) = \frac{r}{x} - \frac{tr}{100x}\)
\(6 * (1 - \frac{q}{100}) = \frac{r}{x} * (1 - \frac{t}{100})\) .................. (3)
Placing value of \(\frac{r}{x}\) (from equation 1) in equation (3)
\(6 * (1 - \frac{q}{100}) = \frac{21q}{t} * (1 - \frac{t}{100})\)
40t = 140q - tq
\(q = \frac{40t}{140-t}\)
Answer = D
Per Piece cost...... 6 .................................. 21 ....
Sales ................. \(x - \frac{qx}{100}\) ...................... \(\frac{qx}{100}\) ....................... x (Assume-Total sales)
Total Revenue ........\(r - \frac{tr}{100}\) ..................... \(\frac{tr}{100}\) ......................... r (Assume - Total revenue)
Two equations would be setup:
\(21 * \frac{qx}{100} = \frac{tr}{100}\)
\(\frac{r}{x} = \frac{21q}{t}\) .................... (1)
\(6 * (x - \frac{qx}{100}) = r - \frac{tr}{100}\) ......... (2)
Divide the complete equation (2) by x
\(6 * (1 - \frac{q}{100}) = \frac{r}{x} - \frac{tr}{100x}\)
\(6 * (1 - \frac{q}{100}) = \frac{r}{x} * (1 - \frac{t}{100})\) .................. (3)
Placing value of \(\frac{r}{x}\) (from equation 1) in equation (3)
\(6 * (1 - \frac{q}{100}) = \frac{21q}{t} * (1 - \frac{t}{100})\)
40t = 140q - tq
\(q = \frac{40t}{140-t}\)
Answer = D
