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Re M1534 [#permalink]
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16 Sep 2014, 00:57
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Official Solution:In the month of November, a company sold 3000 items of model A and 1000 items of model B. Items of model A accounted for 60% of the company's monthly sales in dollars while items of model B accounted for 40% of the monthly sales in dollars. If the company had sold 1000 items of model A less than it actually did, what percent of the total monthly sales in dollars would have been attributed to model A?A. 48 B. 50 C. 52 D. 54 E. 55 Let \(S\) denote the total November sales of the company. The price of an item of model A = \(0.6*\frac{S}{3000}\); the price of an item of model B = \(0.4*\frac{S}{1000}\). If the company had sold 2000 items of model A, the revenue from sales of model A would have amounted to \(0.6*\frac{S}{3000}*2000 = 0.6S*\frac{2}{3} = 0.4S\) which is equal to the revenue from sales of model B. So, in the hypothetical case described in the stem, the two models would have accounted for 50% of the monthly sales each. Alternative explanation  picking numbers Let the total revenue be \($10,000\); then revenue from Model A is \($6,000\) for 3,000 units or \($2\) per unit and Model B's revenue is \($4,000\) for 1,000 units or \($4\) per unit. If we had sold only 2,000 units of model A, the total revenue would have been \($2*2,000 +$4*1,000 =$8,000\). At this point it is clear that model A and B bring equal amounts of revenue (\($4,000\) each) and half of monthly sales are attributed to model A. Answer: B
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Re: M1534 [#permalink]
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14 Jan 2015, 06:28
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A = Qty * Price = 3000 a B = 1000 * b = 1000 b
Now for A, 3000 a / (3000 a + 1000 b) = 3a / ( 3a + b) = 6 / 10 ……. ( i )  As given in the statement
Similarly , b / (3a + b) = 4 / 10 ……….. ( ii )
DIVIDE i / ii = 3a / b = 3 / 2 which gives us b = 2a
NOW As asked in the problem, find 2a / ( 2a + b)
Subtitute b , we get = 2a / 4a = 50%



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Re: M1534 [#permalink]
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11 Jul 2015, 06:05
Dear Bunuel, Could you please confirm that approach used by buddyisraelgmat is correct. I also used the same approach. Regards Vikas



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Re: M1534 [#permalink]
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11 Jul 2015, 08:46



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Re: M1534 [#permalink]
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14 Oct 2015, 14:37
Bunuel wrote: In the month of November, a company sold 3000 items of model A and 1000 items of model B. Items of model A accounted for 60% of the company's monthly sales while items of model B accounted for 40% of the monthly sales. If the company had sold 1000 items of model A less than it actually did, what percent of the total monthly sales would have been attributed to model A?
A. 48 B. 50 C. 52 D. 54 E. 55 Here's one more approach: 60%=20%+20%+20%, 40%=20%+20% > 3000/3=1000=20%, 1000/2=500=20% ==> 1000a=500b means b=2a 2000a/2000a+2000a (=1000b) = 50%
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Re: M1534 [#permalink]
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11 Dec 2015, 04:27
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the question never specifies that monthly sales are in dollars...where as monthly sales can be numbers also... what if 60% of A is 3000 units.. so monthly sales is 5000 units..



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Re: M1534 [#permalink]
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19 Oct 2016, 12:11
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I agree with vijayjec02, the question stem was a bit confusing. I had to assume that may be question intends to mean this. Question asked "A company sold 3000 items of model A and 1000 items of model B" and then says "Items of model A accounted for 60% of the company's monthly sales." Item of model A did not amount for 60% of sales... items amount for 75% of total sales. Money generated from the sale of item A amounted for 60%... Otherwise I think it is a nice question.



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Re: M1534 [#permalink]
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09 Mar 2017, 04:56
Lets say total sales = 100, so for 1000 items of A sales = 20 A: 20 + 20 +20 =60 B: 40
When sales of A is reduced by 1000, total sales = remaining 2000 = 40 There is no change in B sales. Total sale : = 40+40=80
So %age of sale A: = 40/80* 100 = 50%



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Re: M1534 [#permalink]
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10 Sep 2017, 05:14
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The question stem is not clear in this case because monthly sales can be in terms of both money and quantity. Bunuel, I request you to replace this question stem to avoid any ambiguity in the meaning of the stem.



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Re: M1534 [#permalink]
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08 Nov 2017, 02:00
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Scpkulkarni wrote: The question stem is not clear in this case because monthly sales can be in terms of both money and quantity. Bunuel, I request you to replace this question stem to avoid any ambiguity in the meaning of the stem. I too agree. The question stem never specifies the sales in terms of revenue or number of items.



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Re: M1534 [#permalink]
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08 Nov 2017, 23:26



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Let price of item A = a And price of item B = b
A sold 3000 items => total sale/revenue = 3000a B sold 1000 items => total sale/revenue = 1000b
Total sales/revenue = 3000a + 1000b
A revenue accounted for 60% of total revenue => 60% of total = A's revenue =>\(\frac{60}{100}(3000a+1000b) = 3000a\) => 9000a +3000b =15000a => 2a = b
Now A sold 1000 less items = 2000 items => revenue = 2000a Total revenue = 2000a+1000b = 2000a + 1000*2a = 4000a <= as 2a =b
so % of revenue from A = \(\frac{Revenue of A}{Total revenue} * 100\)
= \(\frac{2000a}{4000a}*100\)
= 50%
Answer: B
Here is Bunuel 's solution ( in detail)
Let total sales/revenue from A and B = S
as revenue of A = 60% of S = 0.6*S = items sold by A * price of item A => 0.6*S = 3000 * price of item A
=>price of item A = \(0.6 * \frac{S}{3000}\)
same way : as revenue of B = 40% of S = 0.4*S = items sold by B * price of item B => 0.4*S = 1000 * price of item B
=>price of item B = \(0.4 * \frac{S}{1000}\)
Now as A sold 1000 less items => items sold = 2000
So revenue from A = price of item A * number of items sold = \(0.6 * \frac{S}{3000} * 2000\) = 0.4*S
That is, revenue from A = 0.4*S and revenue from B = 0.4*S
so % of revenue from A = \(\frac{Revenue of A}{Total revenue} * 100\)
= \(\frac{0.4*S}{0.4*S + 0.4*S}*100\)
= 50%



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Re: M1534 [#permalink]
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30 Dec 2017, 04:43
I feel this question is again a 500level question. GMAT Quant has been steadily becoming more challenging and I feel such level questions can no more be incorrectly tagged 600 level.










