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Quick Sell Outlet sold a total of 40 televisions, each of [#permalink]
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14 Oct 2013, 08:51
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Quick Sell Outlet sold a total of 40 televisions, each of which was either a Model P television or a Model Q television. Each Model P television sold for $p and each Model Q sold for $q. The average (arithmetic mean) selling price of the 40 televisions was $141. How many of the 40 televisions were Model P televisions? 1) the Model P televisions sold for $30 less than the Model Q televisions 2) Either p=120 or q=120
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Re: Quick Sell Outlet sold a total of 40 televisions, each of [#permalink]
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14 Oct 2013, 09:13
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Quick Sell Outlet sold a total of 40 televisions, each of which was either a Model P television or a Model Q television. Each Model P television sold for $p and each Model Q sold for $q. The average (arithmetic mean) selling price of the 40 televisions was $141. How many of the 40 televisions were Model P televisions?(average price) = (total sales)/(# of televisions sold) \(141 = \frac{px+q(40x)}{40}\), where x is the number of Model P televisions sold. (1) The Model P televisions sold for $30 less than the Model Q televisions > p=q30. Not sufficient to get x. (2) Either p=120 or q=120. Not sufficient. (1)+(2) If q=120, then p=90, so in this case both prices would be less than the average price ($141) which is not possible. Thus p=120 and q=150. When we substitute these values we'll have only one unknown x, so we can solve for it. Sufficient. Answer: C.
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Re: Quick Sell Outlet sold a total of 40 televisions, each of [#permalink]
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02 Dec 2014, 20:00
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nice conceptual question and well explained by bunuel. 1 and 2 are clearly not sufficient. 1&2. if p = 120 then q = 90 then average cannot be 141 so its has to be p = 150 and q =120. bulletpoint wrote: Quick Sell Outlet sold a total of 40 televisions, each of which was either a Model P television or a Model Q television. Each Model P television sold for $p and each Model Q sold for $q. The average (arithmetic mean) selling price of the 40 televisions was $141. How many of the 40 televisions were Model P televisions?
1) the Model P televisions sold for $30 less than the Model Q televisions
2) Either p=120 or q=120
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Re: Quick Sell Outlet sold a total of 40 televisions, each of [#permalink]
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03 Jun 2015, 09:32
(1)+(2) If q=120, then p=90, so in this case both prices would be less than the average price ($141) which is not possible.
hi bunuel, i don't understand this can you pls elaborate



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Re: Quick Sell Outlet sold a total of 40 televisions, each of [#permalink]
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03 Jun 2015, 09:51
harDill wrote: Quick Sell Outlet sold a total of 40 televisions, each of which was either a Model P television or a Model Q television. Each Model P television sold for $p and each Model Q sold for $q. The average (arithmetic mean) selling price of the 40 televisions was $141. How many of the 40 televisions were Model P televisions?
1) the Model P televisions sold for $30 less than the Model Q televisions
2) Either p=120 or q=120
(1)+(2) If q=120, then p=90, so in this case both prices would be less than the average price ($141) which is not possible.
hi bunuel, i don't understand this can you pls elaborate You need to understand the following first, (average price) = (total sales)/(Number of televisions sold) Because Total Televisions Sold = 40 So Let, Total models of P television sold = x Then, Total models of Q television sold = 40x The Revenue earned by selling x television of type P = $p * x The Revenue earned by selling (40x) television of type Q = $q * (40x) Total Revenue Earned = Average Price per television * No. of televisions sold = $141 * (40) Hence, 141 * 40=px+q(40−x), where x is the number of Model P televisions sold. $5640 = px+q(40−x)Unknown Variable to be calculated = x = ?Statement 1: $p = $q  30i.e. $5640 = px+q(40−x) can be Rewritten as $5640 = (q30)x+q(40−x) But two unknown Variables q and x Hence, NOT SUFFICIENTStatement 2: Either p=120 or q=120@p = 120, q and x are unknown hence not sufficient @q = 120, p and x are unknown hence not sufficient Hence, NOT SUFFICIENTCombining Statement 1 and Statement 2 together:Either p=120 or q=120 AND $p = $q  30 AND $5640 = px+q(40−x) i.e. If p = 120, then q = 150 and then x = 12, A Unique value But if q = 120 then p = 90 and then x = 28 i.e. Negative Value which is IMPOSSIBLE Hence SUFFICIENTAnswer: Option
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Quick Sell Outlet sold a total of 40 televisions, each of [#permalink]
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16 Sep 2016, 16:37
harDill wrote: (1)+(2) If q=120, then p=90, so in this case both prices would be less than the average price ($141) which is not possible.
hi bunuel, i don't understand this can you pls elaborate If you only score 70 and 80 on tests in a class, is it possible for you to get an average test score of 90 in the class?



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Re: Quick Sell Outlet sold a total of 40 televisions, each of [#permalink]
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18 Nov 2017, 05:07
Bunuel wrote: Quick Sell Outlet sold a total of 40 televisions, each of which was either a Model P television or a Model Q television. Each Model P television sold for $p and each Model Q sold for $q. The average (arithmetic mean) selling price of the 40 televisions was $141. How many of the 40 televisions were Model P televisions?
(average price) = (total sales)/(# of televisions sold)
\(141 = \frac{px+q(40x)}{40}\), where x is the number of Model P televisions sold.
(1) The Model P televisions sold for $30 less than the Model Q televisions > p=q30. Not sufficient to get x.
(2) Either p=120 or q=120. Not sufficient.
(1)+(2) If q=120, then p=90, so in this case both prices would be less than the average price ($141) which is not possible. Thus p=120 and q=150. When we substitute these values we'll have only one unknown x, so we can solve for it. Sufficient.
Answer: C. HI, Even here when q=120 and p=150, the average still comes out to be less than 141. Hence how can we take this as a solution?



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Re: Quick Sell Outlet sold a total of 40 televisions, each of [#permalink]
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18 Nov 2017, 05:56
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bulletpoint wrote: Quick Sell Outlet sold a total of 40 televisions, each of which was either a Model P television or a Model Q television. Each Model P television sold for $p and each Model Q sold for $q. The average (arithmetic mean) selling price of the 40 televisions was $141. How many of the 40 televisions were Model P televisions?
1) the Model P televisions sold for $30 less than the Model Q televisions
2) Either p=120 or q=120 Let number of P televisions be m, and Q Televisions be n. p*m + q*n/m+n = 141, m + n = 40 p*m + q*n = 5640 p = q  30 q*(m+n)  30*m = 5640. Now, when q = 120, m will be ve which is not possible. So, only p = 120, q = 150
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Re: Quick Sell Outlet sold a total of 40 televisions, each of [#permalink]
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18 Nov 2017, 06:11
bulletpoint wrote: Quick Sell Outlet sold a total of 40 televisions, each of which was either a Model P television or a Model Q television. Each Model P television sold for $p and each Model Q sold for $q. The average (arithmetic mean) selling price of the 40 televisions was $141. How many of the 40 televisions were Model P televisions? Average = \(\frac{{Total Sales}}{{No. of TVs sold}}\) Let No. of Model P TVs be a => Model Q TVs = (40a) We are given \(\frac{{ap + q(40a)}}{40}\) = 141 We need to find the value of a. Quote: 1) the Model P televisions sold for $30 less than the Model Q televisions
2) Either p=120 or q=120 S1) p = q30 => q(a+40)  a(30+q) = 5640 => Not enough information to find the value of a. Insufficient. S2) p = $120 or q = $120 => Clearly not enough information to find the value of a. Insufficient. Quote: Combining S1+S1 p = q30 if p = $120 then q = $150 if q = $120 then p = $90 => Not possible at the average is $141 [Ignore this case]Inputting values of p & q in \(\frac{{ap + q(40a)}}{40}\) = 141 => 120a + 150(40a)=5640 => we can find the value of a. Sufficient. The answer is C
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Re: Quick Sell Outlet sold a total of 40 televisions, each of [#permalink]
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18 Nov 2017, 06:20
kartzcool wrote: Bunuel wrote: Quick Sell Outlet sold a total of 40 televisions, each of which was either a Model P television or a Model Q television. Each Model P television sold for $p and each Model Q sold for $q. The average (arithmetic mean) selling price of the 40 televisions was $141. How many of the 40 televisions were Model P televisions?
(average price) = (total sales)/(# of televisions sold)
\(141 = \frac{px+q(40x)}{40}\), where x is the number of Model P televisions sold.
(1) The Model P televisions sold for $30 less than the Model Q televisions > p=q30. Not sufficient to get x.
(2) Either p=120 or q=120. Not sufficient.
(1)+(2) If q=120, then p=90, so in this case both prices would be less than the average price ($141) which is not possible. Thus p=120 and q=150. When we substitute these values we'll have only one unknown x, so we can solve for it. Sufficient.
Answer: C. HI, Even here when q=120 and p=150, the average still comes out to be less than 141. Hence how can we take this as a solution? Hi mate, You're assuming that 1 unit of both TVs was sold. if p = 120 and q = 150 then, for the average to be $141, more of model Q would need to be sold as the average is skewed towards 150. Hope this helps.
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Re: Quick Sell Outlet sold a total of 40 televisions, each of [#permalink]
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19 Nov 2017, 04:32
Bunuel wrote: Quick Sell Outlet sold a total of 40 televisions, each of which was either a Model P television or a Model Q television. Each Model P television sold for $p and each Model Q sold for $q. The average (arithmetic mean) selling price of the 40 televisions was $141. How many of the 40 televisions were Model P televisions?
(average price) = (total sales)/(# of televisions sold)
\(141 = \frac{px+q(40x)}{40}\), where x is the number of Model P televisions sold.
(1) The Model P televisions sold for $30 less than the Model Q televisions > p=q30. Not sufficient to get x.
(2) Either p=120 or q=120. Not sufficient.
(1)+(2) If q=120, then p=90, so in this case both prices would be less than the average price ($141) which is not possible. Thus p=120 and q=150. When we substitute these values we'll have only one unknown x, so we can solve for it. Sufficient.
Answer: C. Regarding Statement (2) there are sometimes similar Qs where only the price of one object is given and you can detect by the average that only one combination would fit. Do you have a general approach to this kind of questions, how to can see such a constellation?




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