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Quick Sell Outlet sold a total of 40 televisions, each of
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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
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14 Oct 2013, 09:13
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
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02 Dec 2014, 20:00
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
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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
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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
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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
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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
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18 Nov 2017, 05:56
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
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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
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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 $141, more of model Q would need to be sold as the average is skewed towards 150. Hope this helps.
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22 Oct 2018, 15:12
Bunuel when I try solve questions like this, I always have the fear of missing out on that 1 crazy case where we would be able to solve the equation even with 2 unknowns. In this particular problem for instance, we know the total money sold was 40*141 = 5640$. Now, how do I know that Statement 1 is not sufficient? If I substitute p=q30 into your equation, I get 40q30x = 5640 > And this is when fear comes because x has to be a non negative integer, and q is a price. Those constraints are the ones that make me fear I might be in front of a crazy question where there is actually only one combination of q and x that gives 5640. Is there any way to quickly prove/disprove that I am in front of such crazy case? Lastly, related to the issue above... what is allowed as $ price on GMAT? is 0.00000121 allowed as a price? or do prices need to be of 0.00 form (given cents is the smallest currency)? Thank you. PS. ccooley VeritasKarishma I will be grateful if you could give input too. Thanks a bunch.



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Re: Quick Sell Outlet sold a total of 40 televisions, each of
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22 Oct 2018, 21:51
gmat800live wrote: Bunuel when I try solve questions like this, I always have the fear of missing out on that 1 crazy case where we would be able to solve the equation even with 2 unknowns. In this particular problem for instance, we know the total money sold was 40*141 = 5640$. Now, how do I know that Statement 1 is not sufficient? If I substitute p=q30 into your equation, I get 40q30x = 5640 > And this is when fear comes because x has to be a non negative integer, and q is a price. Those constraints are the ones that make me fear I might be in front of a crazy question where there is actually only one combination of q and x that gives 5640. Is there any way to quickly prove/disprove that I am in front of such crazy case? Lastly, related to the issue above... what is allowed as $ price on GMAT? is 0.00000121 allowed as a price? or do prices need to be of 0.00 form (given cents is the smallest currency)? Thank you. PS. ccooley VeritasKarishma I will be grateful if you could give input too. Thanks a bunch. Yes, I understand your concern and when I looked at stmnt 1, I did ensure that there are multiple solutions possible. But I used a conceptual approach to eliminate (A) as the answer since I am too lazy to use a scratch pad. 141 is the average so I imagined a number line with 141 on it.  (141)  The actual price of P and Q has a difference of 30 so the distance between them on the number line will be 30 p (141) q Now since total number of TVs is 40, I think to myself, can I divide 40 into 2 parts such that their ratio is the distance between p and 141 and 141 and q? (Using scale method of averages) Yes. 40 = 20 + 20 (ratio 1:1) So p could be 15 to the left of 141 and q could be 15 to the right of 141. 40 = 16 + 24 (ratio of 2:3) So p could be 18 to the left of 141 and q could be 12 to the right of 141. Also, even if the second case is hard to arrive at, I know that price can be in decimal. So I just split 40 = 10 + 30 So 30 needs to be split in the ratio 1:3 i.e. 7.50 and 22.50 which is fine. Price would usually be upto 2 decimal places only except in cases in which you buy multiple items together such as you can buy a pack of 6 apples for $2.
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Re: Quick Sell Outlet sold a total of 40 televisions, each of
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23 Oct 2018, 01:52
This is wonderful, thank you! VeritasKarishma wrote: gmat800live wrote: Bunuel when I try solve questions like this, I always have the fear of missing out on that 1 crazy case where we would be able to solve the equation even with 2 unknowns. In this particular problem for instance, we know the total money sold was 40*141 = 5640$. Now, how do I know that Statement 1 is not sufficient? If I substitute p=q30 into your equation, I get 40q30x = 5640 > And this is when fear comes because x has to be a non negative integer, and q is a price. Those constraints are the ones that make me fear I might be in front of a crazy question where there is actually only one combination of q and x that gives 5640. Is there any way to quickly prove/disprove that I am in front of such crazy case? Lastly, related to the issue above... what is allowed as $ price on GMAT? is 0.00000121 allowed as a price? or do prices need to be of 0.00 form (given cents is the smallest currency)? Thank you. PS. ccooley VeritasKarishma I will be grateful if you could give input too. Thanks a bunch. Yes, I understand your concern and when I looked at stmnt 1, I did ensure that there are multiple solutions possible. But I used a conceptual approach to eliminate (A) as the answer since I am too lazy to use a scratch pad. 141 is the average so I imagined a number line with 141 on it.  (141)  The actual price of P and Q has a difference of 30 so the distance between them on the number line will be 30 p (141) q Now since total number of TVs is 40, I think to myself, can I divide 40 into 2 parts such that their ratio is the distance between p and 141 and 141 and q? (Using scale method of averages) Yes. 40 = 20 + 20 (ratio 1:1) So p could be 15 to the left of 141 and q could be 15 to the right of 141. 40 = 16 + 24 (ratio of 2:3) So p could be 18 to the left of 141 and q could be 12 to the right of 141. Also, even if the second case is hard to arrive at, I know that price can be in decimal. So I just split 40 = 10 + 30 So 30 needs to be split in the ratio 1:3 i.e. 7.50 and 22.50 which is fine. Price would usually be upto 2 decimal places only except in cases in which you buy multiple items together such as you can buy a pack of 6 apples for $2.



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Quick Sell Outlet sold a total of 40 televisions, each of
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31 May 2019, 10:28
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? Given: n = 40, Avg = 141, $p, $q, P Tvs, Q Tvs. This is a weighted average question, we can use a number line to visualize as per VeritasKarishma method: 1) the Model P televisions sold for $30 less than the Model Q televisionsP______________________Q ($q30)($141)($q) Not sufficient since we don't have values for the endpoints, can't determine ratio. 2) Either p=120 or q=120P______________________Q ($120)($141)($q) Q______________________P ($120)($141)($p) We only have 1 endpoint. Not sufficient. Note that we don't know which is the smaller value. 3) From 1) $p=$q30, so $p is the smaller value. From 2) $p OR $q = $120 If $q=120, this becomes impossible to do because we can't have a negative amount of TVs. P_____51________21_____Q ($90)($141)($120) If $p=120: P______21__________19____Q ($120)($141)($150) So ratio of P Tv/Q Tv is 19/21 for a total of 40, with P=19 and Q=21. Sufficient. Done algebraically: 1) the Model P televisions sold for $30 less than the Model Q televisions\(\frac{P}{Q}= \frac{$q  $141}{$141  $(q30)}\) > Can't determine P/Q ratio because we don't have a value for $q. 2) Either p=120 or q=120\(\frac{P}{Q}= \frac{$q  $141}{$141  $120}\) OR \(\frac{Q}{P}= \frac{$p  $141}{$141  $120}\) > Can't determine P/Q ratio because we don't have a value for $p. 3) If $q=120 then \(\frac{P}{Q}= \frac{$120  $141}{$141  $90}\) > \(\frac{P}{Q}= \frac{$19}{$51}\) > Not possible to have negative amount of TVs. If $p=120 then \(\frac{P}{Q}= \frac{$150  141}{141  120}\) > \(\frac{P}{Q}= \frac{$21}{$19}\) > So amount of P TVs is 19.




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