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Intern  Joined: 02 Jul 2013
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Quick Sell Outlet sold a total of 40 televisions, each of  [#permalink]

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Question Stats: 64% (02:32) correct 36% (02:39) wrong based on 697 sessions

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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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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) $$141 = \frac{px+q(40-x)}{40}$$, 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=q-30. Not sufficient to get x.

(2) Either p=120 or q=120. Not sufficient.

1) the Model P televisions sold for $30 less than the Model Q televisions 2) Either p=120 or q=120 _________________ Please Help with Kudos, if you like my post. Intern  Joined: 03 Feb 2015 Posts: 13 GMAT 1: 680 Q47 V36 GMAT 2: 720 Q49 V39 Re: Quick Sell Outlet sold a total of 40 televisions, each of [#permalink] Show Tags (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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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 ($141) which is not possible. hi bunuel, i don't understand this can you pls elaborate You need to understand the following first, (average price) = (total sales)/(Number of televisions sold) Because Total Televisions Sold = 40 So Let, Total models of P television sold = x Then, Total models of Q television sold = 40-x The Revenue earned by selling x television of type P =$p * x
The Revenue earned by selling (40-x) television of type Q = $q * (40-x) 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 - 30 i.e.$5640 = px+q(40−x) can be Re-written as
$5640 = (q-30)x+q(40−x) But two unknown Variables q and x Hence, NOT SUFFICIENT Statement 2: Either p=120 or q=120 @p = 120, q and x are unknown hence not sufficient @q = 120, p and x are unknown hence not sufficient Hence, NOT SUFFICIENT Combining Statement 1 and Statement 2 together: Either p=120 or q=120 AND$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 SUFFICIENT

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Quick Sell Outlet sold a total of 40 televisions, each of  [#permalink]

harDill wrote:
(1)+(2) If q=120, then p=90, so in this case both prices would be less than the average price ($141) which is not possible. hi bunuel, i don't understand this can you pls elaborate If you only score 70 and 80 on tests in a class, is it possible for you to get an average test score of 90 in the class? Intern  B Joined: 26 Feb 2017 Posts: 11 Re: Quick Sell Outlet sold a total of 40 televisions, each of [#permalink] Show Tags Bunuel wrote: Quick Sell Outlet sold a total of 40 televisions, each of which was either a Model P television or a Model Q television. Each Model P television sold for$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)

$$141 = \frac{px+q(40-x)}{40}$$, 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=q-30. 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.

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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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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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 = (40-a) We are given $$\frac{{ap + q(40-a)}}{40}$$ = 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 = q-30
=> 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 = q-30
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 $$\frac{{ap + q(40-a)}}{40}$$ = 141 => 120a + 150(40-a)=5640 => we can find the value of a. Sufficient. The answer is C _________________ Put in the work, and that dream score is yours! Senior Manager  G Joined: 06 Jul 2016 Posts: 358 Location: Singapore Concentration: Strategy, Finance Re: Quick Sell Outlet sold a total of 40 televisions, each of [#permalink] Show Tags 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)

$$141 = \frac{px+q(40-x)}{40}$$, 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=q-30. 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.

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 $141, more of model Q would need to be sold as the average is skewed towards 150. Hope this helps. _________________ Put in the work, and that dream score is yours! Intern  B Joined: 06 Sep 2018 Posts: 36 GMAT 1: 760 Q50 V44 GMAT 2: 740 Q48 V44 Quick Sell Outlet sold a total of 40 televisions, each of [#permalink] Show Tags Bunuel when I try solve questions like this, I always have the fear of missing out on that 1 crazy case where we would be able to solve the equation even with 2 unknowns. In this particular problem for instance, we know the total money sold was 40*141 = 5640$. Now, how do I know that Statement 1 is not sufficient?

If I substitute p=q-30 into your equation, I get 40q-30x = 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 $price on GMAT? is 0.00000121 allowed as a price? or do prices need to be of 0.00 form (given cents is the smallest currency)? Thank you. PS. ccooley VeritasKarishma I will be grateful if you could give input too. Thanks a bunch. Veritas Prep GMAT Instructor V Joined: 16 Oct 2010 Posts: 9706 Location: Pune, India Re: Quick Sell Outlet sold a total of 40 televisions, each of [#permalink] Show Tags 2 gmat800live wrote: Bunuel when I try solve questions like this, I always have the fear of missing out on that 1 crazy case where we would be able to solve the equation even with 2 unknowns. In this particular problem for instance, we know the total money sold was 40*141 = 5640$. Now, how do I know that Statement 1 is not sufficient?

If I substitute p=q-30 into your equation, I get 40q-30x = 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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GMAT 1: 760 Q50 V44 GMAT 2: 740 Q48 V44 Re: Quick Sell Outlet sold a total of 40 televisions, each of  [#permalink]

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This is wonderful, thank you!

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$. Now, how do I know that Statement 1 is not sufficient? If I substitute p=q-30 into your equation, I get 40q-30x = 5640 -> And this is when fear comes because x has to be a non negative integer, and q is a price. Those constraints are the ones that make me fear I might be in front of a crazy question where there is actually only one combination of q and x that gives 5640. Is there any way to quickly prove/disprove that I am in front of such crazy case? Lastly, related to the issue above... what is allowed as$ 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 $2. Senior Manager  P Status: Gathering chakra Joined: 05 Feb 2018 Posts: 434 Quick Sell Outlet sold a total of 40 televisions, each of [#permalink] Show Tags 1 bulletpoint wrote: Quick Sell Outlet sold a total of 40 televisions, each of which was either a Model P television or a Model Q television. Each Model P television sold for$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 $30 less than the Model Q televisions P______________________Q ($q-30)----($141)----------($q)

Not sufficient since we don't have values for the endpoints, can't determine ratio.

2) Either p=120 or q=120

P______________________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) $p=$q-30, 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 $$\frac{P}{Q}= \frac{q - 141}{141 - (q-30)}$$ ---> 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. Quick Sell Outlet sold a total of 40 televisions, each of   [#permalink] 31 May 2019, 10:28
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