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Quick Sell Outlet sold a total of 40 televisions, each of

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

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New post 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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New post 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)

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

Answer: C.
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Re: Quick Sell Outlet sold a total of 40 televisions, each of  [#permalink]

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New post 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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New post 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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New post 03 Jun 2015, 09:51
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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

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

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New post 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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New post 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)

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

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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New post 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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New post 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 = (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
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Re: Quick Sell Outlet sold a total of 40 televisions, each of  [#permalink]

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New post 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)

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

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

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New post 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=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.
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Re: Quick Sell Outlet sold a total of 40 televisions, each of  [#permalink]

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New post 22 Oct 2018, 21:51
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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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Re: Quick Sell Outlet sold a total of 40 televisions, each of  [#permalink]

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

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New post 31 May 2019, 10:28
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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?


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.
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Quick Sell Outlet sold a total of 40 televisions, each of   [#permalink] 31 May 2019, 10:28
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