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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] New post 14 Oct 2013, 07:51
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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
[Reveal] Spoiler: OA
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Re: Quick Sell Outlet sold a total of 40 televisions, each of [#permalink] New post 14 Oct 2013, 08:13
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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] New post 10 Nov 2014, 21:40
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Re: Quick Sell Outlet sold a total of 40 televisions, each of [#permalink] New post 02 Dec 2014, 19:00
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 [#permalink] New post 03 Jun 2015, 08:32
(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] New post 03 Jun 2015, 08:51
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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
[Reveal] Spoiler:
C

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Re: Quick Sell Outlet sold a total of 40 televisions, each of   [#permalink] 03 Jun 2015, 08:51
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