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In a certain set of 300 stocks, each is priced at either $50, $100, or

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In a certain set of 300 stocks, each is priced at either $50, $100, or [#permalink]

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In a certain set of 300 stocks, each is priced at either $50, $100, or $150. If an investor bought 95 different stocks for 5,000 dollars, what was the price of the stock in the 80th percentile, when all 300 stocks are ordered by price from least to greatest?

(1) The average price of the 300 stocks is the same as the average number of stocks in the set of 300 that have each price.
(2) The mean, median, mode, and range of the 300 stock prices are all identical.

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Re: In a certain set of 300 stocks, each is priced at either $50, $100, or [#permalink]

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New post 30 Aug 2015, 19:23
In a certain set of 300 stocks, each is priced at either $50, $100, or $150.


An investor bought 95 different stocks for 5,000 dollars.
Let \(x\) be the number of $50 stocks, \(y\) be $100 stocks, and \(z\) be $150 stocks
x + y + z = 95

50x + 100y + 150z = 5000
x + 2y + 3z = 100 --> (x + y + z) + y + 2z = 100
y + 2z = 5
since x, y, and z are integers, y could be 1, 3, or 5.
y = 1, z = 2, x = 92 ---(I)
y = 3, z = 1, x = 91 ---(II)
y = 5, z = 0, x = 90 ---(III)

Statement (1)
The average price of the 300 stocks is the same as the average number of stocks in the set of 300 that have each price.

Case (I) & Case (II)
average price = $100 (average stocks = \(\frac{300}{3}\))
if average price is $100
then number of $50 stocks = $150 stocks in order to get average price of 100 and the rest are $100 stocks
[case (I)--> 92*$50+116*$100+92*$150; case (II)--> 91*$50+118*$100+91*#150]
in either case the 80th percentile (stock #240), when all 300 stocks are ordered by price from least to greatest, is $150

Case (III)
Average price of $150 (average stocks = \(\frac{300}{2}\)) is not possible with only $50 and $100 stocks

Sufficient


Statement (2)
The mean, median, mode, and range of the 300 stock prices are all identical.

Case (III ),
range (50), median (100), mode (100), and mean (between 50 and 100) contradict each other
so case (III) is not possible

Case (I) & (II) (as above),
range (100)= median (100) = mode (100) = mean (100)

Sufficient

Answer D
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Re: In a certain set of 300 stocks, each is priced at either $50, $100, or [#permalink]

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New post 02 Sep 2015, 23:23
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Bunuel wrote:
In a certain set of 300 stocks, each is priced at either $50, $100, or $150. If an investor bought 95 different stocks for 5,000 dollars, what was the price of the stock in the 80th percentile, when all 300 stocks are ordered by price from least to greatest?

(1) The average price of the 300 stocks is the same as the average number of stocks in the set of 300 that have each price.
(2) The mean, median, mode, and range of the 300 stock prices are all identical.

Kudos for a correct solution.



Total number of stocks = 300
Price of each = $50/$100/$150

"If an investor bought 95 different stocks for 5,000 dollars" - Think about this a bit before you move on.
95 is close to 100 and if the investor had bought 100 stocks for $5000, his average price each stock would have been $50. So mostly, he bought the cheapest stocks. So most of the 95 stocks would be $50 stocks.

80th percentile of 300 is (80/100)*300 = 240. So the price of the 240th stock will lie in the 80th percentile. We need to know this price.

(1) The average price of the 300 stocks is the same as the average number of stocks in the set of 300 that have each price.

Average price of 300 stocks = Average number of stocks of each price

Let's calculate "Average number of stocks of each price"
The sum of the number of stocks = 300
The total different prices = 3
Average number of stocks of each price = 300/3 = 100

So average price of 300 stocks = $100

If we have at least about 90 $50 stocks, we must have at least about 90 $150 stocks too to balance out the deficit. So the 240th stock will be $150 stock.
Sufficient alone.

(2) The mean, median, mode, and range of the 300 stock prices are all identical.
Range 0 if all stocks are of the same price: Range cannot be 0 since the mean price cannot be 0.
Range 50 if all stocks are of '$50 and $100' (or '$100 and $150'): Range cannot be 50 since mean price cannot be 50 if there are some $100 stocks too.
So Range must be 100 and there must be stocks of all 3 prices.
This means average price of the stocks = $100

If we have at least about 90 $50 stocks, we must have at least about 90 $150 stocks too to balance out the deficit. So the 240th stock will be $150 stock.
Sufficient alone.

Answer (D)
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In a certain set of 300 stocks, each is priced at either $50, $100, or [#permalink]

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New post 03 Sep 2015, 04:44
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution.

In a certain set of 300 stocks, each is priced at either $50, $100, or $150. If an investor bought 95 different stocks for 5,000 dollars, what was the price of the stock in the 80th percentile, when all 300 stocks are ordered by price from least to greatest?

(1) The average price of the 300 stocks is the same as the average number of stocks in the set of 300 that have each price.
(2) The mean, median, mode, and range of the 300 stock prices are all identical.

Transforming the original codntiion we have the below table of 2by3 case, expanded from 2by2 case that are common in GMAT tests.


a+b+c=95, 50a+100b+150c=5,000 gives us 3 variables (a,b,c) and 2 equations (a+b+c=95, 50a+100b+150c=5,000). Since we need to match the number of variables and equations, we need 1 more equation and since we have 1 each in 1) and 2), D is likely the answer and it turns out that D actually is the answer.
Remember the three things in solving the question
1-st : the remaining questions * 2 minutes
2-nd : 5 minute deadline
3-rd : not want to solve
and we get D as our answer.
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Re: In a certain set of 300 stocks, each is priced at either $50, $100, or [#permalink]

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Re: In a certain set of 300 stocks, each is priced at either $50, $100, or   [#permalink] 21 May 2017, 08:09
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