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