It is currently 24 Feb 2018, 19:39

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 43898
In a certain set of 300 stocks, each is priced at either $50,$100, or [#permalink]

### Show Tags

28 Aug 2015, 01:33
Expert's post
10
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

38% (01:27) correct 63% (03:27) wrong based on 74 sessions

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. [Reveal] Spoiler: OA _________________ Manager Joined: 15 May 2014 Posts: 65 Re: In a certain set of 300 stocks, each is priced at either$50, $100, or [#permalink] ### Show Tags 30 Aug 2015, 18:23 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 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7960 Location: Pune, India Re: In a certain set of 300 stocks, each is priced at either$50, $100, or [#permalink] ### Show Tags 02 Sep 2015, 22:23 4 This post received KUDOS Expert's post 4 This post was BOOKMARKED 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) _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 4908
GPA: 3.82
In a certain set of 300 stocks, each is priced at either $50,$100, or [#permalink]

### Show Tags

03 Sep 2015, 03:44
1
KUDOS
Expert's post
1
This post was
BOOKMARKED
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.