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# Set X consists of 9 positive elements. Set Y consists of the

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Set X consists of 9 positive elements. Set Y consists of the [#permalink]

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12 Jul 2006, 16:37
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Set X consists of 9 positive elements. Set Y consists of the squares of the elements of X. Is the average of the elements in Y greater than that of set X?

(1) The median of X is greater than 1.
(2) The range of X is 2.
[Reveal] Spoiler: OA
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12 Jul 2006, 18:35
kevincan wrote:
Set X consists of 9 positive elements. Set Y consists of the squares of the elements of X. Is the average of the elements in Y greater than that of set X?

(1) The median of X is greater than 1.
(2) The range of X is 2.

first there is no saying if the elements are decimal or integers also it does not mention if the elements are consecutive...

st.1
median = 1, which means that the 5th term once the numbers are put in sequence order, must be 1, but does not limit how big the number can get. depending on the sequence number, we can either answer a yes or no... Not SUFF

st.2
Range=Largest-smallest number in the sequence = 2.

Which means that we can anwer a yes or a no depending on the number sequence... Not SUFF.

Together... I get SUFF...

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12 Jul 2006, 23:11
D.

1) Median > 1

Hence lets say the first integer are 1 (remember we are talking about positive integers) the fifth integer is greater than 1 hence lets say its 2 and the lets say the other 4 (has to be >= 2) is 2

the average of these numbers is obviously less than average of square of these numbers.

2) Range is 2
Lets say eight integers are 1 and the last integer has to be 3 to have a range of 2
Hence the average of these numbers is obviously less than average of square of these numbers.

Hence D
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12 Jul 2006, 23:32
I agree with jaya here, should be D.

for either of the statments, integers and fractions will yield the same result.
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13 Jul 2006, 15:46
jaynayak wrote:
D.

1) Median > 1

Hence lets say the first integer are 1 (remember we are talking about positive integers) the fifth integer is greater than 1 hence lets say its 2 and the lets say the other 4 (has to be >= 2) is 2

the average of these numbers is obviously less than average of square of these numbers.

2) Range is 2
Lets say eight integers are 1 and the last integer has to be 3 to have a range of 2
Hence the average of these numbers is obviously less than average of square of these numbers.

Hence D

how are you assuming the values to be integers?
in the question, it does not mention integers...

shahnandan,
you said: integers and fractions will yield the same result.
for example: if x=0.5, 0.5
y=0.25, 0.25 since y=x^2, the average of y can not be greater than average of x.

If the values of x are integers, then I agree, but not when they are fractions.

I still don't buy the integer part of the explanation. Just like x^2 = 4, therefore x=2. Wrong x=2 or x=-2... we are not supposed to assume too much in DS...
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13 Jul 2006, 21:34
kevincan wrote:
Set X consists of 9 positive elements. Set Y consists of the squares of the elements of X. Is the average of the elements in Y greater than that of set X?

(1) The median of X is greater than 1.
(2) The range of X is 2.

go with E. since we donot know how many elements of x are there in y.
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13 Jul 2006, 22:14
B

St1: In X if 5 numbers are 1.000000001 and 4 are 0.000000001 then X will be greater. If all elements of X are greater than 1 then average of Y is greater : INSUFF

St2: If range is 2 then average of Y will always be greater than that of X. Lets take teh worst case scenario. When 8 elements of X are 0.0000000001 and one element is 2.0000000001 then sum of elements will be approx 2. But sum of elements of Y will be close to 4. If all elements of X are greater than 1 then also average of Y will be greater.: SUFF
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20 Jul 2006, 13:32
You need to look harder for your worst case scenario. Will this change the answer? Who can improve on his great start?
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20 Jul 2006, 13:52
I got A, the reason is

If the median is greater than 1, then for all the numbers above 1, the average will be greater in Y. And in the worst case scenario, the first four numbers are between 0 and 1, but they will have very limited effect on the average.

B is inconclusive.

So I would go with A.
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20 Jul 2006, 15:31
rnachloo wrote:
I got A, the reason is

If the median is greater than 1, then for all the numbers above 1, the average will be greater in Y. And in the worst case scenario, the first four numbers are between 0 and 1, but they will have very limited effect on the average.

B is inconclusive.

So I would go with A.

I will go with A too. since the median has got to be atleast two, the rest of the elements in X has to be atleast 2.01 (or 3 even if they are integers). When squared the increase in value of the last 4 will easily outweigh the decrease in the value of the first four decimals being squared (if they are decimals). So A is suff.
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20 Jul 2006, 15:37
I will go with A too. since the median has got to be atleast two

Why?
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20 Jul 2006, 16:33
It should be B.

stmt1.
counter example
set X : 0.1, 0.1, 0.1, 0.1, 1.01, 1.01, 1.01, 1.01, 1.01
set Y : 0.01, 0.01, 0.01, 0.01, 1.0201, 1.0201, 1.0201, 1.0201, 1.0201
SumX : 1*5 + 0.1*5 + 0.01*5
SumY : 1*5 + 0.01*4+0.0201*5
SumX > SumY
insuff.
Stmt2.
Max (SetX) > 2
Max (SetY) > 4.
Thus
SumY - SumX > 1 (I can prove this using derivative, but not using it, this just become a guess)
Suff.
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21 Jul 2006, 15:10
I don't want to give the OE just yet!
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21 Jul 2006, 21:41
B.

For example a) x=(0.498, 0.499, 0.5, 0.501, 1, 1.001, 1002, 1003, 1004). y corresponding;
[the elements in x could be the same]

but:
x=(0.1, 0.2, 0.3, 0.4, 1, 10, 100, 1000, 1000000000 [just to be sure])

(Note: x^2-x takes a minimum at x=0.5. The decrease from x to x^2 is 0.25 at x=0.5)

Statement (2) is sufficient.

It is easy to find an example where the average of the elements of y is greater than the average of the elements in x).
Could the average of x be greater than the average of the elements in y? No.

Consider [for example] x=(0+epsilon, 0.499, 0.5, 0.5001, 0.5002, 0.5003, 0.50004, 0.50005, 2+epsilon)

the increase from x=0 to x^2=0 is 0; the increase from x=2 to x^2=2 is 2. Hence the total increase is 2, but the decrease from values around 0.5 is approximately 0.25; therefore the maximum total decrease 0.25*7=1.75 is smaller than the total increase. Hence the average of the elements in y must be larger than the average of the elements in x).

[The argument is based on the fact that (x^2-x) + (x+2)^2-(x+2) takes an extremum at x=-0.5 and is strictly increasing for x>-0.5; hence we have a border solution for x=0].

Another interesting question would arise if the set x could consist of negative numbers as well.
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22 Jul 2006, 01:02
Very nice, but this could have been even simpler. The elements of X need not be different! Your reasoning is spot on, though
OA=B

Last edited by kevincan on 22 Jul 2006, 13:46, edited 1 time in total.
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22 Jul 2006, 10:56
Proof of stmt B.
difference between Y 9th term and X 9th term

1. Y9-X9 > 2

let
F(x) = x - x^2
F(x) = -(x-1/2)^2 + 1/4 (parabola, concave down)
Thus, when x = 1/2 we got maximun value of F(x) = 1/4

Assuming that sum of differences between each term of Set X and each term of Set Y, 1st through 8th term, has the maximun value
2. (1/4)*8 = 2

Y9-X9 is always greater than 2. (Y9-X9 >2 and (X1-Y1)+(X2-Y2)..+(X8-Y8) <= 2)

Sufficient.
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Last edited by freetheking on 22 Jul 2006, 14:01, edited 1 time in total.
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22 Jul 2006, 13:43
You're right., pardon my typo, your solution was just what I was looking for- OA=B
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22 Jul 2006, 15:24
I got lost on this one....good explanations though.
22 Jul 2006, 15:24
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