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Set X consists of 9 positive elements. Set Y consists of the [#permalink]
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19 Sep 2006, 13:49
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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.
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Last edited by Bunuel on 12 Jul 2013, 01:53, edited 1 time in total.
Added the OA.



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B for me ....the range indicates you are dealing with fractions and hence the average will be greater for x than y as squaring fractions yields a smaller number?? Please provide the OA/OE on this one.



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Answer: E
X = {a1, a2, a3, ... a9} where an > 0
Y = {a1^2, a2^2 ... a9^2, y1, y2, y3, .. yk} where
yk is the element of Y which we know nothing about...
As X has all +ve elements avg X > 0.
S1:
a5 > 1
If a5 = 2, we can compute average but we still know nothing about the remaining elements of Y (which could be +ve or ve)
If the sum of the remaining elements of Y (which are all ve) = sum of the squares of elements in X then avg Y = 0.
Avg Y = (SumY)/(y+9) where y = number of elements in Y.
Not sufficient.
S2: Range X = 2.
Max Min = 2.
Average > 0.
Again the above condition holds true.. for SumY
S1 & S2:
No new information...
Answer: E



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What if X = {1,1,1,1,1,1,1,1,3}
Range = 2, not fractions...
Matrix02 wrote: B for me ....the range indicates you are dealing with fractions and hence the average will be greater for x than y as squaring fractions yields a smaller number?? Please provide the OA/OE on this one.



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Great point  Thanks Haas....For some reason I though the numbers had to be different (note to self  read the question carefully!).



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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.
From st one
the set could be as follows
{ 4 fractions with large diniminators " so when squared exponantially decrease , 1.0000001 , four elements that could be 1.001} so avearage of y not>x
and we can easily find a counter example....thus one is insuff
from two
if the range of x is 2 this means that the difference between the two extreem values( largest , smallest ) is 2.
ie x,x+2
worst case scenario is all elements = 0 and one element = 2 (but the given says +ve elements and zero is enither +ve nor ve
or all elements are fractions and keeping the range as 2 this means that we have 8 elements approaching zero while one is increasing exponatially and as the no of elements is constant thus this means that the average increase
My answer is B.......hopefully right.......am bad with statistics
Last edited by yezz on 19 Sep 2006, 16:39, edited 1 time in total.



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yezz 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.
From st one
the set could be as follows
{ 4 fractions with large diniminators " so when squared exponantially decrease , 1.0000001 , four elements that could be 1.001} so avearage of y not>x
and we can easily find a counter example....thus one is insuff
from two
if the range of x is 2 this means that the difference between the two extreem values( largest , smallest ) is 2.
ie x,x+2 worst case scenario is all elements = 0 and one element = 2 (but the given says +ve elements and zero is enither +ve nor ve
or all elements are fractions and keeping the range as 2 this means that we have 8 elements approaching zero while one is increasing exponatially and as the no of elements is constant thus this means that the average increase
My answer is B.......hopefully right.......am bad with statistics
Haas  I believe Y is squares of X only, thus there can't be any negatives.
Yezz  you might be mistaken with st 1
My answer D (can't say it was easy though)
From st. 1  a square of any number larger than 1 will be larger than the number. Even if the first 4 members are fractions close to 0, we will still have one member larger than 1 more, i.e. 5 members, thus this one extra member larger than 1 will compensate any differences caused by squaring small fractions; hence the average of the squares will be larger.  suff.
From st 2  as said by Yezz, if the range is 2, this means at least one element is larger than 2, even if its 2.00000001. Thus, in the worst case scenario, where we have 8 'zeros' (i.e. very small numbers) and 2, the average of X is 2.00..01/9, while the average of Y must be larger than (2^2)/8=>4/9, hence the average of Y>X, thus suff.



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I guess that is the correct interpretation of "Y consists of ...".
If there were other elements in Y then it should have said "Y contains... ".
Thanks CaptainZee...
CaptainZeefor700 wrote: yezz 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.
From st one
the set could be as follows
{ 4 fractions with large diniminators " so when squared exponantially decrease , 1.0000001 , four elements that could be 1.001} so avearage of y not>x
and we can easily find a counter example....thus one is insuff
from two
if the range of x is 2 this means that the difference between the two extreem values( largest , smallest ) is 2.
ie x,x+2 worst case scenario is all elements = 0 and one element = 2 (but the given says +ve elements and zero is enither +ve nor ve
or all elements are fractions and keeping the range as 2 this means that we have 8 elements approaching zero while one is increasing exponatially and as the no of elements is constant thus this means that the average increase
My answer is B.......hopefully right.......am bad with statistics Haas  I believe Y is squares of X only, thus there can't be any negatives. Yezz  you might be mistaken with st 1 My answer D (can't say it was easy though) From st. 1  a square of any number larger than 1 will be larger than the number. Even if the first 4 members are fractions close to 0, we will still have one member larger than 1 more, i.e. 5 members, thus this one extra member larger than 1 will compensate any differences caused by squaring small fractions; hence the average of the squares will be larger.  suff. From st 2  as said by Yezz, if the range is 2, this means at least one element is larger than 2, even if its 2.00000001. Thus, in the worst case scenario, where we have 8 'zeros' (i.e. very small numbers) and 2, the average of X is 2.00..01/9, while the average of Y must be larger than (2^2)/8=>4/9, hence the average of Y>X, thus suff.



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Hello Captain
From st. 1  a square of any number larger than 1 will be larger than the number. Even if the first 4 members are fractions close to 0, we will still have one member larger than 1 more, i.e. 5 members, thus this one extra member larger than 1 will compensate any differences caused by squaring small fractions; hence the average of the squares will be larger.  suff.
logically and without calculations i dont think that your assumption is a must be true .
it depends on the values
if mean is equal to 1.000001 ,and
1.0000000001 ^2 = 1.00000000020000000001 the increase is 0.00000000010000000001
and all other elements are 1/2
1/2^2 =1/4 the decrease is thus 0.25 compare the increase and decrease u will find that it depends
I believe answer is b only



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(E) for me.
Stat 1:
o if X:{0,1 ; 1,00000001 ; 1,00000001} Average X > Average Y
o if X:{0,1 ; 1,00000001 ; 100000001} Average Y > Average X
INSUFF
Stat 2:
o if X:{0,1 ; 0,1 ; 0,1 ; 0,1 ; 0,1 ; 2,1} Average X > Average Y
o if X:{2 ; 3 ; 4} Average Y > Average X
INSUFF
When (1) is combined with (2), the 2 cases are still possible:
o if X:{0,9 ; 1,00000001 ; 2,9} Average Y > Average X
o if X:{0,00000000000000000000000000001 ; 0,000000000000000001 ;0,000000000000000001 ; 1,00000000000000000000000000001 ; 1,00000000000000000000000000001 ; 1,00000000000000000000000000001 ; 2,00000000000000000000000000001} Average X > Average Y
All depend on the average of X:
o > 1 : Average Y > Average X
o < 1 : Average X > Average Y
This is due to the property of f(x)=x^2



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Clue: What is the smallest possible value of x^2x ?



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kevincan wrote: Clue: What is the smallest possible value of x^2x ?
For this one, we can search f'(x) = 0 as the constant in front of x^2 is positive.
f'(x)= 2*x1
=> 2*x1=0
<=> x=1/2
The smallest value is f(1/2) = (1/2)^2  1/2 = 1/4
Other way is to observe the roots : x=0 and x=1. We take the mean of the roots : (0+1)/2 = 1/2. Thus, the smallest value is obtained by x=1/2 and so f(1/2) = 1/4

Yes, u are right Kevican... To find the answer, we need to observe the behaviour of the set of X when x are choosen around x=1/2, the minimum value. Actually, it's the farest point between the 2 graph of h(x) = x and g(x) = x^2 when 0<x<1.
It's not simply the average of X < 1.



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Fig my man i know ur Gday is very soon ...but i will appreciate if u can give me some time and elaborate on this function thing and behavior
Because i ve never heard of it except now.
SLO IF U CAN REFER A LINK OR SOMETHING WILL BE VERY HELPFUL
THANKS



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Re: DS: Comparing Means [#permalink]
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10 Oct 2006, 21:21
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I will take a shot at this. I think the answer is B.
First of all, I think for 0<X<1, the equation YX = x^2x will have the minimum at 0.5. If you draw the graph for x^2x, it is a symmetrical curve with the center and minimum at 0.25.
(1) The median of X is greater than 1.
For simplicity, let's take the median of X, i.e X5 to be 1.
Worst case X= {0.5,0.5,0.5,0.5,1,1,1,1,1)
Y = {0.25,0.25,0.25,0.25,1,1,1,1,1)
Thus, average of X > Average of Y
What if X={1,1,1,1,1.01,1.01,1.01,1.01,1.01}
Y = {1,1,1,1,1.02,1.02,1.02,1.02,1.02}
Average Y > Average X.
So 1 is not sufficient.
(2) The range of X is 2
Here again taking the worst case,
X= {0.5,0.5,0.5,0.5,0.5,0.5,0.5,0.5,2.5}
Y = {0.25,0.25,0.25,0.25,0.25,0.25,0.25,0.25,6.25}
Average Y > Average X
Or at the minimum 9th element of Y can be something like 2.0000001
X={0.000001,0.5,0.5,0.5,0.5,0.5,0.5,0.5,2.000001}
Y={0.00000000001,0.25,0.25,0.25,0.25,0.25,0.25,0.25,~4}
The first term can be ignored, and Average Y > Average X
The difference between Y9 and X9 will be at least 2 (2^22).
In order for average of X to be greater than that of Y, the sum of the difference of the first 8 terms of X and Y will need to be at least 2.
Now, any nth term in X can be greater than any nth term in Y only if X is between 0 and 1 and X can be greater than Y by at most 0.25 (if my x^2x graph is correct). As long as the 9th term in 9>2, even if all the first 8 values in X are 0.5, average of Y will still be greater than X.
So, I think statement 2 alone is enough.
What is the OA?



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yezz wrote: Fig my man i know ur Gday is very soon ...but i will appreciate if u can give me some time and elaborate on this function thing and behavior
Because i ve never heard of it except now.
SLO IF U CAN REFER A LINK OR SOMETHING WILL BE VERY HELPFUL
THANKS
Aah...at first I did not understand what Fig meant by f'(x). Now it struck me he is doing differentiation there.
From high school maths, you need to differentiate a function and equate it to 0, in order to find the MIN or MAX of a function.
so, differentiating x^2x , f'(x) = 2x1.
Equating f'(x) to 0, x=1/2.
Now whether x=1/2 is MIN or MAX, we need to differentiate f'(x) again.
f''(x) = 2.
Since f''(x) > 0, x=1/2 is the MIN.



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Re: DS: Comparing Means [#permalink]
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11 Oct 2006, 10:06
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.
When you square a positive number x, you get x^2, which will be less than x if x<1. The difference y(x)= xx^2=x(x1) is a parabola which opens downward, so the maximum value of y can be found by finding y when x=1/2, the midpoint of the two xintercepts. y(1/2)=1/4
(1) Median greater than 1. If 5 elements of X are 1.00001 and the other four are 1/2, the average of the elements in Y will be less. However, if all elements in X are greater than 1, then clearly the opposite will be true. NOT SUFF
(2) If the range is 2 and all elements are positive, then there is at least one element that is greater than 2 i.e. 2+k k>0
The corresponding element in Y is (2+k)^2= 4+4k+k^2. Since k>0, this element in Y is greater than 2 units more than its counterpart in X.
If each of the 8 other elements in X were 1/2 , the sum of Y  sum of X=2+3k+k^28*1/4>0 and thus the average of Y is greater than that of X.
SUFF
OA=B



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e
Statement 1: there are equal number of terms  fractions less than 1 and numbers above 1.
Fractions could be 4 terms of 0.99 and the numbers could be 100 or more.
INSUFF
Statement 2: Range of x is 2 INSUFF
Together,
range of x = 2
It could be .99 (4 terms), 1, and 2.99(4 terms), in which case average of y >average of x
It could be .01(4 terms), 1(5 terms), in which case average of y<average of x
Still INSUFF



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Man those problems are not under timed condition .... I know i am progressing yet i have a lot to work on..........keep it up and lets all fight by each others side to defeat this Gmat.



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Re: DS: Comparing Means [#permalink]
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16 May 2011, 01:49
a x=(0.5,1.1,1.1) y = (0.25,1.21,1.21) avg Y< avg X for other integral positive values of X and Y avg Y > avg X NS. b means avg if Y > X always as greatest element will be 2 or greater than 2. B.
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Re: Set X consists of 9 positive elements. Set Y consists of the [#permalink]
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21 May 2014, 06:27
Could someone please elaborate more on Statement 2 about the range? Thanks Cheers J




Re: Set X consists of 9 positive elements. Set Y consists of the
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21 May 2014, 06:27



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