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Set X consists of 9 positive elements. Set Y consists of the [#permalink]
19 Sep 2006, 12:49

5

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

22% (02:32) correct
78% (01:39) wrong based on 252 sessions

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.

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.

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.

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, 15:39, edited 1 time in total.

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.

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.

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

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

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

For this one, we can search f'(x) = 0 as the constant in front of x^2 is positive.

f'(x)= 2*x-1
=> 2*x-1=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.

Re: DS: Comparing Means [#permalink]
10 Oct 2006, 20:21

I will take a shot at this. I think the answer is B.

First of all, I think for 0<X<1, the equation Y-X = x^2-x will have the minimum at 0.5. If you draw the graph for x^2-x, 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 (2^2-2).
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^2-x 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?

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^2-x , f'(x) = 2x-1.
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.

Re: DS: Comparing Means [#permalink]
11 Oct 2006, 09:06

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)= x-x^2=x(x-1) 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 x-intercepts. 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^2-8*1/4>0 and thus the average of Y is greater than that of X.

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

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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