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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]  10 Mar 2009, 21:30
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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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Need OE;I have the OA
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Last edited by nitya34 on 11 Mar 2009, 03:13, edited 1 time in total.
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Kudos [?]: 1298 [2] , given: 346

Re: DS-Set Q--good one [#permalink]  10 Mar 2009, 23:40
2
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B

1. First of all, let's consider two functions: z=x and z=x^2 at positive x. The function coincides at x=0 and x=1. In the range of (0,1) x > x^2 but in the range of (1,+inf) x<x^2. What relation does it have to our problem? if all numbers lies within 0..1, the average of set X will be greater than that of set Y. And vice-versa, if all numbers lies within 1..+inf, the average of set Y will be greater than that of set X.

2. Let's consider first condition. We can construct two examples:
a) X = {5,5,5,5,5,5,5,5,5} - av(Y)=25>av(X)=5
b) X = {0.5, 0.5, 0.5, 0.5, 1.00001, 1.00001, 1.00001, 1.00001, 1.00001} - av(Y)=6/9<av(X)=7/9
insufficient.

3. Let's consider second condition and try to construct two examples:
a) X = {0.000001,2,2,2,2,2,2,2,2} - av(Y)=32/9>av(X)=16/9
b) Here we have a problem. In order to fit condition we should have 0 and 2 as members of X. (other numbers cannot help us to build example for which av(Y) < av(X)). So, we have X = {0,?????,2}. Now, we need to compensate difference between 2^2 and 2 by adding 0.5 --->
---> x e {0.000001, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 0.5, 2}. But even such set cannot make av(X) greater than av(Y): av(Y)=5.75/9>av(X)=5.5/9
sufficient.

P.S. You may ask me why I use 0.5 but not 0.6 or 0.4 in my way to construct the example?
Let's consider a few numbers:
0: x=0; x^2=0; delta=x^2-x=0
0: x=0.4; x^2=0.16; delta=x^2-x=-0.24
0: x=0.5; x^2=0.25; delta=x^2-x=-0.25
0: x=0.6; x^2=0.36; delta=x^2-x=-0.24
0: x=1; x^2=1; delta=x^2-x=0
So, x=0.5 is the number for which difference between x^2 and x is the least (-0.25).
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Re: DS-Set Q--good one [#permalink]  10 Mar 2009, 23:45
I think the correct answer is A.
As the median is greater than 1, 5 out of 9 numbers (including the meadian) are greater then 1 and therefore have bigger squares. It will overweight 4 numbers that are below 1.
What is the OA?
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Re: DS-Set Q--good one [#permalink]  10 Mar 2009, 23:48
okay, I though there shud be an easier way to attack this problem. Now I see...
So wot is the OA???
Director
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Re: DS-Set Q--good one [#permalink]  11 Mar 2009, 03:12
+1 to awesome Walker!!
OA-B
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Kudos [?]: 1298 [0], given: 346

Re: DS-Set Q--good one [#permalink]  11 Mar 2009, 03:48
You know, it is a good q50+ problem
+1 and post more such problems here
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Re: DS-Set Q--good one   [#permalink] 11 Mar 2009, 03:48
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# Set X consists of 9 positive elements. Set Y consists of the

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