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B
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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.
(1) The median of X is greater than 1.
(2) The range of X is 2.
10 Mar 2009, 23:40
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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).
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).
General Discussion
11 Oct 2006, 10:06
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kevincan
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.
SUFF
OA=B
