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

Question

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.

walker · Accepted Answer

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).

kevincan · Answer

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

jlgdr · Answer

Could someone please elaborate more on Statement 2 about the range? Thanks Cheers J

gmatacequants · Answer

Hi J

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?

Statement 2: The range of X is 2.

If the range is 2 and all elements are positive in X,  range = Maximum element value - Minimum element value = 2 
---> Maximum element value = 2+ Minimum element value = 2 + k , as k >0  
The corresponding element in Y is  (2+k)^2= 4+4*k+k^2  --------------(i)

Also,
The difference f(x)=  x^2 - x  = x (x-1) is a parabola which opens upward, 
so the minimum value of f(x) when x=1/2, the midpoint of the two x-intercepts. 
f(1/2)=1/4 which can also be derived using differentiation eq : 1-2*x = 0 ----------> x= 1/2

Now,taking two cases:
a)when all elements in X > 1
In that case: each elements in Y > each elements in X so, the avg(Y)> avg(X)

b)When at least on element in X < 1
Looking at the boundary case:
If each of the 9 elements in X is 1/2, then this case should give the minimum value of  x^2 - x  as depicted by parabola
then, Sum of Y - Sum of X
=   -    > 0 as k>0
and thus the (avg)Y>avg(X). Sufficient in all cases. So,B.

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rajatjain14 · Answer

Hi Pretzel,

Only when element x lies between 0 & 1,  x^2 < x

Differentiation is not at all required  . In this question, It just helps you to find the maximum/minimum difference between  x  and  x^2  when  x  lies between 0 and 1.

Rgds,
Rajat

kham71 · Answer

Q: Is average of 9 positive numbers smaller than the average of the squares of these numbers? ==> Is the sum of 9 numbers smaller than the sum of their squares? ==> For the sum of squares of the 9 numbers, is the decrease in the value of the sum incurred by squaring all numbers in the set that are smaller 1, less than or greater than the increase in the value of the sum obtained by squaring all numbers in the set that are greater than 1?

Statement 1: Median of the numbers is 1.

Possibility # 1: {.5, .5, .5, .5, 1, 1.2, 1.2, 1.2, 1.2} Each .5 squared(.25) decrease the new sum by .25(.5 - .25 = .25), while each 1.2 squared(1.44) increases the sum by only .24 (1.44 - 1.2 = .24) and 1, the median doesn't affect the sum. Sum of squares in this case is less than sum of original set.

Possibility # 1: {.5, .5, .5, .5, 1, 2, 2, 2, 2} Each .5 squared decrease the sum by .25, while each 2 squared increases the sum by 2 (4 - 2 =2) and 1, the median, doesn't affect the sum. Sum of squares is greater than sum of original digits.

INSUFF

Statement 2: Range is 2: Which also means that the smallest possible value for for the maximum number in the set of 9 positive numbers is just larger than two.{.0000001,...,....,....,...,....,...,...,...., 2.0000001}

Note that the maximum absolute reduction in the value of a fraction when the fraction is squared is for the fraction 1/2 (.5 ==> .25, difference is .25). Any larger or smaller fraction will yield a smaller absolute decrease in value from the original number when squared (.4 ==> .16, difference is .24; .6 ==> .36, difference is .24).

The greatest decrease in the sum of squares for this set would be less than the decrease that occurs if 8 of the 9 numbers are .5 (-.25*8 = -2). Because the set has to include the number 2.0000000001, which when squared increases the sum by just more than 2, the overall effect is an increase in the sum.

Loser94 · Answer

let the X be set of all fraction

Now fro statement :1) The median of X is greater than 1.
Lets assume set in which no are near to 1 on right side and near to zero on left side .0000001 to 1.0000001 in this senario average of X will be more than that of Y set since fraction will become smaller after squaring . and if no near 1 on right side become near to 2 and left side will become near to 1 (0.9999999 to 1.99999) then the average of x will be less than average of Y

(2) The range of X is 2.
Since this legit answer
As it gives us range no can be fraction or integer we will always get our average of Y greater than X

Answer : B
Thanks , Hit for kuddos it helps

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

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