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

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

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

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

Thanks
Cheers
J

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.

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

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.