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

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

Thanks
Cheers
J

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 [given all elements are positive]
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
= \([9*(4+4*k+k^2) ]- [9*1/4]\) > 0 as k>0
and thus the (avg)Y>avg(X). Sufficient in all cases. So,B.

Please press kudos if it helped.

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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

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

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