A set has 4 numbers x^3, x^2, x/2,and x/3 .

Question

A set has 4 numbers :  x^3 ,  x^2 ,  \frac{x}{2} , and  \frac{x}{3} . Is the range of the set " x^3-\frac{x}{3} "?

(1)  \frac{x}{y^2}  <0, where y ≠0
(2)  x^2> x^5

SaquibHGMATWhiz · Answer

Solution:

Pre Analysis:

We are given a set of 4 numbers: \(x^3, x^2, \frac{x}{2}\) and \(\frac{x}{3}\)
We are asked if the range of the above set is \(x^3-\frac{x}{3}\) or not
If the range of the set is \(x^3-\frac{x}{3}\), then the biggest number of the set has to be \(x^3\) and the smallest has to be \(\frac{x}{3}\)

Statement 1: \(\frac{x}{y^2}<0\), where \(y≠0\)

Since \(y≠0\), we can say \(y^2>0\)
Thus, from this statement we can infer that \(x<0\) which means \(x^2>x^3\) and \(x^3\) cannot be the biggest number of the set
We can be sure that the range of the set cannot be \(x^3-\frac{x}{3}\)
Thus, statement 1 alone is sufficient and we can eliminate options B, C and E

Statement 2: \(x^2>x^5\)

\(x^2>x^5\) is valid in the region 2 (\(0<x<1\)), 3 (\(-1<x<0\)) and 4 (\(x<-1\)) of the number line
Or we can say \(x<0\) or \(0<x<1\)
In both the regions above, \(x^2>x^3\) and \(x^3\) cannot be the biggest number of the set
Thus, statement 2 alone is also sufficient

Hence the right answer is Option D

Note: To know more about regions of a number line and respective ranges, please visit here

gmatophobia · Answer

We want to know if the range is x^3-\frac{x}{3} . Now x can be either positive or negative. Case 1 When x is positive, to satisfy the range x^3-\frac{x}{3} , the lowest value in the set should be \frac{x}{3} and the highest value of the set should be x^3 . Hence the set when arranged in ascending order is as follows - \frac{x}{3} __ __ x^3 Case 2 When x is negative, the largest term in the sequence is x^2 , hence the range can NEVER be x^3-\frac{x}{3} Statement 1 \frac{x}{y^2} < 0 We know that y^2 is positive (actually non - negative, but given that y cannot be 0, y^2 is positive) So x is negative. This falls under case 2. We know when x is negative, the highest term in the sequence is x^2 . Hence the range cannot be x^3-\frac{x}{3} . This statement is sufficient and we can eliminate B C and E. Statement 2 x^2 > x^5 x^2 - x^5 > 0 x^2 (1- x^3 ) > 0 We know that x^2 is positive in this case. So 1 - x^3 > 0 - x^3 > -1 x^3 < 1 Thus x < 1 Now there are two regions possible 1) 0 < x < 1 2) x < -1 In both these regions x^3 is not the greatest term. Hence we can conclude that the range will not be x^3-\frac{x}{3} IMO Option D

chetan2u · Answer

x^3 , x^2 , \frac{x}{2} , and \frac{x}{3} Three ranges a) x<0 x^2 is the largest. So, range has to include x^2 . Answer will be NO. b) 0x^3 and \frac{x}{2}>\frac{x}{3} , so surely the largest would be x^2 or \frac{x}{2} Answer will be NO. c) x\geq 1 x^3>x^2>\frac{x}{2}>\frac{x}{3} Answer will be YES. Another point that can be added is x=0, all values will be 0 and range will be 0. Answer will be YES (1) \frac{x}{y^2} <0, where y ≠0 As y^2>0 , x will be negative. Situation (a), the answer will be NO. Sufficient (2) x^2> x^5 x^2-x^5>0…….x^2(1-x^3)>0 As x^2>0 , we can say that 1-x^3>0 or x<1 . Hence, situation (a) and (b) as x is not 0. Each case gives a NO. Sufficient D

A set has 4 numbers x^3, x^2, x/2,and x/3 .

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GMAT Data Sufficiency (DS) Questions

A set has 4 numbers x^3, x^2, x/2,and x/3 .

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