Is x 0 ? (1) x

Question

Is x > 0 ?

(1) x < x^2
(2) x < x^3

Can some one explain how do we eliminate choice "C" for this DS question using plug in values

thanks

Bunuel · Accepted Answer

On DS questions when plugging numbers, goal is to prove that the statement is not sufficient . So we should try to get a YES answer with one chosen number(s) and a NO with another. Question: is x>0 ? When considering statements together we can see that YES answer is easy to get: x=2>0 , so answer YES --> (x=2)<(x^2=4)<(x^3=8) ; Now, for NO answer we should pick negative x --> ANY negative x would be less than x^2 as x^2 would be positive, so satisfying (1) is easy. Next, in order negative x to be less than x^3 , it shoul be a fraction in the range (-1,0) --> x=-\frac{1}{2}<0 , so answer NO --> (x=-\frac{1}{2})<(x^3=-\frac{1}{8})<(x^2=\frac{1}{4}) . We have two different answers, not sufficient. Answer: E.

onedayill · Answer

Can't I write first equation

x < x2 as 1 < x.

Orange08 · Answer

Can I deduce as following : 1) x < x^2 holds true for x > 1 and x < 0 Thus, not sufficient 2) x < x^3 holds true for x > 0 and 0 < x < 1 Not sufficient No unique value is determined when combining 1 & 2. Thus, E. Please correct me if I am incorrect in the above logic.

jallenmorris · Answer

You are correct.

Statement 1: Positive fractions will never make this statement true. The square of a positive fraction is always smaller than the original fraction itself.

Statement 2: Only positive numbers greater than 1 make this statement true. Therefore, by combining them, there is only one possibility to make both true: Numebrs greater than 1, which allows us to answer the original question.

Bunuel · Answer

Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know the sign of it or are not certain that variable (or expression with variable) doesn't equal to zero. So you can not divide both parts of inequality x0 you should write 1x . Hope it helps.

Bunuel · Answer

No your ranges for (2) are not correct, it should be: -11 . Negative fractions from the range (-1,0) also make statement (2) true. Below is complete solution: Is x>0 ? (1) x x*(x-1)>0 --> x<0 or x>1 --> --------- 0----1 ---- . Not sufficient as if x is in the green zone answer is YES but if x is in the red zone answer is NO. (2) x x(x-1)(x+1)>0 --> -11 --> ----(-1) ---- (0)----(1) ---- . Not sufficient as if x is in the green zone answer is YES but if x is in the red zone answer is NO. (1)+(2) Intersection of the ranges from (1) and (2) is: ----(-1) ---- (0)----(1) ---- Again not sufficient as if x is in the green zone answer is YES but if x is in the red zone answer is NO. Answer: E. Hope it helps.

jallenmorris · Answer

You're right. I hadn't thought of negative fractions. -1/2 ^3 = -1/8, which is greater than -1/2. Thanks.

kheba · Answer

No your ranges for (2) are not correct, it should be: -11 . Negative fractions from the range (-1,0) also make statement (2) true. Below is complete solution: Is x>0 ? (1) x x*(1-x)>0 --> x<0 or x>1 --> --------- 0----1 ---- . Not sufficient as if x is in the green zone answer is YES but if x is in the red zone answer is NO. (2) x x(x-1)(x+1)>0 --> -11 --> ----(-1) ---- (0)----(1) ---- . Not sufficient as if x is in the green zone answer is YES but if x is in the red zone answer is NO. (1)+(2) Intersection of the ranges from (1) and (2) is: ----(-1) ---- (0)----(1) ---- Again not sufficient as if x is in the green zone answer is YES but if x is in the red zone answer is NO. Answer: E. Hope it helps. Hi Bunuel, Could you please explain why you're considering only the negative factors while considering x x*(1-x)>0 --> x<0 or x>1 .. From x*(1-x)>0 can't we consider both x and (x-1) to be positive also ? Similarly for x(x-1)(x+1)>0 --> -11 , I seem to be missing your logic. Thanks

Bunuel · Answer

Actually, both cases were considered. x x*(x-1)>0 x>0 and x-1>0 (x>1) --> x>1; x<0 and x-1<0 (x<1) --> x<0. Thus, x1 . Solving inequalities: x2-4x-94661.html#p731476 inequalities-trick-91482.html everything-is-less-than-zero-108884.html xy-plane-71492.html Hope it helps.

Archit143 · Answer

Hi Bunnuel

With ref to your above post i would like to contradictory expalnation from DS section of OG 13 question no 52 
TO make things easier i ll post the question

If X and Y are positive , Is X<10<Y?
1. X<Y and XY = 100
2. X^2 <100< Y^2

In the explanation part 
2nd stem is directly reduced to X<10<Y

Isnt your explanation contradictory to above expalantion of OG
Pls help in getting my concept right.

Bunuel · Answer

First of all, they are not reducing, they are taking square root. Also, notice that the stem says that x and y are positive numbers. P.S. This question is discussed here: if-x-and-y-are-positive-is-x-10-y-139873.html

Archit143 · Answer

Thanks BUnnel 
really missed the statement.Sorry

But is it ok to take square root,considering signs are unknown in above scenario.

Bunuel · Answer

If we were not told that x and y are positive numbers, then from x^2<100 we can square both sides and write: 2^2<4^2 ; 0\leq{x}<{y} --> we can square both sides and write: x^2-2 if we square we'll get 1>4 which is not right. So if given that x>y then we can not square both sides and write x^2>y^2 if we are not certain that both x and y are non-negative. B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality). For example: -2<-1 --> we can raise both sides to third power and write: -2^3=-8<-1=-1^3 or -5<1 --> -5^2=-125<1=1^3 ; x we can raise both sides to third power and write: x^3

mbaiseasy · Answer

1. x < x^2
Test x = 2: 2 < 4 OK!
Test x = -1: -1 < 1 OK!
Test x = -1/4: -1/4 < -1/16 OK!
INSUFFICIENT!

2. x < x^3
Test x = 2: 2 < 8 OK!
Test x = -1: -1 = -1 NOT OK!
Test x = -1/4: -1/4 < -1/64 OK!
INSUFFICIENT!

Together: When x is negative fraction or x is greater than 2, both inequalities work!

Answer: E

MathRevolution · Answer

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution. Is x > 0 ? (1) x < x^2 (2) x < x^3 If the range of the question covers that of the condition, the condition is sufficient There is one variable (x) in the original condition and 2 equations are given by the conditions, so there is high chance (D) will be the answer For condition 1, x x<0, 1 -1

akshayk · Answer

S1 -> x < x^2 x = -2, x^2 = 4 x = 2, x^2 = 4 Insufficient. S2 -> x < x^3 x = 2, x^3 = 8 x = - 1/2, x^3 = - 1/8 Insufficient. (1+2) x can be a negative fraction leading to a negative cube less greater than x, or a positive square greater than x. Insufficient. E.

harshittahuja · Answer

Is x > 0 ?

(1) x < x^2
(2) x < x^3

From A we can say that
x^2>x
if X>0 means
than X is integer or fraction
if X >0 and integer the it satisfy that
if X>0 and fraction then it is not satisfy as x=0.2 then x^2=0.04
so From A we cant say.

From B
If X^3>x
means same happens here as well
so we cant say that X^3>X.

from A and B
we can say that of X lies between 0 and 1 then it is not true
So Answer is E

bumpbot · Answer

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.

Is x > 0 ? (1) x < x^2 (2) x < x^3

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions