Is x 0?

Question

Is x > 0?

(1) x^3 < x^2

(2) x^3 < x^4

twixt · Answer

I would answer E, X^2 or X^4 are always positive whatever X sign

itsprem · Answer

ans D.
- (x^3/x^2)>1(diving by x^2 on both sides,so equality changes)
similiarly (x^4/x^3)>1..
from either of them it can e inferred if x>0.
am i correct?what is the OA?

Please correct me if i'm wrong.

-P
:?:

Epiphany · Answer

The answer is C Tells us 0 < x < 1 in which case x > 0 Or 0 > x Insuff Tells us x > 1 in which case x > 0 Or 0 > x Insuff Only 0 > x satisfies both statements so x is NOT greater than 0

twixt · Answer

You are right epiphany i underestimated this one, though I am quite sure I already solved it.

Answer is C

ComplexVision · Answer

Epiphany is absolutely right:

1. x<1
2. x>1 OR x<0

Neither one is sufficient by itself, but if we take the intersection, we can see that x<0

Nice one... Thx afife76

gayathri · Answer

Prem the inequality does not change when you multiply or divide by a positive number. It only changes when you multiply or divide by a negative number or when you take an inverse. In this case x^2 will always be positive so the direction of inequality does not change.

MA · Answer

OA should be C.

from i, if x=-5, x^3<x^2 holds.
if x=0.5, x^3<x^2 also holds.

from ii, if x=-5, x^3<x^4 holds.
if x=5, x^3<x^4 also holds.

from i and ii, x=-ve.

krisrini · Answer

My answer is B. From 1 x^3/X^2 < 1 or X < 1. This implies that X could be greater than or equal to zero or X could be negative. Hence statement 1 is not sufficient. From 2 X^3/X^4 < 1 => 1/x < 1 ==> 1 < X or X > 1. If x > 1 then X should be greater than zero. Hence Statement 2 is alone sufficient to answer the question. Please let me know if I am right. Thanks

krisrini · Answer

MA, From statement 2, we have 

1/x < 1 , then 1 < X

Or simplifying statement 2 we get X > 1 . So how can x be -ve. Am I making a mistake somewhere, please let me know.

Thanks

MA · Answer

ii. X^3 < X^4. 
if x = +ve, there is no problem.
but x can be negative. suppose x = -5, the inequality X^3 < X^4 holds but x is not more than 0.

ashkg · Answer

C

A. x^3 < x^2

=> x^2(x-1) <0
=> x-1 < 0 since x^2 is always +ve
=> x< 1

x could be -5 or 0.5 so A is insufficient

B. x^4 > x^3
Simplifying we get
x(x-1) < 0 since x^2 is always +ve

this means either 0<x<1 or x>1 and x<0

So B is insufficient.

C.

using both 0<x<1

because x>1 and x<1 cannot happen at the same time

C is the answer

ashkg · Answer

I had to really think hard why you weren't getting it right. Simplifying you got 1/x < 1 Now remember x could be +ve or -ve. The inequation will change signs if x is -ve. if x>0 1/x < 1 => 10 and x>1 if x<0 1/x <1 => 1>x so finally x<0 and x < 1 Looks like i screwed up.....its not the same as what i got earlier :shock:

chunjuwu · Answer

x = 1/2 , -2 , -1/2, 2

x^2= 1/4, 4, 1/4, 4

x^3 = 1/8, -8, -1/8, 8

x^4 = 1/16, 16, 1/16, 16


(1)when x = 1/2 or -2, x^3 < x^2, insufficient

(2) when x=-2 or 2, x^3 < x^4, insufficient

when x=-2 or -1/2, x^3 < x^2 and x^3 < x^4 at the same time

C is sufficient.

sonaketu · Answer

I'll pick B i) X^3 < X^2 => X < 1 doesn't say X<0 or not ... insufficient ii) X^4 > X^3 => X > 1 => X>0 ... sufficient What's the OA?

ashkg · Answer

In B put x=-5......and the inequality still holds true.

HongHu · Answer

Remember, if you can't determine the sign,  you CAN'T multiply or divide both side of an equation or ineqaulity by an algebra expression .


1)X^3 < X^2
x^3-x^2<0
x^2*(x-1)<0
Since x^2 can't be less than zero we get x<1. Not sufficient to determine whether x<0.

2)X^3 < X^4
x^3-x^4<0
x^3*(1-x)<0
x^3<0 or 1-x<0
x<0 or x>1
Not sufficient to determine whether x<0.

Combined, we know that only when x<0 both inequalities are satisfied. So sufficient.

krisrini · Answer

Thanks ashkg and MA, I correct my mistake. Thanks much for pointing it out. I loved the solution from Honghu, her solution was very professional. Sorry for the late replies, I was on a vacation.

Bunuel · Answer

Theory on Inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html

All DS Inequalities Problems to practice: search.php?search_id=tag&tag_id=184
All PS Inequalities Problems to practice: search.php?search_id=tag&tag_id=189

700+ Inequalities problems: inequality-and-absolute-value-questions-from-my-collection-86939.html

klorofill29 · Answer

Statement 1 tells us that:
 a. x is negative (or)
 b. 0<x<1

Statement 2 tells us that:
 a. x is negative (or)
 b. x is positive AND x>1

Combining the 2 statements, x is negative. Therefore, x < 0, which is a definite answer to the problem.

Answer is C.

Is x > 0?

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