Is x greater than 1? (1) 1/x >1 (2) 1/x^5 > 1/x^3

Hi,

Is x > 1?

Using option (1)
\(1/x > -1\)
or \(1/x + 1 >0\)
or \((1+x)/x > 0\)
=> \(x <-1\) or \(x >0\), Not sufficient.

Using option (2)
\(1/x^5 > 1/x^3\)
or \(1/x^5 - 1/x^3 >0\)
or \((1-x^2)/x^5 >0\)
or \((1-x)(1+x)/x^5>0\)
=> \(x <-1\) or\(0<x< 1\), and we have the answer to question "Is x > 1?" as NO

Hence, answer is (B)

Regards,

How did you factor the second statement like that?
1/x^5 > 1/x^3
or 1/x^5 - 1/x^3 >0
or (1-x^2)/x^5 >0
or (1-x)(1+x)/x^5>0
=> x <-1 or0<x< 1, and we have the answer to question "Is x > 1?" as N

Yes now I got it! Thanks! (i was simply cross multiplying for the statement 2 the first time!)

Bunuel
Sorry for asking this again, but I have been trying to do these type of sums only and not getting anywhere.

How can we have these 2 cases for an equation like (1+x)/x>0.

In the post out here it is mentioned is-x-between-0-and-1-1-x-2-is-less-than-x-2-x-3-is-104280.html#p1094206 that the signs should be different i.e. one should be > and the other lesser <. Now in this sum both the cases have the same sign. I did not get this.

The other thing also which I would like to ask is once we have an equation like this (1+x)/x >0 can we automatically write the 2 cases as (1+x) > 0 and x>0 and the other in opposite sign. Is this by rule.

It would be helpful if you could point me to some reading material on these type of sums where we get 2 cases for the root of the equation.

Thanks in advance.

Rahul Goel

It's very basic:
ab>0 means that a and b must have the same sign;
ab<0 means that a and b must have the opposite signs.

Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863
xy-plane-71492.html?hilit=solving%20quadratic#p841486

Hope it helps.

By looking at condition 2, I think this condition is only feasible if x<0, for x>0 it will never be true, for example, x=2; thus from condition 2, we get x<0, thus the question whether x>1 is answered .. NO.

That's not correct.

\(\frac{1}{x^5}> \frac{1}{x^3}\) holds true for \(0<x<1\) and \(x<-1\).

Check this:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
everything-is-less-than-zero-108884.html?hilit=extreme#p868863
xy-plane-71492.html?hilit=solving%20quadratic#p841486

Hope it helps.

Question is x>1?

(1) (1/x)>- 1 --> (1+x)/x>0, two cases: (understood)

A. x>0 and 1+x>0, x>-1 (understood) --> x>0; (how do you get to this step?)

B. x<0 and 1+x<0, x<-1 --> x<-1. (understood)

For A, we have that x>0 AND x>-1, whcih is the same as x>0 (intersection (common range) for x>0 and x>-1 is x>1).

Hope it's clear.

Got it! Thank you for the clarification

I don't understand Case B for statement 1 (why did you set x<0 when the equation sets it to greater than 0, \(\frac{1+x}{x}>0\) ?)

The same applies for Case B in statement 2.

I understand how the equation is constructed, but not why the inequality sign flips.

ALSO – can I confirm that multiplying both sides by x^2 for statement one and x^6 for statement two is possible (as regardless if x is negative, it will be positive as it has been squared?)

Thanks all, apologies if these are straightforward concepts.

Hi,
If you have both numerator and denominator, they both are interlinked to give the SIGN to the fraction...
let me tell you by a easy example...
\(\frac{3}{4} >0\)....... it could be that the denominator is -ive and therefore numerator will also be -ive, so \(\frac{-3}{-4}> 0.\)...
here we are dealing with variable so we cannot say anything about the type of integer it is...
\(\frac{1+x}{x}>0\) .....
so first case is when both numerator and denominator are positive : +/+ = +>0...
second is when both are negative -/- = + >0...

Let's consider the following scenarios :
X is positive then 1/x will always be greater than -1. (any positive number is greater than anything negative number)
X is a negative number whose absolute value is greater than 1 also satisfies (1). ( if X = -2 then 1/X = -0.5 which is greater than -1)

(2) 1/(x^5) > 1/(x^3)

Consider this :
X is and number greater than 1 then X^5 is greater than X^3, therefore, 1/(X^5) is less than 1/(X^3).
X is a positive number less than 1, then X^5 is less than X^3, therefore 1/(X^5) is greater than 1/(X^3)
X is a negative number less -1 then X^5 is less than X^3, therefore, 1/(X^5) is greater than 1/(X^3).
As you can see, the only values that satisfy (2) are the ones in which X<1, therefore (2) is sufficient. We know with absolute certainty that X IS NOT GREAT THAN 1.

answer is B.


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Hi,

yes you are correct...
Both ways it will amount to same..

1+x/x>0.....
same x>0, 1+x>0 OR x<0, 1+x<0..............

multiply by x^2....
(1+x)*x>0...
same cases will come up
same x>0, 1+x>0 OR x<0, 1+x<0..............

This question requires logical reasoning instead of mathematical manipulation.

Statement 1: 1/x > -1. Let's multiply both sides by positive number, say x^2, then we get x>-(x^2). This is true for all positive numbers, since (-(x^2)) is negative. It's also true for all negative numbers whose absolute value is less than the absolute value of its square. Statement 1, therefore, is insufficient.

Statement 2: let's multiply both sides by x^4, we get 1/x > x, if x>1, then 1/x must be less than 1, but that's impossible because x>1. So, the only possible values for x are less than 1. Statement 2 is sufficient.

Sent from my Z988 using GMAT Club Forum mobile app

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 1 variable and 0 equations, D is most likely to be the answer. So, we should consider each of the conditions on their own first.

Condition 1)
\(1/x > -1\)
\(⇔ x > - x^2\) by multiplying by \(x^2\)
\(⇔ x^2 + x > 0\)
\(⇔ x( x + 1 ) > 0\)
\(⇔ x < -1\) or \(x > 0\)

In inequality questions, the law “Question is King” tells us that if the solution set of the question includes the solution set of the condition, then the condition is sufficient
Since the solution set of the question does not include that of condition 1), it is not sufficient.

Condition 2)
\(1/x^5 > 1/x^3\)
\(⇔ 1 > x^2\)
\(⇔ x^2 - 1 < 0\)
\(⇔ (x+1)(x-1) < 0\)
\(⇔ -1 < x < 1\)
The answer is "no".
Since ‘no’ is also a unique answer by CMT (Common Mistake Type) 1, the condition 2) is sufficient.

Therefore, B is the answer.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.

Hello Bunuel,

Thanks for the explanation !!!

When I solved the problem , I used the following way:

Statement 1: 1/x>-1
Multiplying x both the side we get 1>-x;
Multiplying by -1, we get x>-1.
But, this result is inconclusive because x can take value as 0 and 1 as well.Hence, insufficient.

Statement 2:1/x^5>1/x^3
multiplying by x^5 both the side, we get 1>x^2
which can be written as x^2<1.
Since, x^2 is less than 1 so x will be less than 1. Hence, Sufficient.

Is my approach correct?