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

Question

Is x greater than 1?

(1) 1/x >−1

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

timur720 · Answer

I go for E. Both statements are not sufficient to answer the question

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cyberjadugar · Answer

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 1?" as NO Hence, answer is (B) Regards,

Val1986 · Answer

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 1?" as N

Bunuel · Answer

Question is x>1 ? (1) \frac{1}{x}>- 1 --> \frac{1+x}{x}>0 , two cases: A. x>0 and 1+x>0 , x>-1 --> x>0 ; B. x<0 and 1+x<0 , x<-1 --> x<-1 . We got that given inequality holds true in two ranges: x>0 and x<-1 , thus x may or may not be greater than one. Not sufficient. (2) \frac{1}{x^5}> \frac{1}{x^3} --> \frac{1-x^2}{x^5}>0 , two cases: A. x>0 (it's the same as x^5>0 ) and 1-x^2>0 , -1 01 --> x<-1 ; We got that given inequality holds true in two ranges: 0

Val1986 · Answer

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

rggoel9 · Answer

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

Bunuel · Answer

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.

pavanpuneet · Answer

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.

Bunuel · Answer

That's not correct. \frac{1}{x^5}> \frac{1}{x^3} holds true for 0x2-4x-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.

Catcat · Answer

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)

Bunuel · Answer

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.

Catcat · Answer

Got it! Thank you for the clarification

Judy1389 · Answer

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.

chetan2u · Answer

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...

saleck · Answer

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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chetan2u · Answer

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..............

saleck · Answer

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.

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MathRevolution · Answer

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.

pk72 · Answer

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?

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

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