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If \(x>0\) then \(|x|=x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (x - 1) > 0\) --> \(x^2-1>0\) --> \(x^2>1\) --> \(x<-1\) or \(x>1\). Since we consider range when \(x>0\) then we have \(x>1\) for this case;

If \(x\leq{0}\) then \(|x|=-x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (-x - 1) > 0\) --> \(-(x+1) (x+1) > 0\) --> \(-(x+1)^2>0\) --> \((x+1)^2<0\). Now, since the square of a number cannot be negative then for this range given equation has no solution.

So, we have that \((x+1) (|x| - 1) > 0\) holds true only when \(x>1\). Sufficient.

When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\).

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\).

If \(x>0\) then \(|x|=x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (x - 1) > 0\) --> \(x^2-1>0\) --> \(x^2>1\) --> \(x<-1\) or \(x>1\). Since we consider range when \(x>0\) then we have \(x>1\) for this case;

If \(x\leq{0}\) then \(|x|=-x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (-x - 1) > 0\) --> \(-(x+1) (x+1) > 0\) --> \(-(x+1)^2>0\) --> \((x+1)^2<0\). Now, since the square of a number cannot be negative then for this range given equation has no solution.

So, we have that \((x+1) (|x| - 1) > 0\) holds true only when \(x>1\). Sufficient.

(2) |x| < 5 --> \(-5<x<5\). Not sufficient.

Answer: A.

Hope it's clear.

Dear Bunuel, i got (x+1)^2<0 . and further solved to x+1<0 giving x <-1 and ended up in wrong answer E. square of a number cannot be negative. but just confused for (x+1)^2 where x is a variable?

If \(x>0\) then \(|x|=x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (x - 1) > 0\) --> \(x^2-1>0\) --> \(x^2>1\) --> \(x<-1\) or \(x>1\). Since we consider range when \(x>0\) then we have \(x>1\) for this case;

If \(x\leq{0}\) then \(|x|=-x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (-x - 1) > 0\) --> \(-(x+1) (x+1) > 0\) --> \(-(x+1)^2>0\) --> \((x+1)^2<0\). Now, since the square of a number cannot be negative then for this range given equation has no solution.

So, we have that \((x+1) (|x| - 1) > 0\) holds true only when \(x>1\). Sufficient.

(2) |x| < 5 --> \(-5<x<5\). Not sufficient.

Answer: A.

Hope it's clear.

Dear Bunuel, i got (x+1)^2<0 . and further solved to x+1<0 giving x <-1 and ended up in wrong answer E. square of a number cannot be negative. but just confused for (x+1)^2 where x is a variable?

It doesn't matter that \(x\) is a variable, it's still some number and so is \(x+1\). So, \((x+1)^2\) is a square of that number and it cannot be negative.

Also your way of solving is not correct: \((x+1)^2<0\) does not mean \(x+1<0\) it means that \(|x+1|<0\). From that you could deduce the same: since absolute value cannot be negative then this equation has no solution.

Hi Bunuel, I understood your solution but can you please tell me where am I going wrong?

When I see the expression, (x+1)(|x|-1)>0, I immediately think that these two brackets must be either positive or negative.

Hence, if that take both of them are positive, then x>-1 and and x>1 & x<-1

For these two ranges we realize when we start plugging in number that only for x>1 the equation hold true.--------(1)

Similarly, if we take both of them to be negative, then x<-1 and x<1 & x>-1

for x<-1 for example, -2, one of the brackets turn out to be positive, which does not satisfy our initial assumption that both must be negative.

for x<1 ex 0, one bracket is +ve and one is -ve <Not Desired>; for 0, the inequality is not > 0 <not desired>

if we go still less it follows the first case

for x>-1 pretty much follows like the above. Hence nothing desired.--------(2)

Hence, x>1

A is sufficient.

First of all I wouldn't recommend to solve this question the way you are doing.

Next, when you consider both multiples to be negative and get \(x<-1\) from the first one, then the second multiple automatically transformes to \((-x-1)\), since if \(x<-1<0\) then \(|x|=-x\). So, we have that \(-x-1<0\) must also be true or \(x>-1\), which contradicts the case for the first multiple (\(x<-1\)). So, both \(x+1\) and \(|x|-1\) can not be negative.

Also I think you got x<1 & x>-1 from |x|<1, and if yes, then it's not correct: \(|x|<1\) means that \(-1<x<1\). So, again \(x<-1\) (for the first multiple to be negative) and \(-1<x<1\) (for the second multiple to be negative) cannot simultaneously be true.

Statement 1: For (x+1)(lxl-1) > 0, we should have either (x+1)>0 and (lxl-1) > 0 or (x+1)<0 and (lxl-1) < 0 when (x+1)>0 and (lxl-1) > 0 (x+1)>0 => x>-1 (lxl-1) > 0 => x>1 or x <-1 From above two, possible solution is x>1 when (x+1)>0 and (lxl-1) < 0 (x+1)<0 => x<-1 (lxl-1) < 0 => -1<x<1 Both of these can not be satisfied by any value of x. Hence we get only 1 solution, x>1. which is what we wanted to ascertain. Sufficient.

Statement 2: |x| <5 => -5<x<5 Clearly not sufficient to tell whether x>1 or not.

so you just have to find the common region if its done on a number line and we ignore one of the cases, since there isn't a common region? _________________

Click +1 Kudos if my post helped...

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Last edited by fozzzy on 28 Jan 2013, 00:25, edited 1 time in total.

If \(x>0\) then \(|x|=x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (x - 1) > 0\) --> \(x^2-1>0\) --> \(x^2>1\) --> \(x<-1\) or \(x>1\). Since we consider range when \(x>0\) then we have \(x>1\) for this case;

If \(x\leq{0}\) then \(|x|=-x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (-x - 1) > 0\) --> \(-(x+1) (x+1) > 0\) --> \(-(x+1)^2>0\) --> \((x+1)^2<0\). Now, since the square of a number cannot be negative then for this range given equation has no solution.

So, we have that \((x+1) (|x| - 1) > 0\) holds true only when \(x>1\). Sufficient.

(2) |x| < 5 --> \(-5<x<5\). Not sufficient.

Answer: A.

Hope it's clear.

Sorry, I don't understand why we should consider the range where x>0. because of the absolute value?

We need to get rid of the modulus in the expression to solve it and this is the way to do that. Check here: is-x-134652.html#p1097668

If \(x>0\) then \(|x|=x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (x - 1) > 0\) --> \(x^2-1>0\) --> \(x^2>1\) --> \(x<-1\) or \(x>1\). Since we consider range when \(x>0\) then we have \(x>1\) for this case;

If \(x\leq{0}\) then \(|x|=-x\) so \((x+1) (|x| - 1) > 0\) becomes \((x+1) (-x - 1) > 0\) --> \(-(x+1) (x+1) > 0\) --> \(-(x+1)^2>0\) --> \((x+1)^2<0\). Now, since the square of a number cannot be negative then for this range given equation has no solution.

So, we have that \((x+1) (|x| - 1) > 0\) holds true only when \(x>1\). Sufficient.

(2) |x| < 5 --> \(-5<x<5\). Not sufficient.

Answer: A.

Hope it's clear.

Hi,

This is my first post so was little conscious to ask my doubt. In the above question, we took the roots as 1<x>1. However, in this question as mentioned below (unable to post the link as per new member rule)

is ((X-3)^2)^1/2 = 3-X ?

1) X does not = 3

2) -X|X| > 0

the roots are 3<x>3. Can you please explain the difference?

When I try this question for statement 1: (x+1) (|x| - 1) > 0 (x+1)|x|-(x+1)>0 (x+1)|x| >(x+1) |x| > 1

x>1 ; x<-1

Never multiply (or reduce) an inequality by variable (or by an expression with variable) if you don't know its sign.

So you cannot reduce both parts of inequality (x+1)|x|>(x+1) by x+1 as you don't know the sign of x+1: if x+1>0 you should write |x|>1 BUT if x+1<0 you should write |x|<1 (flip the sign).

I just want to present an easier way to prove that I is sufficient.

For I to be positive what are the conditions?

1. x+1 > 0 this means x >-1 2.|x| -1 > 0 this means x > 1 or x < -1

then you must draw the number line ----------------[b]-1[b]-----0------1--------------> then draw this inequalities on this line and look for any unity. then you will fine x > 1

for the other side, I mean, x+1 <0 and |x| - 1 <0 you wont find any unity. so I is sufficient. Hope it clears. cheers

For x≥0 (x+1)(|x|-1)>0 (x+1)(x-1)>0 x^2-1>0 x^2>1 Of course, x could be 2 or negative two for all we know, so it seems like this is insufficient as we are testing for the range of x>1

For x≤0 (x+1)(|x|-1)>0 (x+1)(-x-1)>0 -x^2-1>0 -x^2>-1

Of course, x could be 2 or negative two for all we know, so it seems like this is insufficient as we are testing for the range of x>1

x can never be -2, as because you have assumed that \(x\geq0\). Also, from the inequality you have correctly arrived at, i.e. \(x^2>1\) \(\to\) x>1 OR x<-1. As assumption was \(x\geq0\). thus only x>1 condition is valid. Also,as x>1 automatically makes \(x\geq0\), thus the correct range is x>1. Sufficient.

Quote:

For x≤0 No need to include equality with zero twice. (x+1)(|x|-1)>0 (x+1)(-x-1)>0 This leads to \(-(x+1)^2>0\) and this is not possible for any real value of x. So,there is no solution for this. -x^2-1>0 -x^2>-1

I mean to say, we define |X| as +x when x is greater than equal to zero, and -x when x is less than zero. But I have noticed you at times have taken x as -x based on x less than equal to zero ? I mean equality cant be on both sides right ? Infact I saw you explained definition of mod function keeping equality on both +x and - x, I hope you are able to make out what doubt I have.

I mean to say, we define |X| as +x when x is greater than equal to zero, and -x when x is less than zero. But I have noticed you at times have taken x as -x based on x less than equal to zero ? I mean equality cant be on both sides right ? Infact I saw you explained definition of mod function keeping equality on both +x and - x, I hope you are able to make out what doubt I have.

The point is that |0|=-0=0. So, in that definition we can include = sign in both cases.

The RHS can only be positive if both are positive or both are negative lets take -5 as value of x===> -ve *+ve ===>not possible let's take 0 as value of x ==> 1* (-1) ===>not possible let's take 1 as value (+ve) ===> 2* (0)===> not possible let's take -1 as value (-ve) ====>0 *0 ====>not possible so x != -ve, x != 0, x != 1 conditions apply. only remaining option is x>1 ....sufficient

2> ----(-5)@@@@@@(5)--------- -5<x<5 ...not sufficient since x is less than 1 here

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