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AdmitJA
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Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5 22 Nov 2014, 07:24
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Bunuel,

Can we try this question by evaluating each of these expressions separately? Is this a correct approach, though time consuming?

1) x+1 > 0
2) x-1 > 0
3) -x-1 > 0

If correct, could you please demonstrate on how to proceed?
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Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5 22 Nov 2014, 07:30
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joseph0alexander
Bunuel,

Can we try this question by evaluating each of these expressions separately? Is this a correct approach, though time consuming?

1) x+1 > 0
2) x-1 > 0
3) -x-1 > 0

If correct, could you please demonstrate on how to proceed?
Show more

This would not the best approach. You should consider two cases for |x|: when x<0, then |x| = x and when x>0, then |x| = -x. As well as two cases when (x+1) and (|x| - 1) are both positive and both negative.
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Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5 22 Nov 2014, 07:43
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Bunuel,

Can we try this question by evaluating each of these expressions separately? Is this a correct approach, though time consuming?

1) x+1 > 0 ---> x > -1
2) x-1 > 0 ---> x > 1
3) -x-1 > 0 ---> -x > 1 or x > -1

If correct, could you please demonstrate on how to proceed?

This would not the best approach. You should consider two cases for |x|: when x<0, then |x| = x and when x>0, then |x| = -x. As well as two cases when (x+1) and (|x| - 1) are both positive and both negative.
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Fair enough Bunuel, but just for the sake of understanding and learning the concept.

If we proceed with the 3 equations, I am getting the results in red above. Comparing them, we can take the greatest limiting factor, which is x>1, and prove sufficiency. I find this approach a little easier. Is my arithmetic correct?

Thank you. :-D
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Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5 22 Nov 2014, 07:58
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joseph0alexander
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Bunuel,

Can we try this question by evaluating each of these expressions separately? Is this a correct approach, though time consuming?

1) x+1 > 0 ---> x > -1
2) x-1 > 0 ---> x > 1
3) -x-1 > 0 ---> -x > 1 or x > -1

If correct, could you please demonstrate on how to proceed?

This would not the best approach. You should consider two cases for |x|: when x<0, then |x| = x and when x>0, then |x| = -x. As well as two cases when (x+1) and (|x| - 1) are both positive and both negative.

Fair enough Bunuel, but just for the sake of understanding and learning the concept.

If we proceed with the 3 equations, I am getting the results in red above. Comparing them, we can take the greatest limiting factor, which is x>1, and prove sufficiency. I find this approach a little easier. Is my arithmetic correct?

Thank you. :-D
Show more

Again you should consider 2*2 = 4 cases.

(x+1)(|x| - 1) > 0.

1. x < 0 --> (x+1)(-x - 1) > 0.

(x + 1) > 0 and (-x - 1) > 0 --> x > -1 and x < -1. No solution here.

2. x < 0 --> (x+1)(-x - 1) > 0.

(x + 1) < 0 and (-x - 1) < 0 --> x < -1 and x > -1. No solution here.

3. x > 0 --> (x+1)(x - 1) > 0.

(x + 1) > 0 and (x - 1) > 0 --> x > -1 and x > 1. This gives x > 1.

4. x > 0 --> (x+1)(x - 1) < 0.

(x + 1) < 0 and (x - 1) < 0 --> x < -1 and x < 1. No solution here.

So, (x+1)(|x| - 1) > 0 holds true only when x > 1.

Hope it's clear.
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Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5 27 Nov 2014, 05:05
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Bunuel
Good question. +1.

Is x> 1?

(1) (x+1) (|x| - 1) > 0. Consider two cases:

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.
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Hi Bunuel

Why are we not changing the inequality sign in statement 1 when we assume x is negative-ideally we should. In that case, we get x<-1 from -(x+1)^2<0 as one of (x+1) can be eliminated.
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Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5 27 Nov 2014, 06:30
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sinhap07
Bunuel
Good question. +1.

Is x> 1?

(1) (x+1) (|x| - 1) > 0. Consider two cases:

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 Bunuel

Why are we not changing the inequality sign in statement 1 when we assume x is negative-ideally we should. In that case, we get x<-1 from -(x+1)^2<0 as one of (x+1) can be eliminated.
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\(-(x+1)^2>0\);

Add (x+1)^2 to both sides: \(0>(x+1)^2\), which is the same as \((x+1)^2<0\).
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Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5 06 Dec 2014, 14:52
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Is x > 1?

I would approach it the following way :

(1) (x+1) (|x| - 1) > 0

(+) * (+) > 0 or (-) * (-) > 0

For both of the parts to be positive we can see that x >1 . Just by trying few values you can figure this out. X cant be Zero as then the second part becomes - . X cant be 1 as then second part becomes 0 and hence the whole LHS becomes Zero.

For both of the parts to be negative we try any value less an Zero and see that no value will satisfy the equation. Hence X cannot be negative.. Hence A is Sufficient.

(2) |x| < 5

Clearly not sufficient

Answer is A.
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Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5 12 Mar 2016, 00:45
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Hi Bunuel,

I have a question.

(x+1)(|x|-1)=0 has x=1 and x=-1 as its solution. However, x=-1 is a repeated root. Hence, on the number line, the sign will not change when it passes x=-1.

--------(-1)---------(1)+++++++

Therefore, only when x>1, (x+1)(|x|-1)>0.

Is my solution correct? I remember my highschool math teacher said something like 'the sign doesn't change when it passes a double root'. But, it's years ago and I just want to make sure that the approach is valid.

Thank you!
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Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5 28 May 2018, 13:11
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Tips: The key to unlock confusion with “and” and “or” in inequality:

- Drawing graphs and crossing off all irrelevant values.
- “and” means must; “or” means could

Is x > 1?

(1) (x+1) (|x| - 1) > 0

(2) |x| < 5

(1): There are 2 cases: case 1 or case 2

Case 1: x+1 > 0 AND |x|-1 >0
X+1 >0 → x > -1 (A)
|x|-1 >0 → |x| >1 → x>1 OR x <-1 (B)

→ graph for A: //////////(-1)-------------(1)--------
→ graph for B: ---------(-1)///////////(1)--------

Note about the graphs:
////////////means the values crossed off
----------------- means the values are accepted

Combining the two graphs above: this is the AND case, so we must cross off all values that don’t fit in any graph, represented by “//////////”
So the value of the 2 inequalities of case 1 is: x>1.

Case 2: x+1 <0 AND |x|-1 <0

X+1 < 0 → x <-1 (C)
|x|-1 <0 → |x| < 1 → -1<x<1 (D)

→ graph for C: ---------(-1)/////////(1)///////
→ graph for D: /////////(-1)----------(1)////////

As we see, there is no value of x that satisfy the both graphs. That means there is no solution for the inequalities in case 2.

→ So only case 1 is appropriate to consider. That means (x+1) (|x| - 1) > 0 when x>1. Sufficient.

(2) |x| < 5 → -5<x<5 → graph: //////-5--------1--------5//////

Clearly, x can be bigger than 1 or less than 1. Insufficient.

Answer: A.


P/s: If you are still confused, you can test numbers!
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Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5 03 Nov 2022, 10:45
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Bunuel KarishmaB
Can you please help me understand what mistake I am making in the attachment
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Bunuel KarishmaB
Can you please help me understand what mistake I am making in the attachment
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Note that:

AB = 0 means that either A = 0 or B = 0 or both are 0
AB > 0 means that either Both are positive or Both are negative.
AB < 0 means that exactly one of them is positive and the other is negative.

\((x+1)(|x| - 1) > 0\)

means either (x + 1) > 0 AND (|x| - 1) > 0
or (x + 1) < 0 AND (|x| - 1) < 0

Now review your solution.
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Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5 10 Dec 2025, 02:53
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Hi
How do you see what are the options for A B C D E ? Mine version only shows the questions not the options
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Is x > 1? (1) (x+1)(|x| - 1) > 0 (2) |x| < 5 10 Dec 2025, 04:19
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ZahraHuseynova
Hi
How do you see what are the options for A B C D E ? Mine version only shows the questions not the options
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Hi,

This is a data sufficiency question. Options for DS questions are always the same and usually omitted on the site.

The data sufficiency problem consists of a question and two statements, labeled (1) and (2), in which certain data are given. You have to decide whether the data given in the statements are sufficient for answering the question. Using the data given in the statements, plus your knowledge of mathematics and everyday facts (such as the number of days in July or the meaning of the word counterclockwise), you must indicate whether—

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
D. EACH statement ALONE is sufficient to answer the question asked.
E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

Hope this helps.­
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