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Is |x + 1| > 6? (1) (x - 2)(|x| - 5) > 0 (2) x^2 < 4 Updated on: 09 Oct 2024, 00:06
Originally posted by EgmatQuantExpert on 20 May 2015, 10:36.
Last edited by Bunuel on 09 Oct 2024, 00:06, edited 5 times in total.
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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75% (hard)

Question Stats:

59% (02:32) correct 41% (02:40) wrong based on 898 sessions

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Is \(|x+1| > 6\)?

(1) \((x-2)(|x| - 5) > 0\)
(2) \(x^2 < 4\)

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Question 7 - the Last Question - of the e-GMAT Question Series on Absolute Value.

Previous Question


Provide your solution below. Kudos for participation. Happy Solving! :-D

Best Regards
The e-GMAT Team


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B
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Is |x + 1| > 6? (1) (x - 2)(|x| - 5) > 0 (2) x^2 < 4 21 May 2015, 09:57
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Is \(|x+1| > 6\)?

lets break it down what the question is asking , its basically asking that is x < -5 or x > 5, if x is in that range than x is > 6 (yes), otherwise (no).

Stmt (1) : \((x-2)(|x| - 5) > 0\)
for product of two integers to be > 0 , either both positive or both negative
    both positive
    (x-2) > 0 ; x > 2
    |x| - 5 > 0 ; |x| > 5 ; or -5 > x , x > 5
    combining both x > 5 ; answer to main question |x+1| > 6 = yes

    both negative
    (x-2) < 0 ; x < 2
    |x| - 5 < 0 ; |x| < 5 ; or -5 < x < 5
    combining both -5 < x < 2 ; answer to main question |x+1| > 6 = no
    insufficient

Stmt (2) : \(x^2 < 4\)
    |x| < 2 ; or -2 < x < 2 ; answer to main question |x+1| > 6 = no
    sufficient

Ans B
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Is |x + 1| > 6? (1) (x - 2)(|x| - 5) > 0 (2) x^2 < 4 Updated on: 21 May 2015, 10:34
Originally posted by reto on 20 May 2015, 13:10.
Last edited by reto on 21 May 2015, 10:34, edited 1 time in total.
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Is \(|x+1| > 6\)?

(1) \((x-2)(|x| - 5) > 0\)
(2) \(x^2 < 4\)

This is Question 7 - the Last Question - of the e-GMAT Question Series on Absolute Value.

Provide your solution below. Kudos for participation. The Official Answer and Explanation will be posted on 22nd May.

Till then, Happy Solving! :-D

Best Regards
The e-GMAT Team
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I'll try, but i'm not really sure:

Statement 1:
Too many values are fitting in the formula given (e.g. -4 which would give you a clear NO to: is |x+1| > 6 ? or 6 which would give you a "YES". Therefore Statement 1 is insufficient.

Statement 2
Tells you that x is >-2 and x<2. This statement is sufficent since it gives a clear NO to the question Is \(|x+1| > 6\)

Answer is B.
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Is |x + 1| > 6? (1) (x - 2)(|x| - 5) > 0 (2) x^2 < 4 22 May 2015, 03:33
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Is x>5 or x<-7?
S1) x>2
X>5
X<-5 Not suff
S2) x<2 Not suff
Combined not suff
"E"
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Is |x + 1| > 6? (1) (x - 2)(|x| - 5) > 0 (2) x^2 < 4 22 May 2015, 23:45
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Statement 1
(x-2)(|x|-5)> 0

Suppose x>=0 then |x| = x
statement becomes as

(x-2)(x-5)> 0

which gives x<2, x> 5
As we have assumed x>=0 so range of solution is 0=<x<2 , x>5

Now suppose x <0 then satement becomes
(x-2)(-x-5)>0
-(x+5)(x-2)>0
(x+5)(x-2)<0
-5<x<2
Since we assumed x <0 so -5<x<0

Combining both range of solutions we get
-5<x<2 , x> 5

So if you put x=6 ans to the ques is Yes
If you put x=0 ans to the ques is No
Statement 1 insuff

Statement 2 gives
-2<x<2
So any value in this range of x will give a definite NO as answer to the ques
Hence sufficient


Ans B
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Is |x + 1| > 6? (1) (x - 2)(|x| - 5) > 0 (2) x^2 < 4 24 May 2015, 11:34
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[1]
(x-2)(|x|-5)> 0

for x < 0

(x-2)(-x-5)> 0
or (x-2)(x+5)<0
therefore, -5<x<2.........so, for x<0 |x+1| is not greater than 6

for

Now for x > 0
(x-2)(x-5)>0
therefore, x<2 or x>5
Again, for some values in the above range |x+1|>6 and for some it's not.

So, [1] is insufficient

[2]
Say's X^2 < 4
or x^2-4<0
or (x-2)(x+2) < 0
therefore, -2<x<2....Clearly this range is sufficient to say that for all values in it |x+1| will never be > 0........sufficient


Ans B
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Is |x + 1| > 6? (1) (x - 2)(|x| - 5) > 0 (2) x^2 < 4 25 May 2015, 23:30
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Official Explanation

Correct Answer: B

|x+1| represents the distance of x from -1 on the number line.

So, the question is asking, ‘Is the distance of x from -1 on the number line greater than 6 units or not?’



We see that:

The answer is YES if
i) x < -7 or
ii) x > 5

Else, the answer is NO

Analyzing Statement 1
(x-2)(|x| - 5) > 0

The product of two terms is positive.

This means, either both the terms on Left Hand Side are positive or both are negative.

Case 1: x – 2 is positive AND |x| - 5 is positive

This means, x > 2 AND |x| > 5

Now, |x| is greater than 5 for x < -5 and for x > 5. But x < -5 violates the condition of this case that x – 2 should be positive.

Therefore, the range that satisfies Case 1: x > 5

In this case, the answer to the asked question is YES.

Case 2: x – 2 is negative AND |x| - 5 is negative

That is, x < 2 AND -5 < x < 5

Upon combining the 2 pieces of information, the resultant inequality is:

-5 < x < 2

In this case, the answer to the asked question is NO.

Thus, we have not been able to get a unique answer to the asked question from St. 1. Therefore, Statement 1 is insufficient.

Analyzing Statement 2
\(x^2 < 4\)

That is, \(x^2 - 2^2 < 0\)
\((x+2)(x-2) < 0\)



By using the Wavy Line method, we can easily find that this inequality holds for -2 < x < 2

In this range, the answer is NO.

Since St. 2 does provide us with a unique answer, it is sufficient.
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Is |x + 1| > 6? (1) (x - 2)(|x| - 5) > 0 (2) x^2 < 4 26 May 2015, 00:01
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Statement 1
(x-2)(|x|-5)> 0

Suppose x>=0 then |x| = x
statement becomes as

(x-2)(x-5)> 0

which gives x<2, x> 5
As we have assumed x>=0 so range of solution is 0=<x<2 , x>5

Now suppose x <0 then satement becomes
(x-2)(-x-5)>0
-(x+5)(x-2)>0
(x+5)(x-2)<0
-5<x<2
Since we assumed x <0 so -5<x<0

Combining both range of solutions we get
-5<x<2 , x> 5

So if you put x=6 ans to the ques is Yes
If you put x=0 ans to the ques is No
Statement 1 insuff

Statement 2 gives
-2<x<2
So any value in this range of x will give a definite NO as answer to the ques
Hence sufficient


Ans B
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Dear abhishek05

Your analysis was mostly right. One correction though. Please look at the part in red.

You should completely analyze one case before going to the next case. That is, first completely analyze the case that x is >= 0 (which gives you the pink range of possible values of x in the quoted solution) and only then go to the second case that x is < 0 (which gives the orange range of possible values in the quoted solution)

So, the correct approach would have been:
1. Once you get the (pink) range of values of x in Case 1, check what answer do they give to the question 'Is |x+1| > 6?'
2. After you get the (orange) range of values of x in Case 2, check what answer do they give to the question 'is |x+1| >6?'

Hope this helped! :)

Best Regards

Japinder
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Is |x + 1| > 6? (1) (x - 2)(|x| - 5) > 0 (2) x^2 < 4 13 Jul 2017, 14:35
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abhishek05
Statement 1
(x-2)(|x|-5)> 0

Suppose x>=0 then |x| = x
statement becomes as

(x-2)(x-5)> 0

which gives x<2, x> 5
As we have assumed x>=0 so range of solution is 0=<x<2 , x>5

Now suppose x <0 then satement becomes
(x-2)(-x-5)>0
-(x+5)(x-2)>0
(x+5)(x-2)<0
-5<x<2
Since we assumed x <0 so -5<x<0

Combining both range of solutions we get
-5<x<2 , x> 5

So if you put x=6 ans to the ques is Yes
If you put x=0 ans to the ques is No
Statement 1 insuff

Statement 2 gives
-2<x<2
So any value in this range of x will give a definite NO as answer to the ques
Hence sufficient


Ans B

Dear abhishek05

Your analysis was mostly right. One correction though. Please look at the part in red.

You should completely analyze one case before going to the next case. That is, first completely analyze the case that x is >= 0 (which gives you the pink range of possible values of x in the quoted solution) and only then go to the second case that x is < 0 (which gives the orange range of possible values in the quoted solution)

So, the correct approach would have been:
1. Once you get the (pink) range of values of x in Case 1, check what answer do they give to the question 'Is |x+1| > 6?'
2. After you get the (orange) range of values of x in Case 2, check what answer do they give to the question 'is |x+1| >6?'

Hope this helped! :)

Best Regards

Japinder
Show more


The two different solutions here by the corresponding different method yield solution:

Your method: Gives solution X>5, ( Case 1: x – 2 is positive AND |x| - 5 is positive)
Another method considering |X| = X for X>=0 , yields a broad range ie, 0=<x<2 , x>5

There is a difference right, is this difference of approach going to yield same conclusion every time?
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Is |x + 1| > 6? (1) (x - 2)(|x| - 5) > 0 (2) x^2 < 4 08 Oct 2024, 23:13
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