Is |x + 1| 6? (1) (x - 2)(|x| - 5) 0 (2) x^2

Question

Is  |x+1| > 6 ?

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

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

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The e-GMAT Team

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UJs · Accepted Answer

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

reto · Answer

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.

csirishac · Answer

Is x>5 or x<-7? S1) x>2 X>5 X<-5 Not suff S2) x<2 Not suff Combined not suff "E"

Escalate · Answer

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

rohitd80 · Answer

(x-2)(|x|-5)> 0 for x < 0 (x-2)(-x-5)> 0 or (x-2)(x+5)<0 therefore, -5 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, is insufficient Say's X^2 < 4 or x^2-4<0 or (x-2)(x+2) < 0 therefore, -2 0........sufficient Ans B

EgmatQuantExpert · Answer

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.

EgmatQuantExpert · Answer

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

Supermaverick · Answer

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=5 There is a difference right, is this difference of approach going to yield same conclusion every time?

bumpbot · Answer

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Is |x + 1| > 6? (1) (x - 2)(|x| - 5) > 0 (2) x^2 < 4

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