Is |x - 5| 4 ?

Question

Is |x - 5| > 4 ? (1) x is an integer (2) x^2 < 1 From qs stem: 1 > x >9 Frm statement 1 - no help Frm Statement 2 I get -1 < x < +1 C

shalva · Answer

It seems correct answer to me

flgators519 · Answer

Couldn't statement 2 be sufficient by itself?

I would think it's B, as you said X can only be inbetween -1, and 1, non inclusive and all these values abs(x- 5) are greater than 4

flgators519 · Answer

Putting them together lets you realize that the answer is x=0. however, it doesn't ask what is X. Anyone confirm my answer?

Thanks

Bunuel · Answer

|x - 5|>4 ? x<5 --> -x+5>4 --> x<1 x>5 --> x-5>4 --> x>9 So we get that |x - 5|>4 holds true when x<1 OR x>9 (1) x is an integer --> Clearly insufficient. (2) x^2<1 --> -1 any x from this range will surely be in the range x<1 , hence for ANY x from the range -14 holds true. Sufficient. Answer: B.

shalva · Answer

You're right. We need to determine whether |x-5|>4 or not... Nothing about value of x :oops:

fall2009 · Answer

B. 2) -1 |x-5| = 4 = min But x<1 => |x-5|>4.

Prometoh · Answer

one for B

From the question stem, we derive x<1 or x > 9
(1) is obviously insufficient
(2) -1 < x < 1; thus, (2) is sufficient

ENAFEX · Answer

Bunuel, How does  x^{2}\leq 1 transform to  -1<x<1 ?

Isn't it supposed to be

X< 1 or -1?

Bunuel · Answer

When you are not sure about the ranges when solving inequalities it's a good idea to test some numbers. So, first ask yourself what does x<-1 or x<1 even mean (it does not make any sense). Next, try some numbers from the range you have (whatever it is) to see whether x^2<1 holds true for them. Solution: x^2<{1} , since both parts of the inequality are non-negative then we can take square root: |x|<{1} --> -1<{x}<{1} . Now, other approach would be: x^2-1<{0} --> (x+1)(x-1)<{0} --> the roots are -1 and 1 --> "<" sign indicates that the solution lies between the roots, so -1<{x}<{1} . Solving inequalities: x2-4x-94661.html#p731476 ( check this one first ) inequalities-trick-91482.html data-suff-inequalities-109078.html range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535 everything-is-less-than-zero-108884.html?hilit=extreme#p868863 Hope it helps.

ENAFEX · Answer

hmmm, still missing something. But little better after reading the posts.

x<-1 or x<1 is not valid because these values do not satisfy the condition (x+1)(x-1)<0

Ok, now my question is (x+1)(x-1)<0

is (x+1)<0 AND (x-1)<0

How does (x+1)<0 become x>-1? While moving 1 to the right hand side we are not supposed to change the sign right?

for example x+3<0,
is x<-3 and not x>-3?

Bunuel · Answer

From x+1<0 you can simply move 1 to the other side of the inequality: x<-1. Or: x+1<0 --> add -1 to both sides: x<-1. In this case you are not multiplying inequality by a negative number so there is no need to flip the sign. The same way x+3<0 --> x<-3. It seems that you should brush up fundamentals on inequality.

LogicGuru1 · Answer

Is |x - 5| > 4 ?

(1) x is an integer
(2) x^2 < 1

No need to even solve this one . Just use common sense

(1) x is an integer
X can be any positive or negative number ; INSUFFICIENT.

(2) x^2 < 1
Only decimal between \((-1 AND 1)^2\) and \((0)^2\) are less than 1; all other numbers (+ve or -ve) whenever squared will give a number >1
So based on second statement x= 0.xyz or -0.xyz
Adding 0.xyz to 5 will increase the value of 5 ==>SUFFICIENT
subtracting 0.xyz from 5 will decrease the value of 5 but can never make it less than 4 ; to make 5 less than 4 we need 1 but here we are getting a decimal 0.xyz ==> SUFFICIENT

ANSWER IS B

FlyingCycle · Answer

Bunuel

when you have (x+1)(x-1) < 0. This gives us two cases to test right:
Either,
CASE I:
(x+1) is positive and (x-1) is negative
OR
CASE II
(x+1) is negative and (x-1) is positive

When you solve these:
CASE I:
(x+1) > 0
x > -1
(x-1) < 0
x < 1
so now we have one range -1<x<1 Fair enough. Now my doubt comes when we check CASE II

CASE II
(x+1) is negative and (x-1) is positive
x+1 < 0
x < -1
x-1 > 0
x > 1
now the range is x < -1 and x > 1. Why are we not considering this? Am I missing something?

Bunuel · Answer

CASE II, x < -1 AND x > 1 is not possible. x cannot be simultaneously less than -1 and more than 1. Or, consider this x + 1 is more than x - 1 and you cannot have x + 1 < 0 and x -1 > 0.

FlyingCycle · Answer

Thank you  Bunuel . Yes when I scribbled in the paper then I got it. 
I do not know, while solving, why didn't I opt for curvy method.

Thanks a lot.

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Is |x - 5| > 4 ?

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