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# Solve for x: 0<|x|-4x<5 = ?

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Solve for x: 0<|x|-4x<5 = ?  [#permalink]

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Updated on: 04 Dec 2012, 02:22
3
28
00:00

Difficulty:

65% (hard)

Question Stats:

60% (02:13) correct 40% (02:20) wrong based on 1141 sessions

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Solve for x: 0<|x|-4x<5 = ?

A. x<0
B. 0<x<1
C. -3/5<x<1
D. -3/5<x<0
E. -1<x<0

Originally posted by yezz on 15 Aug 2009, 23:44.
Last edited by Bunuel on 04 Dec 2012, 02:22, edited 2 times in total.
Renamed the topic and edited the question.
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Re: solve for x?  [#permalink]

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14 Jun 2011, 19:24
5
3
puneetj wrote:
Got to the correct answer but took too much time...E

If you get messed up in mods, such a question can be done in under a minute by trying out some values. Now, I am no fan of plugging in numbers, especially not in DS questions, but such questions are perfect for plugging in if you are not comfortable with algebra. Why? Because they have asked for the range of x. If there is even one value in the given range that doesn't satisfy the inequality, it is not the answer and if there is even one value outside the given range that does satisfy the inequality, it is not the answer.

0<|x|-4x<5 = ?

A. x<0
B. 0<x<1
C. -3/5<x<1
D. -3/5<x<0
E. -1<x<0

Say, consider option A. If x = -1, |x|-4x = 5 which doesn't satisfy the inequality so A is out.
If x = 1/2, |x|-4x is negative so B and C are out.
If x = -4/5, |x|-4x = 4 so D is out and E is the answer.
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##### General Discussion
Manager
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Re: solve for x?  [#permalink]

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16 Aug 2009, 01:08
2
1
I am getting E as an answer if x is -ve.
If x is -ve, we get

0 < -5x < 5

Dividing both sides by -5, we flip both the sides and we land up on

0 > x > -1
ie
-1 < x < 0.

Is that the correct method to solve this problem? Please explain.
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Re: solve for x?  [#permalink]

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16 Aug 2009, 12:54
3
IMO E:

5 + 4x > |x|
5 + 4x > x first cndtn
5 > - 3x
x>-5/3

or

5 + 4x > -x
x > -1

using

4x < |x|
or 4x < x
x<0

or 4x < -x
x<0

thus range lies between -1 to 0

correct me if i m wrong!!
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Joined: 05 Jul 2006
Posts: 1713
Re: solve for x?  [#permalink]

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16 Aug 2009, 13:10
0<|x|-4x<5 = ?

A. x<0
B. 0<x<1
C. -3/5<x<1
D. -3/5<x<0
E. -1<x<0

i did it this way

split the inquality into 2

a) /x/-4x<5 ie: -5<x-4x<5 ie: -5<-3x<5 ie: 5/3>x>-5/3....1

b) /x/-4x>0 thus /x/>4x thus eithe x>4x ie: -3x>0 ie: x <0 or -x<4x ie: x>-4x ie: x>0

and i get lost here??
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Re: solve for x?  [#permalink]

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03 Apr 2011, 12:40
0<-5x<5

=> -1<x<0

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Joined: 16 Nov 2010
Posts: 1407
Location: United States (IN)
Concentration: Strategy, Technology
Re: solve for x?  [#permalink]

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03 Apr 2011, 17:15
I'm also getting E.

2 cases x > 0 or x < 0

0< -x - 4x < 5

=> 0< -5x < 5

=> x > -1 and x < 0

x - 4x < 5

=> -3x < 5

=> x > -5/3

so x > 0 as x is +ve

So -1 < x < 0

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Re: solve for x?  [#permalink]

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14 Jun 2011, 14:22
hi,
apply definition absolute value obtain two possibilites:
A. if x≥0 then 0<|x|-4x<5 go 0<x-4x<5 go 0<-3x<5, go 0>x>-5/3. The intersection is empty
B. if x<0 then 0<|x|-4x<5 go 0<-x-4x<5 go 0<-5x<5 go 0>x>-1. the intersection is -1<x<0

Bye...
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Re: solve for x?  [#permalink]

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03 Dec 2012, 20:52
0<|x|-4x<5

A. x<0
Test: x=-5
$$|-5|-4(-5) = 5 + 20$$ is not less than 5
FALSE!

B. 0<x<1
Test: x=$$\frac{1}{4}$$
$$\frac{1}{4}-4(\frac{1}{4})=-\frac{3}{4} is not greater than 0$$
FALSE!

C. -3/5<x<1
x<1 as tested with B
FALSE!

D. -3/5<x<0
E. -1<x<0

We see that D and E are almost the same except for E. covers -3/5 unlike D.
Let x=-3/5
$$|-\frac{3}{5}|-4(-\frac{3}{5})=\frac{15}{5}=3$$

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Re: solve for x?  [#permalink]

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10 May 2013, 07:39
VeritasPrepKarishma wrote:
puneetj wrote:
Got to the correct answer but took too much time...E

If you get messed up in mods, such a question can be done in under a minute by trying out some values. Now, I am no fan of plugging in numbers, especially not in DS questions, but such questions are perfect for plugging in if you are not comfortable with algebra. Why? Because they have asked for the range of x. If there is even one value in the given range that doesn't satisfy the inequality, it is not the answer and if there is even one value outside the given range that does satisfy the inequality, it is not the answer.

0<|x|-4x<5 = ?

A. x<0
B. 0<x<1
C. -3/5<x<1
D. -3/5<x<0
E. -1<x<0

Say, consider option A. If x = -1, |x|-4x = 5 which doesn't satisfy the inequality so A is out.
If x = 1/2, |x|-4x is negative so B and C are out.
If x = -4/5, |x|-4x = 4 so D is out and E is the answer.

I believe the choice of -4/5 to execlude D is wrong -4/5 is not in the range of -3/5<x<0 ????? accordingly i think both D and E could solve as the right range in my opinion is -5/3 < x < 0??? am i right or wrong plz advise!
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Re: Solve for x: 0<|x|-4x<5 = ?  [#permalink]

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10 May 2013, 22:46
yezz wrote:
Solve for x: 0<|x|-4x<5 = ?

A. x<0
B. 0<x<1
C. -3/5<x<1
D. -3/5<x<0
E. -1<x<0

|x|-4x>0 = |x|>4x =$$\frac{x}{|x|}$$<1 --> x<0. Again, |x|-4x<5 = -x-4x<5 = -5x<5-->x>-1.
E.
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Re: solve for x?  [#permalink]

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11 May 2013, 03:33
2
yezz wrote:
VeritasPrepKarishma wrote:
puneetj wrote:
Got to the correct answer but took too much time...E

If you get messed up in mods, such a question can be done in under a minute by trying out some values. Now, I am no fan of plugging in numbers, especially not in DS questions, but such questions are perfect for plugging in if you are not comfortable with algebra. Why? Because they have asked for the range of x. If there is even one value in the given range that doesn't satisfy the inequality, it is not the answer and if there is even one value outside the given range that does satisfy the inequality, it is not the answer.

0<|x|-4x<5 = ?

A. x<0
B. 0<x<1
C. -3/5<x<1
D. -3/5<x<0
E. -1<x<0

Say, consider option A. If x = -1, |x|-4x = 5 which doesn't satisfy the inequality so A is out.
If x = 1/2, |x|-4x is negative so B and C are out.
If x = -4/5, |x|-4x = 4 so D is out and E is the answer.

I believe the choice of -4/5 to execlude D is wrong -4/5 is not in the range of -3/5<x<0 ????? accordingly i think both D and E could solve as the right range in my opinion is -5/3 < x < 0??? am i right or wrong plz advise!

Focus on
"and if there is even one value outside the given range that does satisfy the inequality, it is not the answer."
given above.

-4/5 is a value which satisfies 0 < |x|-4x < 5 since |-4/5|-4(-4/5) = 4.
Since -4/5 does not lie in the range -3/5<x<0 so (D) cannot be the answer. The correct range needs to cover all possible values of x.
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Re: Solve for x: 0<|x|-4x<5 = ?  [#permalink]

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11 May 2013, 05:31
@Karishma ... U r absolutely right , been away for years from GMAT ... E is the perfect answer
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Re: Solve for x: 0<|x|-4x<5 = ?  [#permalink]

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12 May 2013, 06:44
we ve got 3 critical values that are ( 0 from modulus , -5/3& -1 from /x/ = 5+4x) , draw on the number line and test values

.....-5/3........-1...........0....................

only -1<x<0 is the range where all values of x satisfy the compound inequality 0</x/-4x<5
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Re: Solve for x: 0<|x|-4x<5 = ?  [#permalink]

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01 Jul 2013, 14:08
1
Solve for x: 0<|x|-4x<5 = ?

A. x<0
B. 0<x<1
C. -3/5<x<1
D. -3/5<x<0
E. -1<x<0

There are two options here - plugging in values given to us in the answer choices or simplifying the inequality.

0<|x|-4x<5

x>0: 0<x-4x<5 0<-3x<5 0<x<-5/3 -5/3<x<0 INVALID as x does not fall within the range of x>0
OR
x<0: 0<(-x)-4x<5 0<-5x<5 0<x<-1 -1<x<0 VALID as x falls within the range of x<0

There is only one valid solution: -1<x<0.
(E)
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Re: Solve for x: 0<|x|-4x<5 = ?  [#permalink]

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29 Jan 2014, 07:05
Would be happy to hear some comments on whether this approach is correct

|x|-4x>0

So we have two cases

If x>0 then x-4x>0
-3x>0
x<0, this contradicts and hence is not a valid solution

If x<0 then -5x>0
x<0, this solution is valid

So we get that -1<x<0 replacing in the original inequality

E
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Re: Solve for x: 0<|x|-4x<5 = ?  [#permalink]

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29 Jan 2014, 20:03
2
1
jlgdr wrote:
Would be happy to hear some comments on whether this approach is correct

|x|-4x>0

So we have two cases

If x>0 then x-4x>0
-3x>0
x<0, this contradicts and hence is not a valid solution

If x<0 then -5x>0
x<0, this solution is valid

So we get that -1<x<0 replacing in the original inequality

E

Knowing only x < 0, how do you choose between (A), (D) and (E)?
You need to consider |x| - 4x < 5 too
When x < 0, -x -4x < 5
-5x < 5
x > -1
That's how you get -1 < x < 0

Or work on the whole inequality in one go
0 < |x| - 4x < 5
When x < 0,
0< -x - 4x < 5
0 < -5x < 5
0 < -x < 1
0 > x > -1

which is the same as -1 < x < 0
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Re: Solve for x: 0<|x|-4x<5 = ?  [#permalink]

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25 Apr 2015, 20:50
Hey,
So I had a doubt. For the equaltiy: 0<|x|-4x<5 if I try and solve it algebraically, i first take x<0
In that case won't this equality be: 0<-x-4(-x)<5. I can't just substitute mod x with -x and leave the other x be can I? Please help!
Thanks!
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Re: Solve for x: 0<|x|-4x<5 = ?  [#permalink]

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26 Apr 2015, 20:49
1
1
natashakumar91 wrote:
Hey,
So I had a doubt. For the equaltiy: 0<|x|-4x<5 if I try and solve it algebraically, i first take x<0
In that case won't this equality be: 0<-x-4(-x)<5. I can't just substitute mod x with -x and leave the other x be can I? Please help!
Thanks!

Say you have an inequality: 4x < 5
and you know that x must be negative. How will you solve the inequality?
Will you say that the inequality becomes -4x < 5? No. You are given that 4x < 5. Without changing the inequality, you can write this as x < 5/4.
x needs to be negative. All negative values will be less than 5/4.

Why do you substitute -x in place of |x|? You cannot solve an equation/inequality with |x| in it. You need to remove the absolute value sign.

You know that |x| = x if x is positive and |x| = -x if x is negative.

Since you know that x is negative, you can write -x in place of |x| without changing the inequality.
If you change the simple x to -x in the inequality, the inequality changes.

Check out this post for a more detailed explanation: http://www.veritasprep.com/blog/2014/06 ... -the-gmat/
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Re: Solve for x: 0<|x|-4x<5 = ?  [#permalink]

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06 Dec 2015, 23:33
Solve for x: 0<|x|-4x<5 = ?

A. x<0
B. 0<x<1
C. -3/5<x<1
D. -3/5<x<0
E. -1<x<0

Soln: We'll consider two cases: x>0 and $$x<0$$ where $$|x|$$ will become $$x$$ and $$-x$$ respectively.
x>0: 0 < x - 4x < 5, or, 0 < -3x < 5, or -5/3 < x < 0 (dividing by -3 and thereby changing the direction of the inequality)
BUT, if $$x>0$$ then x cannot lie in $$-5/3 < x < 0$$. No solution here.

$$x<0: 0 < -x - 4x < 5,$$ or, $$0 < -5x < 5,$$ or $$-1 < x < 0$$ (dividing by -3 and thereby changing the direction of the inequality)
This range is GOOD. Anything in this range will be permissible.

Choice E is the same range: $$-1 < x < 0$$
Re: Solve for x: 0<|x|-4x<5 = ?   [#permalink] 06 Dec 2015, 23:33

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# Solve for x: 0<|x|-4x<5 = ?

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