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0<|x|-4x<5 = ? A. x<0 B. 0[#permalink ]
27 Jul 2006, 22:04

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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

try this one..

It's easy but I've seen many ppl. have problem with |x| in this forum

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Will go with E.
Let say x = -0.9
|x| - 4x = 0.9 + 3.6 = 4.5
X = -0.1
|x| - 4x = 0.1 + 0.4 = 0.5
E satisifies the condition

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E by plugging numbers.

x = 0 fails.

x = 1/2 fails

x = 1 fails

x = 2 fails

x = -1 fails

x = 0.5 true.

hence -1<x<0

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Plugging number is a good way to solve this. On the day of exam, I, sure, plug in some numbers to solve this.

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

1. 0<|x|-4x

if x<0

0<-x-4x

x<0

if x>=0

0<= x-4x

x<=0 (X, it contradicts assumption)

from 1

x<0

2.|x|-4x <5

if x<0

-x-4x < 5

x > -1

thus, -1<x<0

if x>= 0

x-4x < 5

x > -5/3

thus, x>=0

from 2

-1<x<=0

intersection from 1 and 2

-1<x<0

E is OA.

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Last edited by

freetheking on 28 Jul 2006, 22:18, edited 1 time in total.

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Re: PS:another |x| [#permalink ]
28 Jul 2006, 22:10

freetheking wrote:

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

0<|x|-4x<5

since lxl is always +ve, 0<|x|-4x<5 is equal to 0< -3x < 5.

lets divide the nequality by -3.

0> x > -5/3.

your answer in D is incorrect. so only choice that fits to the above range is E.

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Re: PS:another |x| [#permalink ]
28 Jul 2006, 22:12

MA wrote:

freetheking wrote:

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

0<|x|-4x<5

since lxl is always +ve, 0<|x|-4x<5 is equal to 0< -3x < 5. lets divide the nequality by -3. 0> x > -5/3.

your answer in D is incorrect. so only choice that fits to the above range is E.

It's typo.. OA is E..

but MA I don't think you can do that way..

1. if x<0

0<-x-4x<5

-1<x<0

2. if x>=0

-5/3<x<0 but it contradicts assumtion, thus trivial soulution..

from 1 and 2.. -1<x<0

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Re: PS:another |x| [#permalink ]
29 Jul 2006, 01:59

freetheking wrote:

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 try this one.. It's easy but I've seen many ppl. have problem with |x| in this forum

Part 1. |x|-4x<5 == > X > -5/3 or x >-1.

part 2. 0<|x|-4x == > x<0

Combining 1 and 2 ==> E.

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Re: PS:another |x| [#permalink ]
29 Jul 2006, 12:47

freetheking wrote:

1. if x<0 0<-x-4x<5 -1<x<02. if x>=0 -5/3<x<0 but it contradicts assumtion, thus trivial soulution.. from 1 and 2.. -1<x<0

but from the information as provided in the question, x cannot be +ve or 0. the only value that satisfies x is it is -ve. so E makes sense.

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one can also try not plugging nos. but straightforward solving of the equations:
original equation can be broken into 2 standard equations
1. 0 < x-4x <5 <-> 0 < -3x < 5 <-> -5/3 < x < 0
2. 0 > x-4x > 5 (change signs as this is the -ve condition)
0 > 3x > -5 <--> 0 > x > -5/3
(same as 1)
Only solution that satisfies above is E

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Still, many ppl. seem to have wrong concept on |x|

0<|x|-4x<5 is exactaly -1<x<0
-5/3< x < 0 is definitely wrong

plug in -4/3 0<|-4/3|-4*(-4/3)<5 0< 4/3+16/3 <5 0< 20/3 < 5 0< 6.666.. < 5 _________________

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