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# 0<|x|-4x<5 = ? A. x<0 B. 0<x<1 C.

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Senior Manager
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0<|x|-4x<5 = ? A. x<0 B. 0[#permalink]

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27 Jul 2006, 23:04
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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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27 Jul 2006, 23:16
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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28 Jul 2006, 00:05
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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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Senior Manager
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28 Jul 2006, 22:17
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

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, 23:18, edited 1 time in total.
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28 Jul 2006, 23:10
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.
Senior Manager
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28 Jul 2006, 23:12
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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Senior Manager
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29 Jul 2006, 02:59
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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29 Jul 2006, 13:47
freetheking wrote:
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

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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29 Jul 2006, 20:52
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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29 Jul 2006, 22:08
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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29 Jul 2006, 22:08
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