What is x? (1) |x| < 2 (2) |x| = 3x 2 : Quant Question Archive [LOCKED]
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# What is x? (1) |x| < 2 (2) |x| = 3x 2

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Director
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What is x? (1) |x| < 2 (2) |x| = 3x 2 [#permalink]

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20 Oct 2007, 20:50
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What is x?

(1) |x| < 2

(2) |x| = 3x – 2
Director
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20 Oct 2007, 21:09
eyunni wrote:
What is x?

(1) |x| < 2

(2) |x| = 3x – 2

i get E

(1)
for x > 0 => x < 2
for x < 0 => x > -2
so range of x = -2 < x < 2
not sufficient

(2)
for x > 0 => x = 3x - 2
2x = 2 : x = 1
for x < 0 => -x = 3x - 2
4x = 2 : x = 1/2
x could be 1 or 1/2, not sufficient

putting (1) and (2) together, both 1 and 1/2 fits in -2 < x < 2
so both not sufficient
Director
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20 Oct 2007, 21:11
eyunni wrote:
What is x?

(1) |x| < 2
(2) |x| = 3x – 2

E.

From 1, x could be anything smaller than 2 but grater than -2. nsf
From 2, x = 1/2 and 1. nsf

from 1 and 2 also nsf.

Last edited by Fistail on 20 Oct 2007, 21:14, edited 3 times in total.
VP
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20 Oct 2007, 21:14
Fistail wrote:
trivikram wrote:
yes it should be B

how???

Mistake....I messed a calculation here..

squaring both sides in (B) and applyingA would lead us to E
SVP
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21 Oct 2007, 00:27
Perfect explanation from beckee529 .... Nothing to add
Manager
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21 Oct 2007, 08:38
Statement 1 is an absolute so we have 2 possibilities. When X is positive and when X is negative.

So, |x| < 2
1. x<2 or
-x<2>-2

So, when x > 0 => x < 2
And if x <0> x > -2
Manager
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21 Oct 2007, 09:05
Statement 1 is an absolute so we have 2 possibilities. When X is positive and when X is negative.

So, |x| < 2
1. x<2 or
-x<2>-2

So, when x > 0 => x < 2
And if x <0> x > -2

But from stmt 2
for x > 0 => x = 3x - 2
2x = 2 : x = 1
for x <0> -x = 3x - 2
4x = 2 : x = 1/2

here with x<0 we r gettin x=1/2 ????
Manager
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21 Oct 2007, 09:07
Statement 1 is an absolute so we have 2 possibilities. When X is positive and when X is negative.

So, |x| < 2
1. x<2 or
-x<2>-2

So, when x > 0 => x < 2
And if x <0> x > -2

But from stmt 2
for x > 0 => x = 3x - 2
2x = 2 : x = 1
for x <0> -x = 3x - 2
4x = 2 : x = 1/2

here with x<0 we r gettin x=1/2 ????
Manager
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21 Oct 2007, 09:24
Yes you're right.
Because when you take the absolute into consideration, the equations is:
-x=3x-2 --> -4x=-2--> x=1/2
Manager
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22 Oct 2007, 09:27
But How can B alone be sufficient? We get two values - x=1 or x=1/2

Eyunni- could you please share the source of this question?
Manager
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22 Oct 2007, 10:04
The question probably assumed X was an integer.
Manager
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22 Oct 2007, 10:23
But in DS how can we ASSUME x is an integer? I thought we had to be told if this was the case. Am i wrong?
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22 Oct 2007, 11:10
(B) it is ... I didnt look in all details of the explanations given precedently...

What is x?
(1) |x| < 2
(2) |x| = 3x – 2

Stat1
|x| < 2
<=> -2 < x < 2

INSUFF.

Stat1
|x| = 3x – 2

o If x >= 0, then
|x| = 3x – 2
<=> x = 3x - 2
<=> x = 1 >>>> Solution ok as 1 > 0 (x>=0)

o If x < 0, then
|x| = 3x – 2
<=> -x = 3x - 2
<=> x = 1/2 >>>> Solution out as x must be negative 0

So, we have unique solution for x : 1

SUFF.
SVP
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22 Oct 2007, 11:12
But in DS how can we ASSUME x is an integer? I thought we had to be told if this was the case. Am i wrong?

Actually u are right. If not stated, we can never assume that a variable is an integer.

Here, it's just because x = 1/2 is ruled out as x < 0.
Director
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22 Oct 2007, 11:43
But How can B alone be sufficient? We get two values - x=1 or x=1/2

Eyunni- could you please share the source of this question?

That was a touch tricky. I don't remember the source exactly but I guess it is from MGMAT/Kaplan.
Manager
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25 Nov 2008, 11:15
B

1--> -2<x<2 insuff

2-->
x=3x-2, x>0
x=2-3x, x<0

for x<0, 2-3x is allways +ve--> thus x>0

sufficient
VP
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25 Nov 2008, 20:24
Fig wrote:
(B) it is ... I didnt look in all details of the explanations given precedently...

What is x?
(1) |x| < 2
(2) |x| = 3x – 2

Stat1
|x| < 2
<=> -2 < x < 2

INSUFF.

Stat1
|x| = 3x – 2

o If x >= 0, then
|x| = 3x – 2
<=> x = 3x - 2
<=> x = 1 >>>> Solution ok as 1 > 0 (x>=0)

o If x < 0, then
|x| = 3x – 2
<=> -x = 3x - 2
<=> x = 1/2 >>>> Solution out as x must be negative 0

So, we have unique solution for x : 1

SUFF.

Whew! What the Q did is lot of trickery. Lets us start with x < 0 and arrive at x > 0 and then rule out that case as it leads to a contradiction.
Re:   [#permalink] 25 Nov 2008, 20:24
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