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What is x? (1) |x| < 2 (2) |x| = 3x 2 [#permalink ]
20 Oct 2007, 20:50

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

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Re: MGMAT Math Modulus [#permalink ]
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

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Re: MGMAT Math Modulus [#permalink ]
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.

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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

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Perfect explanation from beckee529

.... Nothing to add

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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

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ColumbiaDream wrote:

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 ????

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ColumbiaDream wrote:

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 ????

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Yes you're right.
Because when you take the absolute into consideration, the equations is:
-x=3x-2 --> -4x=-2--> x=1/2

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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?

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The question probably assumed X was an integer.

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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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(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.

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ColumbiaDream wrote:

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.

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ColumbiaDream wrote:

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 .

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Re: MGMAT Math Modulus [#permalink ]
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

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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.