Is x > 0? ; (2) |x – 3| = |2x – 3|

[jp.jprasanna]

Hi Stacey -

Is x > 0?

(1) |x + 3| = 4x - 3
(2) |x - 3| = |2x - 3|
(1) Sufficient
CONDITION :: x+3>0
x + 3 = 4x - 3 or x = 2 ... valid solution (Substituting x=2 makes x+3>0 )

CONDITION :: x+3<0
-(x + 3) = 4x - 3 or x = 0 ... invalid solution. (Substituting x=0 DOESN'T satisfy x+3<0 )

We know that 2 is the only solution possible and we can say that x is definitely positive.

(2) |x - 3| = |2x - 3| - INSUFFICIENT

Im not sure how to check here can u please help.

My guess work -

X-3> 0
X-3 = 2X -3 => x = 0 (Substituting x=0 DOESN'T satisfy X-3> 0)

Is this correct how do i proceed?
Correct Ans A

[njospe2]

Hey,

I would do the following to solve for (2)

CONDITION 1: x+3>0

x - 3 = 2x - 3 or x = 0 ---which is a valid solution

CONDITION 2: x+3<0

-( x - 3) = 2x - 3 or x=2 ---which is also a valid solution

Thus is x > 0 not always - INSUFFICIENT

[jp.jprasanna]

Hey,

I would do the following to solve for (2)

CONDITION 1: x+3>0

x - 3 = 2x - 3 or x = 0 ---which is a valid solution

CONDITION 2: x+3<0

-( x - 3) = 2x - 3 or x=2 ---which is also a valid solution

Thus is x > 0 not always - INSUFFICIENT

Hey- thanks
But can u please let me know why would take the condition x+3>0 while the mods in option B says |x - 3| = |2x - 3|?

Cheers

[bineeshnair4u]

(1) Sufficient
CONDITION :: x+3>0
x + 3 = 4x - 3 or x = 2 ... valid solution (Substituting x=2 makes x+3>0 )

CONDITION :: x+3<0
-(x + 3) = 4x - 3 or x = 0 ... invalid solution. (Substituting x=0 DOESN'T satisfy x+3<0 )

We know that 2 is the only solution possible and we can say that x is definitely positive.

this is correct.. so 1 is sufficient..

(2) |x - 3| = |2x - 3| - INSUFFICIENT

Im not sure how to check here can u please help.

My guess work -

X-3> 0
X-3 = 2X -3 => x = 0 (Substituting x=0 DOESN'T satisfy X-3> 0)

I would go with taking combination of signs. We have 4 choices
A. +=+
B. -=-
C. +=-
D. -=+
A and B are same
C and D are same
so we have to solve only 2 equations

for A,B => x-3 = 2x - 3 => x=0 which is a valid solution
when x=0 then x-3 = 2x-3 => -3 = -3 still valid solution also i eliminates the possibility of A

for C,D => x-3 = -(2x-3) => x=2 which is valid solution
by substituting it back u get
2-3 = 2.2 - 3 => C is invalid but D is valid
still we have 2 option left with us. so 2) is insufficient to find whether x>0 or not

Hence answer A......

[RonPurewal]

hi,
is this problem actually from the official gmat prep software?

if so, please post a screenshot from the software. if not, please cite the correct source, so that we can move the thread to the appropriate folder.

thanks.