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# Is x > 0? (1) |x + 3| = 4x 3 (2) |x 3| = |2x 3

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Director
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Is x > 0? (1) |x + 3| = 4x 3 (2) |x 3| = |2x 3 [#permalink]

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24 Nov 2007, 05:37
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x – 3| = |2x – 3|
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GMAT the final frontie!!!.

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24 Nov 2007, 06:11
Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x – 3| = |2x – 3|

D for me.
I will explain if correct.
Director
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24 Nov 2007, 07:10
Any more takers. ??
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Director
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24 Nov 2007, 08:26
Ravshonbek wrote:
Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x – 3| = |2x – 3|

D for me.
I will explain if correct.

Yes, I think you are right.. Here is my reasoning.. Please let me know what you think and then we can sync up whose strategy is better ..

I am using the concept of critical point as mentioned in one of the Doc that I have..

St1 : |x + 3| = 4x – 3

the critical point of x here is -3. hence the value of x could be in the range x < -3, -3 < x <3> 3.

a) when x < -3, (x+3) is -ve. hence s1 can be written as
-(x+3) = 4x -3
-x +3 = 4x -3
5x = 0
x = 0 which is not true since x < 3.

b) When -3 < x <3> -3 and <3> 3. Not needed as we already got the value of 2 and this will anyway not hold true since x cannot be equal to 2 and greater than 3 at the same time. hence x = 2 which is greater than 0. hence S1 is suff..

S2 : |x – 3| = |2x – 3|

|x – 3| - |2x – 3| = 0
As above we will check the range of x.
the CP is 3 and 3/2.

Hence 3 scenarios.
a) x > 3, b) x < 3/2 and c) 3/2 < x <3> 3, (x-3) and (2x-3 ) +ve.
(x-3) - (2x-3 ) = 0
=> x = 0 which is not true.

b) when x <3> x =0 Which again is not true..

c) When 3/2 < x < 3, (x-3) -ve and (2x-3) +ve
-(x-3) - (2x -3) = 0
-x + 3 -2x +3 = 0
-3x +6 = 0
x =2. Which is true ..
Hence s2 is also suff.

Hence D.

What say Ravshonbek ??
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24 Nov 2007, 08:27
Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x – 3| = |2x – 3|

The only value I felt that both 1 and 2 worked for me was
X = 2....I guess its D for me too!!!!
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24 Nov 2007, 11:11
Amit05 wrote:
Ravshonbek wrote:
Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x – 3| = |2x – 3|

D for me.
I will explain if correct.

Yes, I think you are right.. Here is my reasoning.. Please let me know what you think and then we can sync up whose strategy is better ..

I am using the concept of critical point as mentioned in one of the Doc that I have..

St1 : |x + 3| = 4x – 3

the critical point of x here is -3. hence the value of x could be in the range x < -3, -3 < x <3> 3.

a) when x < -3, (x+3) is -ve. hence s1 can be written as
-(x+3) = 4x -3
-x +3 = 4x -3
5x = 0
x = 0 which is not true since x < 3.

b) When -3 < x <3> -3 and <3> 3. Not needed as we already got the value of 2 and this will anyway not hold true since x cannot be equal to 2 and greater than 3 at the same time. hence x = 2 which is greater than 0. hence S1 is suff..

S2 : |x – 3| = |2x – 3|

|x – 3| - |2x – 3| = 0
As above we will check the range of x.
the CP is 3 and 3/2.

Hence 3 scenarios.
a) x > 3, b) x < 3/2 and c) 3/2 < x <3> 3, (x-3) and (2x-3 ) +ve.
(x-3) - (2x-3 ) = 0
=> x = 0 which is not true.

b) when x <3> x =0 Which again is not true..

c) When 3/2 < x < 3, (x-3) -ve and (2x-3) +ve
-(x-3) - (2x -3) = 0
-x + 3 -2x +3 = 0
-3x +6 = 0
x =2. Which is true ..
Hence s2 is also suff.

Hence D.

What say Ravshonbek ??

what i normally do with absolutes is simplly drawing it on a scratch paper. it takes the same time to do it algebraically.

first one is simple no drawing is needed x=2.

second, i drew two function on xy plane and found that y=|x – 3|, y=|2x – 3| intersect or are equal when x=2 so suff.

hence D
Director
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24 Nov 2007, 11:15
Ravshonbek wrote:
Amit05 wrote:
Ravshonbek wrote:
Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x – 3| = |2x – 3|

D for me.
I will explain if correct.

Yes, I think you are right.. Here is my reasoning.. Please let me know what you think and then we can sync up whose strategy is better ..

I am using the concept of critical point as mentioned in one of the Doc that I have..

St1 : |x + 3| = 4x – 3

the critical point of x here is -3. hence the value of x could be in the range x < -3, -3 < x <3> 3.

a) when x < -3, (x+3) is -ve. hence s1 can be written as
-(x+3) = 4x -3
-x +3 = 4x -3
5x = 0
x = 0 which is not true since x < 3.

b) When -3 < x <3> -3 and <3> 3. Not needed as we already got the value of 2 and this will anyway not hold true since x cannot be equal to 2 and greater than 3 at the same time. hence x = 2 which is greater than 0. hence S1 is suff..

S2 : |x – 3| = |2x – 3|

|x – 3| - |2x – 3| = 0
As above we will check the range of x.
the CP is 3 and 3/2.

Hence 3 scenarios.
a) x > 3, b) x < 3/2 and c) 3/2 < x <3> 3, (x-3) and (2x-3 ) +ve.
(x-3) - (2x-3 ) = 0
=> x = 0 which is not true.

b) when x <3> x =0 Which again is not true..

c) When 3/2 < x < 3, (x-3) -ve and (2x-3) +ve
-(x-3) - (2x -3) = 0
-x + 3 -2x +3 = 0
-3x +6 = 0
x =2. Which is true ..
Hence s2 is also suff.

Hence D.

What say Ravshonbek ??

what i normally do with absolutes is simplly drawing it on a scratch paper. it takes the same time to do it algebraically.

first one is simple no drawing is needed x=2.

second, i drew two function on xy plane and found that y=|x – 3|, y=|2x – 3| intersect or are equal when x=2 so suff.

hence D

Can you elaborate more on it or have any Doc that you can send me ..??
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24 Nov 2007, 15:58
Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x – 3| = |2x – 3|

In stat2 when x is less than 3 / 2 we have x = 0 which is an acceptable solution. I didnt understand why you eliminated this possibility.

VP
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24 Nov 2007, 17:12
ronneyc wrote:
Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x – 3| = |2x – 3|

In stat2 when x is less than 3 / 2 we have x = 0 which is an acceptable solution. I didnt understand why you eliminated this possibility.

that's a good point, i did not notice it when drawing the function on the plane.

does that leave us with A then?
Amit, i will send you a doc. later
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24 Nov 2007, 20:57
Ravshonbek wrote:
ronneyc wrote:
Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x – 3| = |2x – 3|

In stat2 when x is less than 3 / 2 we have x = 0 which is an acceptable solution. I didnt understand why you eliminated this possibility.

that's a good point, i did not notice it when drawing the function on the plane.

does that leave us with A then?
Amit, i will send you a doc. later

Thanks man.. I sent you a PM on where to send the Doc...
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25 Nov 2007, 01:36
I think the ans. is A
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25 Nov 2007, 02:30
Ravshonbek wrote:
ronneyc wrote:
Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x – 3| = |2x – 3|

In stat2 when x is less than 3 / 2 we have x = 0 which is an acceptable solution. I didnt understand why you eliminated this possibility.

that's a good point, i did not notice it when drawing the function on the plane.

does that leave us with A then?
Amit, i will send you a doc. later

I'm sorry for interposing myself here, may I ask to have the same document sent?
If it's not a problem, I will appreciate it very much. Thank you.
Re: Absolute Value   [#permalink] 25 Nov 2007, 02:30
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