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Is x > 0

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Is x > 0

by mehravikas » Sat Aug 01, 2009 10:27 pm
Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x + 1| = 2x – 1

Please provide explanation.

Source: MGMAT
OA: D


I answered it correctly but I am confused after going through MGMAT explanation. MGMAT says that, consider x to be negative, to do that multiple (x + 3) by -1 -> -1(x + 3) = 4x – 3

From what I remember, if we have to consider x as negative, then the equation should be: -x + 3 = 4x - 3

Please correct me if I am wrong.
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Solve step by step

by john555 » Sun Aug 02, 2009 6:40 am
statement -1
first consider : (x+3) = 4x-3 ; solve, X = 2
then -(x+3) = 4x-3 ; solve , X =0
So x is either 2 or 0
similarly for stament -2
X is either 2 or 0
therefore, the answer is D
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Absolute values

by ahmad.kadry » Sun Aug 02, 2009 7:08 am
Hi, i believe it wasn't meant to set X to a negative value. usually when you solve an absolute value problem, the full expression within the absolute value might equal a positive or a negative value. for example:

|3| = |-3| = 3

So in this case: |(x+3)| = |-(x+3)| = x+3, Hence, you will have two cases for the original value for the expression within the absolute value:

1 - |x+3| = +(x+3) -> x+3 = 4x - 3
2 - |x+3| = -(x+3) -> -x-3 = 4x - 3

And in both cases solve for x

I hope that helped.
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by shahdevine » Sun Aug 02, 2009 7:22 am
don't like this question.

in both statements, x=0 or 2 or x=0 or 1. How could each statement alone be sufficient if each statement has two possibilities?

Isn't it after combining that we know x=0 definitively and so answer is C.

Someone pls explain.
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by ahmad.kadry » Sun Aug 02, 2009 7:37 am
Hi shahdevine,

In my opinion the full solution will look like this.

For statement 1, we have A the +ve case and B the -ve case.
1-A: x + 3 = 4x - 3 -> 3x = 6 -> x = 2
1-B: -x - 3 = 4x - 3 -> 5x = 0 -> x = 0

So, statement 1 gives us x = {0,2), however, after solving the absolute value you should validate the output by replacing the value of x by each of the outputs. In this case,
- if we replace x = 0 -> |0 + 3| = 4(0) – 3 -> 3 = -3 ...so apparently 0 is not a correct value
- if we replace x = 2 -> |2 + 3| = 4(2) - 3 -> 5 = 5 ...so 2 is a correct value... and as a result ..this statement is sufficient.

For statement 2, we have A the +ve case and B the -ve case.
1-A: x + 1 = 2x - 1 -> x = 2
1-B: -x - 1 = 2x - 1 -> 3x = 0 -> x = 0

So, statement 2 also gives us x = {0.2} .. validation will be
- replacing x = 0 -> |0 + 1| = 2(0) - 1 -> 1 = -1 ..so incorrect
- replacing x = 2 -> |2 + 1| = 2(2) - 1 -> 3 = 3 ...correct

So x can be only 2...sufficient.

The key is to validate the output.
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Re: Absolute values

by mehravikas » Sun Aug 02, 2009 4:09 pm
So does that mean, when evaluating negative value of X, the following are:

Expression: |x + 3|

1. -x + 3 - is incorrect
2. -x - 3 - is correct

ahmad.kadry wrote:Hi, i believe it wasn't meant to set X to a negative value. usually when you solve an absolute value problem, the full expression within the absolute value might equal a positive or a negative value. for example:

|3| = |-3| = 3

So in this case: |(x+3)| = |-(x+3)| = x+3, Hence, you will have two cases for the original value for the expression within the absolute value:

1 - |x+3| = +(x+3) -> x+3 = 4x - 3
2 - |x+3| = -(x+3) -> -x-3 = 4x - 3

And in both cases solve for x

I hope that helped.
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by tohellandback » Sun Aug 02, 2009 10:18 pm
IMO D,
|X-a| is the distance between point X and point a and distance cannot be negative

(1) |x + 3| = 4x – 3
4x-3>=0, x>=3/4
sufficient, X>0

2)|x + 1| = 2x – 1
2x-1>=0
x>=1/2
SUFFICIENT
answer D

the question could be made complicated if right side involved modulus
Last edited by tohellandback on Mon Aug 03, 2009 1:09 am, edited 1 time in total.
The powers of two are bloody impolite!!
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by mehravikas » Sun Aug 02, 2009 11:58 pm
Sorry but its not clear. Can you elaborate please?
tohellandback wrote:IMO D,
|X-a| is the distance between point X and point a and distance cannot be negative

(1) |x + 3| = 4x – 3
4x-3>0, x>3/4
sufficient, X>0

2)|x + 1| = 2x – 1
2x-1>0
x>1/2
SUFFICIENT
answer D

the question could be made complicated if right side involved modulus
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by tohellandback » Mon Aug 03, 2009 12:03 am
mehravikas wrote:Sorry but its not clear. Can you elaborate please?
tohellandback wrote:IMO D,
|X-a| is the distance between point X and point a and distance cannot be negative

(1) |x + 3| = 4x – 3
4x-3>0, x>3/4
sufficient, X>0

2)|x + 1| = 2x – 1
2x-1>0
x>1/2
SUFFICIENT
answer D

the question could be made complicated if right side involved modulus
what I mean is:
|5-3|=2, which is the distance between 5 and 3 on the number line
|5-(-3)|=8 is the distance between 5 and -3.
|x| is the distance of any point X from the origin
|X-a| is the distance of any point X from a.

when we say |X| is always positive, we are basically saying that the distance between two points cannot be negative. Of course. Distance is a quantity and it cannot be negative.
and that is my solution
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by ahmad.kadry » Mon Aug 03, 2009 12:34 am
Hi tohellandback,

You are correct about the distance designation, however, IMO, the distance can be either positive or negative but the absolute value of that distance is only positive.

In more details, consider x = 5, a point that is 2 units away from x on the number line can be either "7" or "3", so the distance can be either:
5 - 7 = -2 or
5 - 3 = +2

however to indicate the absolute value -how many units away- we use the absolute value operator. so |-2| = |2| = 2.

Thats is why (x + 3) which is the value between the || operator can be evaluated to either (x+3) or -(x+3).. the absolute value for both expressions |x+3| or |-(x+3)| will result into the same number of units.
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by tohellandback » Mon Aug 03, 2009 12:38 am
ahmad.kadry wrote:Hi tohellandback,

You are correct about the distance designation, however, IMO, the distance can be either positive or negative but the absolute value of that distance is only positive.

In more details, consider x = 5, a point that is 2 units away from x on the number line can be either "7" or "3", so the distance can be either:
5 - 7 = -2 or
5 - 3 = +2

however to indicate the absolute value -how many units away- we use the absolute value operator. so |-2| = |2| = 2.

Thats is why (x + 3) which is the value between the || operator can be evaluated to either (x+3) or -(x+3).. the absolute value for both expressions |x+3| or |-(x+3)| will result into the same number of units.
ummm no.
distance is not x-a, but |x-a| which can never be negative.
I think its getting complicated. Let's just say that "|X-a| is the distance of any point X from any point a and it is always positive". That should solve all the problems
The powers of two are bloody impolite!!
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by ahmad.kadry » Mon Aug 03, 2009 12:57 am
tohellandback wrote:
ahmad.kadry wrote:Hi tohellandback,

You are correct about the distance designation, however, IMO, the distance can be either positive or negative but the absolute value of that distance is only positive.

In more details, consider x = 5, a point that is 2 units away from x on the number line can be either "7" or "3", so the distance can be either:
5 - 7 = -2 or
5 - 3 = +2

however to indicate the absolute value -how many units away- we use the absolute value operator. so |-2| = |2| = 2.

Thats is why (x + 3) which is the value between the || operator can be evaluated to either (x+3) or -(x+3).. the absolute value for both expressions |x+3| or |-(x+3)| will result into the same number of units.
ummm no.
distance is not x-a, but |x-a| which can never be negative.
I think its getting complicated. Let's just say that "|X-a| is the distance of any point X from any point a and it is always positive". That should solve all the problems

:) no, we have to distinguish between the value of an expression and the output of the absolute value containing that expression....the distance can be +ve or -ve, because the distance here is an amount in a certain direction, and thats why we use the abs || .. the "output" of the absolute value operator represents the quantity as u mentioned.. that quantity is the distance in terms of "how far" regardless of the direction.

consider the simple case of |x|, if x =2 -> |2| = 2 ...and if x = -2 -> |-2| = 2 as well.

The following links elaborate on that explanation.

https://www.purplemath.com/modules/absolute.htm (explains the concept)
https://www.purplemath.com/modules/solveabs.htm (contains a similar example)
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by tohellandback » Mon Aug 03, 2009 1:00 am
ahmad.kadry wrote:
tohellandback wrote:
ahmad.kadry wrote:Hi tohellandback,

You are correct about the distance designation, however, IMO, the distance can be either positive or negative but the absolute value of that distance is only positive.

In more details, consider x = 5, a point that is 2 units away from x on the number line can be either "7" or "3", so the distance can be either:
5 - 7 = -2 or
5 - 3 = +2

however to indicate the absolute value -how many units away- we use the absolute value operator. so |-2| = |2| = 2.

Thats is why (x + 3) which is the value between the || operator can be evaluated to either (x+3) or -(x+3).. the absolute value for both expressions |x+3| or |-(x+3)| will result into the same number of units.
ummm no.
distance is not x-a, but |x-a| which can never be negative.
I think its getting complicated. Let's just say that "|X-a| is the distance of any point X from any point a and it is always positive". That should solve all the problems

:) no, we have to distinguish between the value of an expression and the output of the absolute value containing that expression....the distance can be +ve or -ve, because the distance here is an amount in a certain direction, and thats why we use the abs || .. the "output" of the absolute value operator represents the quantity as u mentioned.. that quantity is the distance in terms of "how far" regardless of the direction.

consider the simple case of |x|, if x =2 -> |2| = 2 ...and if x = -2 -> |-2| = 2 as well.

The following links elaborate on that explanation.

https://www.purplemath.com/modules/absolute.htm (explains the concept)
https://www.purplemath.com/modules/solveabs.htm (contains a similar example)
buddy I think you are getting me wrong..
when I say distance it is |X-a|. Now (X-a) can be negative or positive. But the distance that is |X-a| is never negative.
The powers of two are bloody impolite!!
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by ahmad.kadry » Mon Aug 03, 2009 1:05 am
tohellandback wrote:
ahmad.kadry wrote:
tohellandback wrote:
ahmad.kadry wrote:Hi tohellandback,

You are correct about the distance designation, however, IMO, the distance can be either positive or negative but the absolute value of that distance is only positive.

In more details, consider x = 5, a point that is 2 units away from x on the number line can be either "7" or "3", so the distance can be either:
5 - 7 = -2 or
5 - 3 = +2

however to indicate the absolute value -how many units away- we use the absolute value operator. so |-2| = |2| = 2.

Thats is why (x + 3) which is the value between the || operator can be evaluated to either (x+3) or -(x+3).. the absolute value for both expressions |x+3| or |-(x+3)| will result into the same number of units.
ummm no.
distance is not x-a, but |x-a| which can never be negative.
I think its getting complicated. Let's just say that "|X-a| is the distance of any point X from any point a and it is always positive". That should solve all the problems

:) no, we have to distinguish between the value of an expression and the output of the absolute value containing that expression....the distance can be +ve or -ve, because the distance here is an amount in a certain direction, and thats why we use the abs || .. the "output" of the absolute value operator represents the quantity as u mentioned.. that quantity is the distance in terms of "how far" regardless of the direction.

consider the simple case of |x|, if x =2 -> |2| = 2 ...and if x = -2 -> |-2| = 2 as well.

The following links elaborate on that explanation.

https://www.purplemath.com/modules/absolute.htm (explains the concept)
https://www.purplemath.com/modules/solveabs.htm (contains a similar example)
buddy I think you are getting me wrong..
when I say distance it is |X-a|. Now (X-a) can be negative or positive. But the distance that is |X-a| is never negative.
hmmm.. okay.. that is right..i guess i got you wrong... probably because i was just thinking in the frame of explaining to mehravikas why MGMAT interpreted the expression by multiplying -1(4+3).
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