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Updated on: 11 Jul 2013, 00:52
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What is x?

(1) |x| < 2
(2) |x| = 3x – 2

Originally posted by rohitgoel15 on 31 Jan 2012, 18:48.
Last edited by Bunuel on 11 Jul 2013, 00:52, edited 1 time in total.
Edited the question.
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31 Jan 2012, 18:56
17
15
rohitgoel15 wrote:
What is x?
(1) |x| < 2
(2) |x| = 3x – 2

Can anyone justify the OA?

What is x?

(1) |x| < 2 --> $$-2<x<2$$. Not sufficient.

(2) |x| = 3x – 2 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: $$3x-2\geq{0}$$ --> $$x\geq{\frac{2}{3}}$$ --> $$x$$ positive, so $$|x|=x$$ --> $$x=3x-2$$ --> $$x=1$$. Sufficient

Hope it's clear.
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Re: What is x? (1) |x| < 2 (2) |x| = 3x – 2  [#permalink]

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31 Jan 2012, 20:38
Vow! that is a very good solution It opened my eyes on how to read modulus based problems...
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01 Feb 2012, 03:42
2
Bunuel wrote:
rohitgoel15 wrote:
What is x?
(1) |x| < 2
(2) |x| = 3x – 2

Can anyone justify the OA?

What is x?

(1) |x| < 2 --> $$-2<x<2$$. Not sufficient.

(2) |x| = 3x – 2 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: $$3x-2\geq{0}$$ --> $$x\geq{\frac{2}{3}}$$ --> $$x$$ positive, so $$|x|=x$$ --> $$x=3x-2$$ --> $$x=1$$. Sufficient

Hope it's clear.

Hi Bunuel,

Thanks for reply .. But this is not clear to me.
Generally in modulus questions we do as below:
(1) |x| < 2
I agree with u. -2<x<2
(2) |x| = 3x – 2
In this case i did
a. x is positive so x = 3x - 2 , upon solving x = 1
a. x is negative so -x = 3x - 2 , upon solving x = 1/2

Why have you considered the bold part in this particular question. for eg. if |x| = 2 we always consider x = +2/-2.
I am confused !
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01 Feb 2012, 04:05
7
5
rohitgoel15 wrote:
Bunuel wrote:
rohitgoel15 wrote:
What is x?
(1) |x| < 2
(2) |x| = 3x – 2

Can anyone justify the OA?

What is x?

(1) |x| < 2 --> $$-2<x<2$$. Not sufficient.

(2) |x| = 3x – 2 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: $$3x-2\geq{0}$$ --> $$x\geq{\frac{2}{3}}$$ --> $$x$$ positive, so $$|x|=x$$ --> $$x=3x-2$$ --> $$x=1$$. Sufficient

Hope it's clear.

Hi Bunuel,

Thanks for reply .. But this is not clear to me.
Generally in modulus questions we do as below:
(1) |x| < 2
I agree with u. -2<x<2
(2) |x| = 3x – 2
In this case i did
a. x is positive so x = 3x - 2 , upon solving x = 1
a. x is negative so -x = 3x - 2 , upon solving x = 1/2

Why have you considered the bold part in this particular question. for eg. if |x| = 2 we always consider x = +2/-2.
I am confused !

Your way is more common, orthodox. Though you should discard x=1/2 as you get this value when considering x<0 and 1/2 is not less than 0. Also, if x=1/2 then 3x-2<0 and you get that |x|=negative, which is never true.

Next, the way I'm doing is based on the fact that absolute value is always non negative, thus 3x-2 which equals to some absolute value must also be non-negative: $$3x-2\geq{0}$$ --> $$x\geq{\frac{2}{3}}$$ --> $$x$$ positive, so $$|x|=x$$ and no need to consider $$|x|=-x$$ --> $$x=3x-2$$ --> $$x=1$$.

Check Absolute Values chapter of Math Book for more: math-absolute-value-modulus-86462.html

Hope it helps.
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Re: What is x? (1) |x| < 2 (2) |x| = 3x – 2  [#permalink]

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11 Jul 2013, 00:51
1
Think of statement 2 this way:

If we consider x to be negative, then -x=3x-2 which translates to x=1/2. Now when we test this situation 1/2=3*1/2-2 the outcome is =3/2-2= -1/2. The RHS is not equal to the LHS. Hence x has to be 1, because when we test x=3x-2 we get x=1.
and the RHS and LHS are equal to each other
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11 Jul 2013, 11:22
4
What is x?

(1) |x| < 2

x<2 x>-2
Invalid, x could be any number greater than 2 or less than 2.
INSUFFICIENT

(2) |x| = 3x – 2
x>=0
|x| = 3x – 2
x = 3x - 2
-2x = -2
x = 1 Valid
OR
x<0
-x = 3x - 2
-4x = -2
x = (1/2) Invalid as 1/2 is not less than zero.
SUFFICIENT

(B)
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28 Nov 2013, 18:35
Bunuel wrote:
rohitgoel15 wrote:
What is x?
(1) |x| < 2
(2) |x| = 3x – 2

Can anyone justify the OA?

What is x?

(1) |x| < 2 --> $$-2<x<2$$. Not sufficient.

(2) |x| = 3x – 2 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: $$3x-2\geq{0}$$ --> $$x\geq{\frac{2}{3}}$$ --> $$x$$ positive, so $$|x|=x$$ --> $$x=3x-2$$ --> $$x=1$$. Sufficient

Hope it's clear.

Hi Bunuel,
I have a question.
(2) |x| = 3x – 2 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: 3x-2\geq{0} --> x\geq{\frac{2}{3}} --> x positive, so |x|=x --> x=3x-2 --> x=1. Sufficient

If it is true as above then why we are taking as -2<X<2 for |x| ,i mean why we are considering negative value (<0) for |X|?

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28 Nov 2013, 19:53
sunita123 wrote:
Bunuel wrote:
rohitgoel15 wrote:
What is x?
(1) |x| < 2
(2) |x| = 3x – 2

Can anyone justify the OA?

What is x?

(1) |x| < 2 --> $$-2<x<2$$. Not sufficient.

(2) |x| = 3x – 2 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: $$3x-2\geq{0}$$ --> $$x\geq{\frac{2}{3}}$$ --> $$x$$ positive, so $$|x|=x$$ --> $$x=3x-2$$ --> $$x=1$$. Sufficient

Hope it's clear.

Hi Bunuel,
I have a question.
(2) |x| = 3x – 2 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: 3x-2\geq{0} --> x\geq{\frac{2}{3}} --> x positive, so |x|=x --> x=3x-2 --> x=1. Sufficient

If it is true as above then why we are taking as -2<X<2 for |x| ,i mean why we are considering negative value (<0) for |X|?

We never considered negative value for |x|.... we considered negative value for just x...... By definition, mod never has a negative value....
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25 Mar 2014, 21:38
rohitgoel15 wrote:
What is x?

(1) |x| < 2
(2) |x| = 3x – 2

When absolute values are given I guess it's better to go for number substitution.

1) |x| < 2 implies x can be either 1 or -1. 0 won't come here as it is neither positive nor negative and there is no need to take it here.
2) |x| = 3x - 2 is possible only when x = 1.

So, the final answer will be B
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15 Jul 2015, 10:53
Hi Bunuel

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

Can anyone justify the OA?

Your way is more common, orthodox. Though you should discard x=1/2 as you get this value when considering x<0 and 1/2 is not less than 0. Also, if x=1/2 then 3x-2<0 and you get that |x|=negative, which is never true.

Next, the way I'm ding is based on the fact that absolute value is always non negative, thus 3x-2 which equals to some absolute value must also be non-negative: 3x-2>=0 --> x>=2/3 --> x positive, so |x|=x and no need to consider |x|=-x --> x=3x-2 --> x=1.

ok my concern is if by cosidering |x| as positive and solving equation , if we get x as -negative value then , will we put |x| = -x and solve equation as we did in case of +x. please clarify
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15 Jul 2015, 11:16
2
Hi Bunuel

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

Can anyone justify the OA?

Your way is more common, orthodox. Though you should discard x=1/2 as you get this value when considering x<0 and 1/2 is not less than 0. Also, if x=1/2 then 3x-2<0 and you get that |x|=negative, which is never true.

Next, the way I'm ding is based on the fact that absolute value is always non negative, thus 3x-2 which equals to some absolute value must also be non-negative: 3x-2>=0 --> x>=2/3 --> x positive, so |x|=x and no need to consider |x|=-x --> x=3x-2 --> x=1.

ok my concern is if by cosidering |x| as positive and solving equation , if we get x as -negative value then , will we put |x| = -x and solve equation as we did in case of +x. please clarify

I dont think so. You need to follow the following 2 step procedure for ALL absolute value questions:

Treat |x| as following:

1. Let x $$\geq$$0 ----> |x| = x ---> x=3x-2 ---> x=1 and as x=1 satisfies x $$\geq$$0, this is a valid answer.

2. Let |x| < 0 ----> |x| = -x ----> -x=3x+2 ---> x = 1/2 and as x= 1/2 does not satisfy x<0, this solution is rejected. Thus we only 1 value, x = 1 and thus combining the 2 statements is sufficient. Thus C is the correct answer.

ok my concern is if by cosidering |x| as positive and solving equation , if we get x as -negative value then , will we put |x| = -x and solve equation as we did in case of +x. please clarify

If you consider |x| as positive (or >0) and you get x <0 , then this solution is not possible as you have 2 conflicting scenarios.

Hope this helps.
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15 Jul 2015, 22:48
HI Enge2012
I completely agree what you have mentioned, and that is way we normally solve mod equation, but if you see the Bunuel solution, he consider |x| as positive first and put equation >= 0 through which he gets +ve X and gain he solve the equation. Now my concern is over above equation what if we get -ve value of X example
instead of 3x-2>=0 if we have equation as 3x +2 >=0 now we get x as negative. hope i am cleared with doubt.
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23 Aug 2015, 10:07
OA:

(1) INSUFFICIENT: This expression provides only a range of possible values for x.

(2) SUFFICIENT: Absolute value problems often -- but not always -- have multiple solutions because the expression within the absolute value bars can be either positive or negative even though the absolute value of the expression is always positive. For example, if we consider the equation |2 + x| = 3, we have to consider the possibility that 2 + x is already positive and the possibility that 2 + x is negative. If 2 + x is positive, then the equation is the same as 2 + x = 3 and x = 1. But if 2 + x is negative, then it must equal -3 (since |-3| = 3) and so 2 + x = -3 and x = -5.

So in the present case, in order to determine the possible solutions for x, it is necessary to solve for x under both possible conditions.

For the case where x > 0:

x = 3x – 2
-2x = -2
x = 1

When dealing with absolute values that contain variables, you should always check that the solution is valid. Plug x = 1 into the original equation.

|1| = 3(1) – 2
1 = 3 – 2

The equation is true, so the solution is valid.

For the case when x < 0:

x = -1(3x – 2) We multiply by -1 to make x equal a negative quantity.
x = 2 – 3x
4x = 2
x = 1/2

Once again, we need to test whether the solution is valid. Plug x = 1/2 into the original equation.

|1/2| = 3(1/2) – 2
1/2 = 3/2 – 2
1/2 = -1/2

This equation is not true, so x = 1/2 is actually not a valid solution. That means that there is only one solution to the equation. x = 1. The statement is sufficient.

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07 May 2016, 08:06
OA:

(1) INSUFFICIENT: This expression provides only a range of possible values for x.

(2) SUFFICIENT: Absolute value problems often -- but not always -- have multiple solutions because the expression within the absolute value bars can be either positive or negative even though the absolute value of the expression is always positive. For example, if we consider the equation |2 + x| = 3, we have to consider the possibility that 2 + x is already positive and the possibility that 2 + x is negative. If 2 + x is positive, then the equation is the same as 2 + x = 3 and x = 1. But if 2 + x is negative, then it must equal -3 (since |-3| = 3) and so 2 + x = -3 and x = -5.

So in the present case, in order to determine the possible solutions for x, it is necessary to solve for x under both possible conditions.

For the case where x > 0:

x = 3x – 2
-2x = -2
x = 1

When dealing with absolute values that contain variables, you should always check that the solution is valid. Plug x = 1 into the original equation.

|1| = 3(1) – 2
1 = 3 – 2

The equation is true, so the solution is valid.

For the case when x < 0:

x = -1(3x – 2) We multiply by -1 to make x equal a negative quantity.
x = 2 – 3x
4x = 2
x = 1/2

Once again, we need to test whether the solution is valid. Plug x = 1/2 into the original equation.

|1/2| = 3(1/2) – 2
1/2 = 3/2 – 2
1/2 = -1/2

This equation is not true, so x = 1/2 is actually not a valid solution. That means that there is only one solution to the equation. x = 1. The statement is sufficient.

Shouldn't the statement 2 have a true definite answer since it consist of 2 answer x=1 and x=1/2
could you please provide a further explanation on this
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Updated on: 07 May 2016, 08:57
satishchaudharygmat Statement 2 gives us two answers but, only one satisfies the equation. Whenever, you find answers in modulus questions, always check whether the value satisfies the equation. In this question 1/2 will not satisfy the equation from which 1/2 is derived.

Hope this helped.

Originally posted by bimalr9 on 07 May 2016, 08:17.
Last edited by bimalr9 on 07 May 2016, 08:57, edited 2 times in total.
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07 May 2016, 08:20
rohitgoel15 wrote:
What is x?

(1) |x| < 2
(2) |x| = 3x – 2

statement 1 gives you value of x b/w 1 to - 1 not sufficient.
statement 2 gives you 2 values x=1 and 1/2. but you will 1/2 in the equation LHS will not be equal to RHS. so we can cancel this value. so X=1
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30 Aug 2016, 18:41
St1: |x| < 2

This implies that -2 < x < 2.
Insufficient.

St2: |x| = 3x – 2

Using one of Bunuel's legendary tricks:

Since absolute value is always >= 0, the expression on right hand side will be >= 0.

3x - 2 >= 0
x >= 2/3, meaning x is positive.

So, we can only check for one condition.

If x > 0

x = 3x - 2
2x = 2
x = 1 -> satisfies the condition that x is positive.
Sufficient.

The trick is legendary because it saves us from checking the condition x is negative. Let's see what will happen if we do check:

If x < 0

x = 2 - 3x
4x = 2
x = 1/2 -> does not satisfy the condition that x is negative.

As you can see, the trick saves time and can make a big difference in the actual exam.
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30 Aug 2016, 20:30
neverplayd wrote:
What is x?

(1) |x| < 2

(2) |x| = 3x – 2

Question : x = ?

Statement 1: |x| < 2
i.e. -2 < x < 2
NOT SUFFICIENT

Statement 1: |x| = 3x – 2
i.e. x = 3x – 2 and -x = 3x – 2
i.e. x = 1 and x = 1/2
But 1/2 doesn't satisfy the given statement hence that is not acceptable. Therefore,
x=2 is the only acceptable solution

SUFFICIENT

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30 Aug 2016, 22:45
neverplayd wrote:
What is x?

(1) |x| < 2

(2) |x| = 3x – 2

Spoiler: :: Solution
(1) INSUFFICIENT: This expression provides only a range of possible values for x.

(2) SUFFICIENT: Absolute value problems often -- but not always -- have multiple solutions because the expression within the absolute value bars can be either positive or negative even though the absolute value of the expression is always positive. For example, if we consider the equation |2 + x| = 3, we have to consider the possibility that 2 + x is already positive and the possibility that 2 + x is negative. If 2 + x is positive, then the equation is the same as 2 + x = 3 and x = 1. But if 2 + x is negative, then it must equal -3 (since |-3| = 3) and so 2 + x = -3 and x = -5.

So in the present case, in order to determine the possible solutions for x, it is necessary to solve for x under both possible conditions.

For the case where x > 0:

x = 3x – 2
-2x = -2
x = 1

When dealing with absolute values that contain variables, you should always check that the solution is valid. Plug x = 1 into the original equation.

|1| = 3(1) – 2
1 = 3 – 2

The equation is true, so the solution is valid.

For the case when x < 0:

x = -1(3x – 2) We multiply by -1 to make x equal a negative quantity.
x = 2 – 3x
4x = 2
x = 1/2

Once again, we need to test whether the solution is valid. Plug x = 1/2 into the original equation.

|1/2| = 3(1/2) – 2
1/2 = 3/2 – 2
1/2 = -1/2

This equation is not true, so x = 1/2 is actually not a valid solution. That means that there is only one solution to the equation. x = 1. The statement is sufficient.

As donnie mentioned, the primary condition for he problem to be solved is abs(x)>=0, so in the LHS the min value is 0. Applying x=2/3 results LHS as "0" , which implies that x>2/3 and for x=1 alone, the condition answers.
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Re: What is x? &nbs [#permalink] 30 Aug 2016, 22:45

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