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Re: |x-2| > 3 [#permalink]
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good question

Originally posted by boxing506 on 04 Mar 2018, 02:29.
Last edited by boxing506 on 15 Mar 2018, 23:26, edited 1 time in total.
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Re: |x-2| > 3 [#permalink]
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correct : C
|X -2 | > 3 means either x-2 > 3 or x-2< -3
so
x - 2 > 3 -> x > 5 or
x -2 < -3 -> x < -1
if the numbers are integers then we can't choose x =5 and x = -1 :
the minimum possible value of |x - 3.5 | is when x = 6, in this case |x-3.5| equals 2.5
the minimum possible value of |x - 1.5 | is when x = -1, in this case |x-3.5| equals 2.5 again
so considering numbers as integers these are equal.
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Re: |x-2| > 3 [#permalink]
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tigran wrote:
Why can't we put o as x at that time it will be the minimum value of x


0 is not a valid value for |x-2|>3....
Substitute x as 0... |0-2|=2 which is NOT greater than 3
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Re: |x-2| > 3 [#permalink]
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FatemehAsgarinejad wrote:
correct : C
|X -2 | > 3 means either x-2 > 3 or x-2< -3
so
x - 2 > 3 -> x > 5 or
x -2 < -3 -> x < -1
if the numbers are integers then we can't choose x =5 and x = -1 :
the minimum possible value of |x - 3.5 | is when x = 6, in this case |x-3.5| equals 2.5
the minimum possible value of |x - 1.5 | is when x = -1, in this case |x-3.5| equals 2.5 again
so considering numbers as integers these are equal.


nope.
"the minimum possible value of |x - 1.5 | is when x = -1, in this case |x-3.5| equals 2.5 again"

if x = -1, then |-1-3.5| = 4.5
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Re: |x-2| > 3 [#permalink]
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Carcass wrote:

This question is part of GREPrepClub - The Questions Vault Project



\(|x-2| > 3\)

Quantity A
Quantity B
The minimum possible
value of |x - 3.5|
The minimum possible
value of | x - 1.5 |


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.


\(|x-2| > 3\)

we can find the values of x in three ways...

(I) Since both sides are positive, square both sides
\(|x-2|^2 > 3^2.......x^2-4x+4>9.........x^2-4x-5>0.....x^2-5x+x-5>0........(x+1)(x-5)>0\)
a) either both are positive.. x+1>0 and x-5>0, so x>-1 and x>5...x>5
b) or both are negative .. x+1<0 and x-5<0, so x<-1 and x<5 ... so x<-1
x<-1 and x>5

(II) critical point..
x<=2..... -(x-2)>3.......x-2<-3....x<-1
x>2...... (x-2)>3......x>5

(III) logical approach via number line..
\(|x-2| > 3\) means the distance of x from 2 is greater than 3 units..
so if x is on the left side of 2, it will be < (2-3) or <-1
if x is on the right side of 2, it will be > (2+30 or >5

you can learn more of this from
https://gre.myprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html

so we have range of x as x<-1 and x>5
The minimum possible value of |x - 3.5| ... again which value in range is closer to 3.5, it is 5, so value is just > |5-3.5| or just > 1.5
The minimum possible value of | x - 1.5 |... again which value in range is closer to 1.5, it is -1, so value is just > |-1-1.5| or just > 2.5
so B>A
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Re: |x-2| > 3 [#permalink]
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Answer: B

3 < |x-2|

** If a < |b+c| Then a < (b+c) or (b+c) < -a
So we can conduct that:
3 < x-2 or (x-2) < -3
x > 5 or x < -1

A: The minimum possible value of |x-3.5|
First we should consider that an absolute value can’t be negative. At least it equals to 0.
We have x > 5 or x < -1
For x>5, |x-3.5| will be bigger than 1.5
For x<-1, |x-3.5| will be bigger than 4.5
So the minimum amount for A is 1.5

B: The minimum possible value of |x-1.5|
We have x > 5 or x < -1
For x>5, |x-1.5| will be bigger than 3
For x<-1, |x-1.5| will be bigger than 2.5
So the minimum amount for A is 2.5


So B is bigger than A.
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Re: |x-2| > 3 [#permalink]
Can't we take any decimal value in the place of x? As it is not said that x is an integer?

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Re: |x-2| > 3 [#permalink]
Fatemeh wrote:
correct : C
|X -2 | > 3 means either x-2 > 3 or x-2< -3
so
x - 2 > 3 -> x > 5 or
x -2 < -3 -> x < -1
if the numbers are integers then we can't choose x =5 and x = -1 :
the minimum possible value of |x - 3.5 | is when x = 6, in this case |x-3.5| equals 2.5
the minimum possible value of |x - 1.5 | is when x = -1, in this case |x-3.5| equals 2.5 again
so considering numbers as integers these are equal.


Since x > 5 or x < -1 why did you pick x = -1? I think if x is an integer it's minimum value should be -2. Therefore the minimum possible value of |x - 1.5 | is when x = -2, in this case |x-3.5| equals 3.5 which is bigger than the other value.

Can someone please explain?
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|x-2| > 3 [#permalink]
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The question does not say that the numbers are integers.

So a number < -1 could be even 1.00000000000000000000000000000000000000000000000000001 OR 1,000,000


Regards
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Re: |x-2| > 3 [#permalink]
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We need to first find the value of x,

now there are two cases possible,
\(x - 2 > 3\) and \(x - 2 < -3\)
\(x > 5\) and \(x < -1\)
Minimum value of \(|x-3.5|\) is when x = 5 => |5-3.5| = 1.5
Minimum value of \(| x - 1.5 |\) is when x = -1 => |-1-1.5| = 2.5

Hence, B is the answer
Carcass wrote:

This question is part of GREPrepClub - The Questions Vault Project



\(|x-2| > 3\)

Quantity A
Quantity B
The minimum possible
value of |x - 3.5|
The minimum possible
value of | x - 1.5 |


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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|x-2|>3 [#permalink]
\(|x-2|>3\)

Quantity A
Quantity B
The minimum possible value of |x-3.5|
The minimum possible value of |x-1.5|


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.

Originally posted by ohanish on 09 Mar 2020, 18:27.
Last edited by Carcass on 21 Mar 2020, 02:26, edited 2 times in total.
Updated
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Re: Algebra QC [#permalink]
Mod of number will be positive.
Suppose x= 6, then in first case, |6-3.5| = 2.5 and |6-1.5|= 4.5. Hence choice B.
Suppose x=-6, then, |-6-3.5| = |-9.5|= 9.5 and |-6-1.5| = |-7.5| = 7.5. Hence choice A.
Thus choice D.
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Re: Algebra QC [#permalink]
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ohanish wrote:
Hello everyone,

Need some help with one question that is not making any sense to me right now.

|x-2|>3

Quantity A
The minimum possible value of |x-3.5|

Quantity B
The minimum possible value of |x-1.5|





We have: |x - 2| > 3
=> x - 2 > 3 OR x - 2 < -3
=> x > 5 OR x < -1

Quantity A: The minimum possible value of |x-3.5|
Ideally, the minimum value of any 'absolute value' term is zero

However, in this case, x cannot have the value 3.5 and hence, the above minimum cannot be zero
We need to find the closest x can be to 3.5, and that value is just exceeding 5 (not -1):

Thus, the minimum value of |x - 3.5| just exceeds |5 - 3.5| i.e. just more than 1.5


Quantity B: The minimum possible value of |x-1.5|

In this case too, x cannot have the value 1.5 and hence, the above minimum cannot be zero
We need to find the closest x can be to 1.5, and that value is just less than -1 (not 5):

Thus, the minimum value of |x - 1.5| just exceeds |-1 - 1.5| i.e. just more than 2.5

Thus, Quantity B is greater


Note: We need to find the minimum value, so plugging in ANY value of x is not going to work!

There is a bit of confusion, however, since we have x > 5 or x < -1. The question would have been much better if it had been: x ≥ 5 or x ≤ -1
In that case, Quantity A would have been 1.5 and Quantity B would have been 2.5 and we could have concluded that Quantity B is greater.

However, ideally speaking, in the present question, we have:
Quantity A is greater than 1.5 while Quantity B is greater than 2.5
Hence, the quantities cannot ideally be compared --- Option D
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Re: |x-2| > 3 [#permalink]
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|x-2| > 3

It's better to look at the Quantities to know how to manipulate this equation. Since we need |x-3.5| and |x-1.5|, we can subtract both sides by 1.5 and add both sides by 0.5 separately to transform it in terms of the 2 quantities. Because 1.5 and 0.5 are scalar values (absolute quantities), we can freely add or subtract them to the equality and insert them inside the modulus without any significant issues.

|x-2| - 1.5 > 3 - 1.5
|x - 3.5| > 1.5 => Quantity A > 1.5 so we can assume minimum value as 1.500000001

|x-2| + 0.5 > 3 + 0.5
|x - 1.5| > 3.5 => Quantity B > 3.5 so we can assume minimum value as 3.500000001

Hence Quantity B is greater than Quantity A.
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Re: |x-2| > 3 [#permalink]
I could not understand the last part of your solution how you choose x= 5 and 1.5.
IlCreatore wrote:
Solving the disequality in the header, \(|x-2|>3\), we get \(x<-1, x>5\).

Then, the smallest value for \(|x-3.5|\) is when x = 5, i.e. \(|5-3.5| = 1.5\). The smallest value for \(|x-1.5|\) is when x = -1, i.e. \(|-1-1.5| = 2.5\).

Thus, since 2.5>1.5, quantity B is greater!
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Re: |x-2| > 3 [#permalink]
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Hey,

I hope you are clear till \(|x-2|>3\), we get \(x<-1, x>5\)

Now we are given that \(x > 5\) , so the smallest value for \(|x-3.5|\) will be when \(x = 5\), which is \(1.5\)
if you take let's say \(x=5.5\) then \(|x-3.5|\) will be \(2\) & if \(x=6\) then \(|x-3.5|\) will be \(2.5\)

Same goes with the smallest value for \(|x-1.5|\) when \(x = -1\)
You can try \(x = -1.5\) or \(-2\)

Ask in case of any doubts.

kumarneupane4344 wrote:
I could not understand the last part of your solution how you choose x= 5 and 1.5.
IlCreatore wrote:
Solving the disequality in the header, \(|x-2|>3\), we get \(x<-1, x>5\).

Then, the smallest value for \(|x-3.5|\) is when x = 5, i.e. \(|5-3.5| = 1.5\). The smallest value for \(|x-1.5|\) is when x = -1, i.e. \(|-1-1.5| = 2.5\).

Thus, since 2.5>1.5, quantity B is greater!
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Re: |x-2| > 3 [#permalink]
IlCreatore wrote:
Solving the disequality in the header, \(|x-2|>3\), we get \(x<-1, x>5\).

Then, the smallest value for \(|x-3.5|\) is when x = 5, i.e. \(|5-3.5| = 1.5\). The smallest value for \(|x-1.5|\) is when x = -1, i.e. \(|-1-1.5| = 2.5\).

Thus, since 2.5>1.5, quantity B is greater!



Can someone please elaborate why he has taken the extreme cases to plug in the value of 'X' like 5, -1, whereas x>5 or x<-1, so why not x = 6 or -2. I tried choosing x=6 wherein A<B, and if x=-2 A>B, it should be 'D'. Can someone else resolve my query here
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