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M24-03

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 01:20
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Difficulty:

55% (hard)

Question Stats:

58% (00:50) correct 43% (01:05) wrong based on 120 sessions

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Is $$x$$ negative?

(1) $$|1 - x| \lt 1$$

(2) $$|1 + x| \gt 1$$
[Reveal] Spoiler: OA

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16 Sep 2014, 01:20
Official Solution:

(1) $$|1-x| \lt 1$$. From this statement it follows that $$-1 \lt 1-x \lt 1$$. Add -1 to all parts: $$-2 \lt -x \lt 0$$. Now multiply by -1: $$2 \gt x \gt 0$$. So, $$x$$ is positive. Sufficient.

(2) $$|1+x| \gt 1$$. From this statement it follows that $$x \gt 0$$ or $$x \lt -2$$. So, $$x$$ may or may not be positive. Not sufficient.

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29 Jan 2015, 04:31
Bunuel wrote:
Official Solution:

(1) $$|1−x| \lt 1$$. From this statement it follows that $$-1 \lt 1-x \lt 1$$. Add -1 to all parts: $$-2 \lt -x \lt 0$$. Now multiply by -1: $$2 \gt x \gt 0$$. So, $$x$$ is positive. Sufficient.

(2) $$|1+x| \gt 1$$. From this statement it follows that $$x \gt 0$$ or $$x \lt -2$$. So, $$x$$ may or may not be positive. Not sufficient.

|1-x|<1
will have two parts to the solution

1-x<1 =====>x-1>-1==========> x>0
and -(1-x)<1======>x-1<1 =====> x<2

So does this mean that x lies between 0 and 2???

Or does it mean that either X is greater than 0 or X is less than 2.

Because if it means that x is less than 2 or x is greater than 0, then x is less than 2 could also signify that x is negative.
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Joined: 02 Sep 2009
Posts: 44600

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29 Jan 2015, 05:30
mayankpant wrote:
Bunuel wrote:
Official Solution:

(1) $$|1−x| \lt 1$$. From this statement it follows that $$-1 \lt 1-x \lt 1$$. Add -1 to all parts: $$-2 \lt -x \lt 0$$. Now multiply by -1: $$2 \gt x \gt 0$$. So, $$x$$ is positive. Sufficient.

(2) $$|1+x| \gt 1$$. From this statement it follows that $$x \gt 0$$ or $$x \lt -2$$. So, $$x$$ may or may not be positive. Not sufficient.

|1-x|<1
will have two parts to the solution

1-x<1 =====>x-1>-1==========> x>0
and -(1-x)<1======>x-1<1 =====> x<2

So does this mean that x lies between 0 and 2???

Or does it mean that either X is greater than 0 or X is less than 2.

Because if it means that x is less than 2 or x is greater than 0, then x is less than 2 could also signify that x is negative.

$$|1−x| \lt 1$$ translates to 0<x<2 as written above.
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02 Mar 2016, 08:26
Bunuel wrote:
Official Solution:

(1) $$|1-x| \lt 1$$. From this statement it follows that $$-1 \lt 1-x \lt 1$$. Add -1 to all parts: $$-2 \lt -x \lt 0$$. Now multiply by -1: $$2 \gt x \gt 0$$. So, $$x$$ is positive. Sufficient.

(2) $$|1+x| \gt 1$$. From this statement it follows that $$x \gt 0$$ or $$x \lt -2$$. So, $$x$$ may or may not be positive. Not sufficient.

in statement 1 cant we change the equation to |x-1|>1 and then solve ?
im confused~!
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03 Mar 2016, 07:27
nishantdoshi wrote:
Bunuel wrote:
Official Solution:

(1) $$|1-x| \lt 1$$. From this statement it follows that $$-1 \lt 1-x \lt 1$$. Add -1 to all parts: $$-2 \lt -x \lt 0$$. Now multiply by -1: $$2 \gt x \gt 0$$. So, $$x$$ is positive. Sufficient.

(2) $$|1+x| \gt 1$$. From this statement it follows that $$x \gt 0$$ or $$x \lt -2$$. So, $$x$$ may or may not be positive. Not sufficient.

in statement 1 cant we change the equation to |x-1|>1 and then solve ?
im confused~!

No. |x-1| is the distance between x and 1, so is |1-x|. Thus those two are the same.
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06 Mar 2016, 22:10
Bunuel wrote:
nishantdoshi wrote:
Bunuel wrote:
Official Solution:

(1) $$|1-x| \lt 1$$. From this statement it follows that $$-1 \lt 1-x \lt 1$$. Add -1 to all parts: $$-2 \lt -x \lt 0$$. Now multiply by -1: $$2 \gt x \gt 0$$. So, $$x$$ is positive. Sufficient.

(2) $$|1+x| \gt 1$$. From this statement it follows that $$x \gt 0$$ or $$x \lt -2$$. So, $$x$$ may or may not be positive. Not sufficient.

in statement 1 cant we change the equation to |x-1|>1 and then solve ?
im confused~!

No. |x-1| is the distance between x and 1, so is |1-x|. Thus those two are the same.

so you mean

if |1-x|<1 then |x-1|<1 , it wont be |x-1|>1

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Joined: 02 Sep 2009
Posts: 44600

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07 Mar 2016, 10:48
nishantdoshi wrote:

so you mean

if |1-x|<1 then |x-1|<1 , it wont be |x-1|>1

______________
Yes, that's correct.
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09 Nov 2016, 22:32
Hi bunuel,

I understand that when we open a mod, we have to do things:

a. Solve without considering the mod (i.e normally)
b. Change the signs of any one side of the equation.

I believe thus same process is applicable both when = sign is used and when >= or <= signs are used.

Getting very confused here.

Request you to kindly clarify.

Many Thanks.
Math Expert
Joined: 02 Sep 2009
Posts: 44600

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10 Nov 2016, 07:34
dsheth7 wrote:
Hi bunuel,

I understand that when we open a mod, we have to do things:

a. Solve without considering the mod (i.e normally)
b. Change the signs of any one side of the equation.

I believe thus same process is applicable both when = sign is used and when >= or <= signs are used.

Getting very confused here.

Request you to kindly clarify.

Many Thanks.

Here are the topics that will help you to brush up fundamentals on absolute values:
Theory on Abolute Values: math-absolute-value-modulus-86462.html
Absolute value tips: absolute-value-tips-and-hints-175002.html

DS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=37
PS Abolute Values Questions to practice: search.php?search_id=tag&tag_id=58

Hard set on Abolute Values: inequality-and-absolute-value-questions-from-my-collection-86939.html

Hope it helps.
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10 Apr 2018, 03:04
+1 for option A
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Re: M24-03   [#permalink] 10 Apr 2018, 03:04
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