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M04-11

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New post 29 Jul 2018, 04:09
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adkikani wrote:
Bunuel niks18 gmatbusters KarishmaB amanvermagmat chetan2u


Quote:
If \(j \ne 0\), what is the value of \(j\) ?


(1) \(|j| = j^{-1}\)

(2) \(j^j = 1\)


Is it correct to write: |j| * j = 1 as \(j^2\) = 1


No, you cannot because you do not know what is j, is it positive or negative?
If \(j^2=1\), only thing that you can say is |j|=1 that is j=1 or j=-1
If j=1 |j|*j=|1|*1=1 but if j=-1, |j|*j=|-1|*-1=1*-1=-1
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New post 29 Jul 2018, 05:10
Thanks chetan2u for your prompt response.

You wrote:
Quote:
No, you cannot because you do not know what is j, is it positive or negative?


Quote:
but if j=-1, |j|*j=|-1|*-1=1*-1=-1


But RHS of equation : |j| * j = 1 is Positive 1 and this excludes
possibility of -1 in your steps. Am I and Bunuel not on same page with
slight change in perception?
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New post 29 Jul 2018, 05:18
adkikani wrote:
Thanks chetan2u for your prompt response.

You wrote:
Quote:
No, you cannot because you do not know what is j, is it positive or negative?


Quote:
but if j=-1, |j|*j=|-1|*-1=1*-1=-1


But RHS of equation : |j| * j = 1 is Positive 1 and this excludes
possibility of -1 in your steps. Am I and Bunuel not on same page with
slight change in perception?


I started the steps with
'if j^2=1"
You cannot write |j|*j=1 because j can be positive or negative.

Then I gave two cases
if j =1, yes |j|*j=1...(I)
But if j=-1, |j|*j =1 is not correct...(II)

Say we were given |j|*j=1, then case (I) that is j=1
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New post 19 May 2019, 15:59
When it comes to Modulus, this always confuses me.
Various places I have read |x|=-x if x<0.

But when should one apply that?
In this, though I know |-5| = 5 or |-8| = 8 or so on, but I applied the above mentioned statement and came to a wrong conclusion.
i.e.

if j=-1
|j|=-1 as if |x|=-x if x<0
then |j|*j=1(-1*-1)
-1 also satisfies this

Can you help me out with this?
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New post 19 May 2019, 20:52
PuneetSinghAnand wrote:
When it comes to Modulus, this always confuses me.
Various places I have read |x|=-x if x<0.

But when should one apply that?
In this, though I know |-5| = 5 or |-8| = 8 or so on, but I applied the above mentioned statement and came to a wrong conclusion.
i.e.

if j=-1
|j|=-1 as if |x|=-x if x<0
then |j|*j=1(-1*-1)
-1 also satisfies this

Can you help me out with this?


Your doubt regarding j being -1 is already answered in this thread. Please re-read the whole discussion.

Also, absolute value CANNOT be negative: \(|anything| \geq 0\). So, |j|=-1 cannot be true.

One should brush up basics before attempting questions (especially harder ones).

10. Absolute Value



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.
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New post 20 May 2019, 01:42
Bunuel wrote:
If \(j \ne 0\), what is the value of \(j\) ?


(1) \(|j| = j^{-1}\)

(2) \(j^j = 1\)



#1
\(|j| = j^{-1}\)
ljl*j =1
j = +1 only possible
sufficient
#2
\(j^j = 1\)
j=1 sufficient
IMO D
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New post 02 Aug 2019, 16:54
Made a stupid mistake on this one and ignored the cannot be zero constraint.

Test 1 and -1

Statement 1
|-1| = -1^-1
1 = 1/-1
1=/= -1
-1 is not possible
The only other number that suits the constraints is 1.

Statement 2
Apart from 1, the only other number that would result in 1 would be 0, but 0 isn't an option. Therefore j =1
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New post 09 Aug 2019, 00:54
Just want to confirm on the step by step method of statement 1

|j|=j^-1.
|j| = 1 / J
|j| * J = 1
Only positive 1, works since pos. * pos. = pos. 1

If someone can confirm this as the correct step that would be great
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New post 09 Aug 2019, 01:29
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New post 09 Aug 2019, 13:00
Bunuel wrote:
kapstone1996 wrote:
Just want to confirm on the step by step method of statement 1

|j|=j^-1.
|j| = 1 / J
|j| * J = 1
Only positive 1, works since pos. * pos. = pos. 1

If someone can confirm this as the correct step that would be great

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Yes, that's correct.


Thanks, that is actually a big help
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Re: M04-11   [#permalink] 09 Aug 2019, 13:00

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