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# M04-11

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Math Expert
Joined: 02 Aug 2009
Posts: 8619

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29 Jul 2018, 03:09
1
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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29 Jul 2018, 04:10
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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Math Expert
Joined: 02 Aug 2009
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29 Jul 2018, 04:18
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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Joined: 18 May 2019
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19 May 2019, 14:59
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?
Math Expert
Joined: 02 Sep 2009
Posts: 64250

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19 May 2019, 19:52
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?

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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20 May 2019, 00:42
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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Joined: 14 Feb 2017
Posts: 1369
Location: Australia
Concentration: Technology, Strategy
GMAT 1: 560 Q41 V26
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GMAT 5: 650 Q48 V31
GMAT 6: 600 Q38 V35
GMAT 7: 710 Q47 V41
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02 Aug 2019, 15:54
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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08 Aug 2019, 23:54
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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Joined: 02 Sep 2009
Posts: 64250

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09 Aug 2019, 00:29
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

____________________
Yes, that's correct.
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09 Aug 2019, 12:00
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

____________________
Yes, that's correct.

Thanks, that is actually a big help
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12 Apr 2020, 07:49
Bunuel wrote:
If $$j \ne 0$$, what is the value of $$j$$ ?

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

(2) $$j^j = 1$$

(2) $$j^j = 1$$

Since j is not 0 let's try other values.
-1 won't work.
j=1 works

Upon thinking further no other value suffices. We are down to B and D.

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

|j|=1/j

Again -1 wont work but 1 will work.

No other value works.

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Re: M04-11   [#permalink] 12 Apr 2020, 07:49

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# M04-11

Moderators: chetan2u, Bunuel