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M04-11

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Bunuel
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M04-11 [#permalink] New post   16 Sep 2014, 00:22
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If \(j \ne 0\), what is the value of \(j\) ?


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

(2) \(j^j = 1\)
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D
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Bunuel
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Re: M04-11 [#permalink] New post   19 Feb 2016, 01:55
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NandishSS wrote:
Hi Bunuel,

Not getting where am I going wrong with statement 1

|j|=j^-1

j=1/j

j^2=1

And j=-(1/j)

j^2=-1

Can you please help me.


j^2=-1 has no real solutions.

j^2=1 has two solutions j=-1 and j=1 but j=-1 does not satisfy |j|=j^(-1), so we are left with j=1 only.

Hope it's clear.
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Re: M04-11 [#permalink] New post   19 Sep 2014, 02:06
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earnit wrote:
As per ii) j^j = 1.

So 1 can also be equal to j^0.

Therefore, j^j = j^0. so same base power is equated which brings us to: j=0.

But 0^0 = 0 so that is not our solution.

hence j=1, as any number raised to same power if = 1 then that number is 1.

Am i right in assuming the following rule (highlighted one) based on the above calculations ?

Kindly shed some light on it.


Sorry, but not following you...

Anyway, 0^0, in some sources equals to 1, some mathematicians say it's undefined. But you won't need this for the GMAT because the case of 0^0 is not tested on the GMAT. So on the GMAT the possibility of 0^0 is always ruled out (in our question it's also rules out, notice that we are given that j is not 0).
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Re M04-11 [#permalink] New post   16 Sep 2014, 00:22
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Official Solution:


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

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

\(|j| = \frac{1}{j}\);

\(|j|*j=1\).

Notice that \(j\) cannot be negative since in this case we would have \(|j|*j=\text{positive}*\text{negative}=\text{negative} \ne 1\). Therefore, \(j\) is positive. \(j > 0\) implies that \(|j| = j\), hence we get:

\(j^2=1\);

Since we know that \(j\) is positive, then \(j=1\).

Sufficient.

(2) \(j^j = 1\). Again only one solution: \(j=1\). Sufficient.


Answer: D
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Re: M04-11 [#permalink] New post   25 Feb 2016, 10:21
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sinhap07 wrote:
Bunuel wrote:
If \(j \ne 0\), what is the value of \(j\) ?


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

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


Hi

For stmt 1, what I did was I squared both sides to get rid of the mod sign. So :

j^2=1/j^2

j^4=1
Hence j takes both 1 0r -1. not sufficient.

Pls advise. tx


When you square any equation, you are invariably increasing the number of solutions to twice the actual number of solution. A linear equation has 1 solution while a quadratic equation has 2 etc.

Statement 1 is linear in j and hence only 1 solution should be possible. When you square, make sure to check back the solutions by plugging them into the main equation and see which one actually satisfies the original linear equation.

After you got 1 and -1 as your solutions, -1 is rejected as it does not satisfy \(|j| = j^{-1}\). Thus, only j=1 satisfies the given conditions.

Hope this helps.
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Re: M04-11 [#permalink] New post   19 Sep 2014, 01:57
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As per ii) j^j = 1.

So 1 can also be equal to j^0.

Therefore, j^j = j^0. so same base power is equated which brings us to: j=0.

But 0^0 = 0 so that is not our solution.

hence j=1, as any number raised to same power if = 1 then that number is 1.

Am i right in assuming the following rule (highlighted one) based on the above calculations ?

Kindly shed some light on it.
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Re: M04-11 [#permalink] New post   23 Sep 2017, 07:19
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susheelh wrote:
Hello Bunuel,

Though I agree with the solution, I am not able to see where I am going wrong when solving for S1.Can you please check my work and suggest where I am going wrong?

S1: \(|J| = J^{-1}\)

Opening the Modulus we have

Case 1:

\(J = J^{-1}\)

\(J^2 = 1\)

\(J = ± 1\)

Case 2 :

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

\(-J^2 = 1\) (Undefined on GMAT).

Substituting Both J = 1 and J = -1 into the equation we can also see the equation is satisfied.

So, should we not conclude that J = ± 1 ?


j = -1 does NOT satisfy |j| = j^(-1):

LHS = |-1| = 1 while RHS = (-1)^(-1) = -1.
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Re: M04-11 [#permalink] New post   16 Mar 2023, 12:38
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joe123x wrote:
Bunuel wrote:
If \(j \ne 0\), what is the value of \(j\) ?


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

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


in statement 1,
is it possible to do it like this:

|j| = j^-1
j^2 = j^-2
j = 1 or j = 0, in this case j=1. ?


j^2 = j^(-2);

j^2 = 1/j^2;

j^4 = 1;

j = 1 or j = -1.

Since squaring can in some cases create false roots, you should plug back the solutions to check their validly.

If j = 1, then |j| = j^(-1) = 1. Hence this root is valid.
If j = -1, then |j| = 1, while j^(-1) = -1. Hence this root is not valid.

P.S. j = 0, is not the root at all.
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Re: M04-11 [#permalink] New post   13 Apr 2015, 17:02
What if J=-1?
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Re: M04-11 [#permalink] New post   14 Apr 2015, 04:41
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holyrage wrote:
What if J=-1?


J = -1 does not satisfy any of the statements:

(1) |J| = 1 and J^(-1) = (-1)^(-1) = -1.

(2) (-1)^(-1) = -1 not 1.
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Re: M04-11 [#permalink] New post   18 Feb 2016, 22:12
Hi Bunuel,

Not getting where am I going wrong with statement 1

|j|=j^-1

j=1/j

j^2=1

And j=-(1/j)

j^2=-1

Can you please help me.
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Re: M04-11 [#permalink] New post   25 Feb 2016, 09:59
Bunuel wrote:
If \(j \ne 0\), what is the value of \(j\) ?


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

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


Hi

For stmt 1, what I did was I squared both sides to get rid of the mod sign. So :

j^2=1/j^2

j^4=1
Hence j takes both 1 0r -1. not sufficient.

Pls advise. tx
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Re: M04-11 [#permalink] New post   23 Sep 2017, 06:15
Hello Bunuel,

Though I agree with the solution, I am not able to see where I am going wrong when solving for S1.Can you please check my work and suggest where I am going wrong?

S1: \(|J| = J^{-1}\)

Opening the Modulus we have

Case 1:

\(J = J^{-1}\)

\(J^2 = 1\)

\(J = ± 1\)

Case 2 :

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

\(-J^2 = 1\) (Undefined on GMAT).

Substituting Both J = 1 and J = -1 into the equation we can also see the equation is satisfied.

So, should we not conclude that J = ± 1 ?
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Re: M04-11 [#permalink] New post   10 Apr 2018, 10:52
cant j be an imaginary number for the first statement?
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Re: M04-11 [#permalink] New post   10 Apr 2018, 11:07
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s12345 wrote:
cant j be an imaginary number for the first statement?


1. On the GMAT all numbers used are real numbers.

2. If this were not the case, still |j| = j^(-1) has no complex solution.
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Re: M04-11 [#permalink] New post   03 Jan 2021, 01:57
I think this is a high-quality question and I agree with explanation.
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Re: M04-11 [#permalink] New post   04 Mar 2023, 04:52
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M04-11 [#permalink] New post   16 Mar 2023, 11:21
Bunuel wrote:
If \(j \ne 0\), what is the value of \(j\) ?


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

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


in statement 1,
is it possible to do it like this:

|j| = j^-1
j^2 = j^-2
j = 1 or j = 0, in this case j=1. ?
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Re: M04-11 [#permalink]
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