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(1) \(|j| = j^{-1}\). Rewrite as \(|j|*j=1\). From that we have \(j=1\) (here \(j\) cannot be a negative number, since in this case we would have \(|j|*j=\text{positive}*\text{negative}=\text{negative} \ne 1\)). Sufficient.
(2) \(j^j = 1\). Again only one solution: \(j=1\). Sufficient.
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). _________________
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).
Thanks for pointing that out. Yes, I was confused because 0^0 = 1 (actually several answers to this solution), some say its 0, some say its undefined.
But, yes the question says "j is not equal to 0".
_________________
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.
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 ?
_________________
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.
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.
I understand that when you square both sides of an equation you create another solution that is potentially invalid due to some other parameter (equation, stipulation etc). But how is that happening here? I just multiplied both sides by j, I didn't square anything.
|j| = 1\j |j| * j = 1
I then considered both cases to remove the modulus and found j = +/- 1.
Now, I am sure the best practice here is just always to check too see that both solutions work, but can someone explain to me here why I ended up with two solutions (one of which does not hold) without squaring both sides of the equation?
close
58% (00:51) correct 
<Slapping myself now!>
