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If z >1
[#permalink]
07 Sep 2021, 07:22

Expert Reply

Question Stats:

If \(z>1\), then \(\frac{2z^2(z-1)+z-z^2}{z(z-1)} = \)

(A) \(2z^2 − z\)

(B) \(2z + 1\)

(C) \(2z\)

(D) \(2z − 1\)

(E) \(z − 2\)

(A) \(2z^2 − z\)

(B) \(2z + 1\)

(C) \(2z\)

(D) \(2z − 1\)

(E) \(z − 2\)

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If z >1
[#permalink]
07 Sep 2021, 07:44

3

Carcass wrote:

If \(z>1\), then \(\frac{2z^2(z-1)+z-z^2}{z(z-1)} = \)

(A) \(2z^2 − z\)

(B) \(2z + 1\)

(C) \(2z\)

(D) \(2z − 1\)

(E) \(z − 2\)

(A) \(2z^2 − z\)

(B) \(2z + 1\)

(C) \(2z\)

(D) \(2z − 1\)

(E) \(z − 2\)

APPROACH #1: Algebra

Given: \(\frac{2z^2(z-1)+z-z^2}{z(z-1)}\)

Rearrange the last two terms in the numerator: \(\frac{2z^2(z-1)-z^2+z}{z(z-1)}\)

Factor out \(-z\) from the last two terms of numerator: \(\frac{2z^2(z-1)-z(z-1)}{z(z-1)}\)

Rewrite the numerator as follows: \(\frac{(2z^2-z)(z-1)}{z(z-1)}\)

Simplify: \(\frac{2z^2-z}{z}\)

Finally, we can divide numerator and denominator by \(z\) to get: \(2z-1\)

Answer: D

APPROACH #2: Test a value of z

Since we are looking for an expression that's equivalent to \(\frac{2z^2(z-1)+z-z^2}{z(z-1)}\), let's first evaluate this expression for a certain value of \(z\), and then look for an answer choice that has the same value for that value of \(z\)

Let's see what happens when we plug \(z = 2\) into the given expression:

\(\frac{2z^2(z-1)+z-z^2}{z(z-1)} = \frac{2(2^2)(2-1)+2-2^2}{2(2-1)}\) \(= \frac{8+2-4}{2}\) \(= \frac{6}{2} = 3\)

So, the given expression evaluates to equal \(3\) when \(z = 2\).

We can now plug \(z = 2\) into each answer choice to see which one evaluates to \(3\)

(A) \(2(2^2) − 2 =6\). ELIMINATE

(B) \(2(2) + 1=5\). ELIMINATE

(C) \(2(2)=4\). ELIMINATE

(D) \(2(2) − 1=3\). KEEP

(E) \(2 − 2=0\). ELIMINATE

Answer: D

If z >1
[#permalink]
07 Sep 2021, 07:36

1

\((2* z^2 (z-1)+z - z^2)/(z(z-1))\)

=\((2* z^2 (z-1)- z^2 + z)/(z(z-1))\)

=\((2* z^2 (z-1)- z(z-1))/(z(z-1))\)

dividing both num and den by \(z(z-1)\)

= \(2z-1\)

=\((2* z^2 (z-1)- z^2 + z)/(z(z-1))\)

=\((2* z^2 (z-1)- z(z-1))/(z(z-1))\)

dividing both num and den by \(z(z-1)\)

= \(2z-1\)

Re: If z >1
[#permalink]
22 Jan 2024, 15:23

Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.

gmatclubot

Re: If z >1 [#permalink]

22 Jan 2024, 15:23
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