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If k#0 and k - (3 -2k^2)/k = x/k, then x =

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If k#0 and k - (3 -2k^2)/k = x/k, then x = [#permalink]

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The Official Guide For GMAT® Quantitative Review, 2ND Edition

If \(k\neq{0}\) and \(k - \frac{3 -2k^2}{k} = \frac{x}{k}\), then x =

(A) -3 - k^2
(B) k^2 -3
(C) 3k^2 - 3
(D) k - 3 - 2k^2
(E) k - 3 + 2k^2

Problem Solving
Question: 111
Category: Algebra Second-degree equations
Page: 76
Difficulty: 500


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Re: If k#0 and k - (3 -2k^2)/k = x/k, then x = [#permalink]

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Re: If k#0 and k - (3 -2k^2)/k = x/k, then x = [#permalink]

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k - \frac{3 -2k^2}{k} = \frac{x}{k};
After simplifying and canceling k in the denominator, we get: k^2 - 3 + 2k^2 = x;
3k^2 - 3 = x;

Ans is (C).

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Re: If k#0 and k - (3 -2k^2)/k = x/k, then x = [#permalink]

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New post 26 Feb 2014, 01:10
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Option C.
k-(3-2k^2)/k=x/k
Multiply both sides by k
k^2-3+2k^2=x
3k^2-3=x

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Re: If k#0 and k - (3 -2k^2)/k = x/k, then x = [#permalink]

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New post 08 Jul 2014, 21:17
Bunuel wrote:
SOLUTION

If \(k\neq{0}\) and \(k - \frac{3 -2k^2}{k} = \frac{x}{k}\), then x =

(A) -3 - k^2
(B) k^2 -3
(C) 3k^2 - 3
(D) k - 3 - 2k^2
(E) k - 3 + 2k^2

\(k - \frac{3 -2k^2}{k} = \frac{x}{k}\);

Multiply by k: \(k^2 - (3 -2k^2) = x\);

\(x=3k^2-3\).

Answer: C.


I just added a step after the highlighted one to remove -ve sign :) . Rest all did same

\(k - \frac{3 -2k^2}{k} = \frac{x}{k}\);

\(k + \frac{2k^2 - 3}{k} = \frac{x}{k}\)
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Re: If k#0 and k - (3 -2k^2)/k = x/k, then x = [#permalink]

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New post 08 Jul 2014, 22:55
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

If \(k\neq{0}\) and \(k - \frac{3 -2k^2}{k} = \frac{x}{k}\), then x =

(A) -3 - k^2
(B) k^2 -3
(C) 3k^2 - 3
(D) k - 3 - 2k^2
(E) k - 3 + 2k^2





You can also put in k = 1 in \(k - \frac{3 -2k^2}{k} = \frac{x}{k}\) to get x = 0.
When you put k = 1 in options, only (C) and (E) give x = 0.

Now put k = -1 in \(k - \frac{3 -2k^2}{k} = \frac{x}{k}\) to get x = 0.
When you put k = -1 in options, only (C) gives x = 0
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Re: If k#0 and k - (3 -2k^2)/k = x/k, then x = [#permalink]

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New post 10 Mar 2015, 19:07
Why do we automatically assume there are parentheses around (3-2k^2)? I didn't assume that so then I didn't distribute the negative on the outside which of course then gave me answer A.

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Re: If k#0 and k - (3 -2k^2)/k = x/k, then x = [#permalink]

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soniasawhney wrote:
Why do we automatically assume there are parentheses around (3-2k^2)? I didn't assume that so then I didn't distribute the negative on the outside which of course then gave me answer A.


hi soniasawhney..
it does not require any parenthesis as any sign in front of a fraction modifies the entire fraction irrespective of the term..
just an example..
let the fraction be - \(\frac{7-9}{2}\)..
two ways to do it ..
first change signs first and then find answer...
\(\frac{-7+9}{2}\)=\(\frac{2}{2}\)=1...

second simplify and then change the signs...
- \(\frac{7-9}{2}\)=- \(\frac{-2}{2}\)=-(-1)=1..

so both ways answer is same, which means parentheses is not required and a sign in front of a fraction automatically means that the entire fraction is inside bracket....
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Re: If k#0 and k - (3 -2k^2)/k = x/k, then x = [#permalink]

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soniasawhney wrote:
Why do we automatically assume there are parentheses around (3-2k^2)? I didn't assume that so then I didn't distribute the negative on the outside which of course then gave me answer A.


To clarify further:
There are 3 ways in which you can make a fraction negative
\(\frac{-3}{5}\)

\(-\frac{3}{5}\) Take this to mean \(\frac{-3}{5}\)

and \(\frac{3}{-5}\)

Each one of these fractions is the same as \(\frac{-3}{5}\)

Now what happens in case you have multiple terms in numerator:

\(\frac{-x + 2}{4}\) The negative is only in front of x.

\(-\frac{x+2}{4}\) The negative is in effect, in front of the entire numerator. So this is the same as \(\frac{-(x+2)}{4}\)
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Re: If k#0 and k - (3 -2k^2)/k = x/k, then x = [#permalink]

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New post 12 Mar 2015, 22:12
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

If \(k\neq{0}\) and \(k - \frac{3 -2k^2}{k} = \frac{x}{k}\), then x =

(A) -3 - k^2
(B) k^2 -3
(C) 3k^2 - 3
(D) k - 3 - 2k^2
(E) k - 3 + 2k^2

Problem Solving
Question: 111
Category: Algebra Second-degree equations
Page: 76
Difficulty: 500


GMAT Club is introducing a new project: The Official Guide For GMAT® Quantitative Review, 2ND Edition - Quantitative Questions Project

Each week we'll be posting several questions from The Official Guide For GMAT® Quantitative Review, 2ND Edition and then after couple of days we'll provide Official Answer (OA) to them along with a slution.

We'll be glad if you participate in development of this project:
1. Please provide your solutions to the questions;
2. Please vote for the best solutions by pressing Kudos button;
3. Please vote for the questions themselves by pressing Kudos button;
4. Please share your views on difficulty level of the questions, so that we have most precise evaluation.

Thank you!


Given that,
k - (3 - 2k^2)/k = x/k
Take LCM of denominators,
So, [k^2 - (3 - 2k^2)]/k = x/k
So, [k^2 - 3 + 2k^2]/k = x/k
Canceling out k from the denominators of both sides,
So, [k^2 - 3 + 2k^2] = x
So, 3k^2 - 3 = x
Hence option C.

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Re: If k#0 and k - (3 -2k^2)/k = x/k, then x = [#permalink]

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New post 07 Apr 2016, 15:29
I'm not sure whats wrong here:

1. k - [3-2k^2][/k] = [x][/k]

2. k - 3 + 2k^2 = x

I am assuming I can't move the denominator to the right side because of the first term (k), just not sure why.

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Re: If k#0 and k - (3 -2k^2)/k = x/k, then x = [#permalink]

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Apollon wrote:
I'm not sure whats wrong here:

1. k - [3-2k^2][/k] = [x][/k]

2. k - 3 + 2k^2 = x

I am assuming I can't move the denominator to the right side because of the first term (k), just not sure why.


1. k - [3-2k^2][/k] = [x][/k]

2. k - 3 + 2k^2 = x

Hi Apollon,

If you see, only two of the three terms have the denominator k.
In order to cancel out the denominator from all the terms, you need to have the same denominator in each term.

In the given case,

1. k - (3-2k^2)/ k = x / k
We need to make sure that the denominator of the first term is also the same.

Hence
k^2/k - (3-2k^2)/ k = x / k
From here on, we can cancel the terms.
k^2 - 3 + 2k^2 = x
Or x = 3k^2 - 3
Option C

Does this help?

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Re: If k#0 and k - (3 -2k^2)/k = x/k, then x = [#permalink]

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New post 10 Apr 2016, 22:10
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Apollon wrote:
I'm not sure whats wrong here:

1. k - [3-2k^2][/k] = [x][/k]

2. k - 3 + 2k^2 = x

I am assuming I can't move the denominator to the right side because of the first term (k), just not sure why.


Yes, you re right. To move the denominator to the other side, it must be the denominator of the entire expression.

Look at it from a very basic viewpoint:

1/2 = x
1 = 2x
Since x is 1/2, twice of x will be 1.

1 + 1/2 = x
1 + 1 = 2x ????
x is actually 3/2 or (1.5). If you double it, will you get 2? No.

On the other hand,
(1 + 3+ 5)/2 = x
Then (1 + 3 + 5) = 2x is correct.
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Re: If k#0 and k - (3 -2k^2)/k = x/k, then x = [#permalink]

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New post 30 Apr 2017, 00:33
VeritasPrepKarishma wrote:
soniasawhney wrote:
Why do we automatically assume there are parentheses around (3-2k^2)? I didn't assume that so then I didn't distribute the negative on the outside which of course then gave me answer A.


To clarify further:
There are 3 ways in which you can make a fraction negative
\(\frac{-3}{5}\)

\(-\frac{3}{5}\) Take this to mean \(\frac{-3}{5}\)

and \(\frac{3}{-5}\)

Each one of these fractions is the same as \(\frac{-3}{5}\)

Now what happens in case you have multiple terms in numerator:

\(\frac{-x + 2}{4}\) The negative is only in front of x.

\(-\frac{x+2}{4}\) The negative is in effect, in front of the entire numerator. So this is the same as \(\frac{-(x+2)}{4}\)


Thank you, VeritasPrepKarishma

I too fell for this rule.

So basically in this exercise.

\(-\cfrac { 3-2{ k }^{ 2 } }{ k } \neq \cfrac { -3-2{ k }^{ 2 } }{ k } \\ \\ -\cfrac { 3-2{ k }^{ 2 } }{ k } \rightarrow \cfrac { -\left( 3-2{ k }^{ 2 } \right) }{ k }\)
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Re: If k#0 and k - (3 -2k^2)/k = x/k, then x =   [#permalink] 30 Apr 2017, 00:33
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