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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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24 Feb 2014, 23:50
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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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24 Feb 2014, 23:50
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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.
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Re: If k#0 and k - (3 -2k^2)/k = x/k, then x = [#permalink]

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25 Feb 2014, 05:10
After Solving the equation , I got A
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Re: If k#0 and k - (3 -2k^2)/k = x/k, then x = [#permalink]

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25 Feb 2014, 23:26
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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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26 Feb 2014, 02:10
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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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26 Feb 2014, 02:28
Thank You for kudos,WoundedTiger!
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Re: If k#0 and k - (3 -2k^2)/k = x/k, then x = [#permalink]

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01 Mar 2014, 04:12
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.
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Re: If k#0 and k - (3 -2k^2)/k = x/k, then x = [#permalink]

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08 Jul 2014, 22:17
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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08 Jul 2014, 23:55
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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Current Student Joined: 14 Oct 2013 Posts: 48 Followers: 0 Kudos [?]: 8 [0], given: 120 Re: If k#0 and k - (3 -2k^2)/k = x/k, then x = [#permalink] ### Show Tags 10 Mar 2015, 20:07 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. Verbal Forum Moderator Joined: 02 Aug 2009 Posts: 4515 Followers: 394 Kudos [?]: 4200 [1] , given: 109 Re: If k#0 and k - (3 -2k^2)/k = x/k, then x = [#permalink] ### Show Tags 10 Mar 2015, 20:28 1 This post received KUDOS Expert's post 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.... _________________ Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372 Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7377 Location: Pune, India Followers: 2288 Kudos [?]: 15127 [2] , given: 224 Re: If k#0 and k - (3 -2k^2)/k = x/k, then x = [#permalink] ### Show Tags 10 Mar 2015, 21:24 2 This post received KUDOS Expert's post 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 $$\frac{-(x+2)}{4}$$ _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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

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12 Mar 2015, 19:40
Mountain14 wrote:
After Solving the equation , I got A

Because you didn't distribute the first negative sign on the first fraction. I made the same mistake. 8th grade rules but still find them hard to remember.
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Re: If k#0 and k - (3 -2k^2)/k = x/k, then x = [#permalink]

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12 Mar 2015, 23:12
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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GMAT On Demand Course $299 Free Online Trial Hour Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7377 Location: Pune, India Followers: 2288 Kudos [?]: 15127 [1] , given: 224 Re: If k#0 and k - (3 -2k^2)/k = x/k, then x = [#permalink] ### Show Tags 10 Apr 2016, 23:10 1 This post received KUDOS Expert's post 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. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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

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18 Apr 2016, 06:31
I get it. Thank you for both of the explanations!
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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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30 Apr 2017, 01:33
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, 01:33
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