If x ≠ 0, what is the value of k? : Data Sufficiency (DS)

2018-04-03T21:16:52-07:00

If x ≠ 0, what is the value of k? (1) [m][fraction]3/4x[/fraction] + [fraction]k/3x[/fraction] = [fraction]1/x[/fraction][/m] (2) [m][fraction]k/x[/fraction] = 2[/m]

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If x ≠ 0, what is the value of k?

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If x ≠ 0, what is the value of k? [#permalink]

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03 Apr 2018, 21:16

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28% (01:45) wrong

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If x ≠ 0, what is the value of k?

(1) \(\frac{3}{4x} + \frac{k}{3x} = \frac{1}{x}\)

(2) \(\frac{k}{x} = 2\)

Spoiler: OA

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Re: If x ≠ 0, what is the value of k? [#permalink]

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03 Apr 2018, 21:54

Bunuel wrote:

If x ≠ 0, what is the value of k?

(1) \(\frac{3}{4x} + \frac{k}{3x} = \frac{1}{x}\)

(2) \(\frac{k}{x} = 2\)

From1:since x ≠ 0.
dividing x from both side
\(\frac{3}{4} + \frac{k}{3} = \frac{1}{}\)

one variable one equation sufficient.

from2:
two variable one equation ,insufficient.

hence A

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If x ≠ 0, what is the value of k? [#permalink]

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04 Apr 2018, 15:19

Bunuel wrote:

If x ≠ 0, what is the value of k?

(1) \(\frac{3}{4x} + \frac{k}{3x} = \frac{1}{x}\)

(2) \(\frac{k}{x} = 2\)

(1) \(\frac{3}{4x} + \frac{k}{3x} = \frac{1}{x}\)

As x ≠ 0 &and can't be zero actually), then we can cancel out x

\(\frac{3}{4} + \frac{k}{3} =1\)

We have an equation with 1 unknown (K)

Sufficient

(2) \(\frac{k}{x} = 2\)

This a ratio so we can't determine value of k

insufficient

Answer: A

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Re: If x ≠ 0, what is the value of k? [#permalink]

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28 Nov 2018, 09:19

Mo2men wrote:

Bunuel wrote:

If x ≠ 0, what is the value of k?

(1) \(\frac{3}{4x} + \frac{k}{3x} = \frac{1}{x}\)

(2) \(\frac{k}{x} = 2\)

Can you provide some clarity on how you eliminated X from the equation without simply assuming x = 1? Thx.

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Re: If x ≠ 0, what is the value of k? [#permalink]

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06 Dec 2018, 08:14

chowarth wrote:

Mo2men wrote:

Bunuel wrote:

If x ≠ 0, what is the value of k?

(1) \(\frac{3}{4x} + \frac{k}{3x} = \frac{1}{x}\)

(2) \(\frac{k}{x} = 2\)

Can you provide some clarity on how you eliminated X from the equation without simply assuming x = 1? Thx.

\(\frac{3}{4x} + \frac{k}{3x} = \frac{1}{x}\) If you now multiply by \(x\) then: \(\frac{3}{4x}*x + \frac{k}{3x}*x = \frac{1}{x}*x\) --> Then the \(x\) in each denominator chancels out \(\frac{3}{4} + \frac{k}{3} = 1\)
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