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# If x ≠ 0, what is the value of k?

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Math Expert
Joined: 02 Sep 2009
Posts: 52108
If x ≠ 0, what is the value of k?  [#permalink]

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03 Apr 2018, 20:16
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Difficulty:

25% (medium)

Question Stats:

73% (00:49) correct 27% (01:17) wrong based on 98 sessions

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If x ≠ 0, what is the value of k?

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

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

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Joined: 05 Feb 2016
Posts: 144
Location: India
Concentration: General Management, Marketing
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Re: If x ≠ 0, what is the value of k?  [#permalink]

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03 Apr 2018, 20:54
1
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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Joined: 26 Mar 2013
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If x ≠ 0, what is the value of k?  [#permalink]

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04 Apr 2018, 14:19
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

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Joined: 21 Jan 2013
Posts: 7
Re: If x ≠ 0, what is the value of k?  [#permalink]

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28 Nov 2018, 08:19
Mo2men wrote:
Bunuel wrote:
If x ≠ 0, what is the value of k?

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

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

(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

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, 07:14
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) $$\frac{k}{x} = 2$$

(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

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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Re: If x ≠ 0, what is the value of k? &nbs [#permalink] 06 Dec 2018, 07:14
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