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Bunuel
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




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




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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Hello from the GMAT Club BumpBot!

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