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Re: If x >= 2400, then the value of 2x/(4x + 24) is approximately [#permalink]
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QZ wrote:
If \(x\geq {2400}\), then the value of \(\frac{2x}{4x + 24}\) is approximately

A. \(\frac{1}{7}\)

B. \(\frac{2}{7}\)

C. \(\frac{1}{2}\)

D. \(\frac{3}{5}\)

E. \(\frac{3}{2}\)


Plug in 2400 in the equation

\(\frac{2x}{4x + 24}\) = \(\frac{200}{401}\)

It is almost 1/2

Answer: C
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Re: If x >= 2400, then the value of 2x/(4x + 24) is approximately [#permalink]
Asked: If \(x\geq {2400}\), then the value of \(\frac{2x}{4x + 24}\) is approximately

Since x>>24, 24 may be ignored.

IMO C

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Re: If x >= 2400, then the value of 2x/(4x + 24) is approximately [#permalink]
AkshdeepS wrote:
If \(x\geq {2400}\), then the value of \(\frac{2x}{4x + 24}\) is approximately

A. \(\frac{1}{7}\)

B. \(\frac{2}{7}\)

C. \(\frac{1}{2}\)

D. \(\frac{3}{5}\)

E. \(\frac{3}{2}\)

­\(x\geq {2400}\),  \(\frac{2x}{4x + 24}\)

Or, \(\frac{x}{2x + 12}\)

So, \(\frac{2400}{4800 + 12}\)

Or, ~ \(\frac{2400}{4800}=\frac{1}{2}\), Answer must be (C)­
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Re: If x >= 2400, then the value of 2x/(4x + 24) is approximately [#permalink]
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