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Carcass
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x>0 [#permalink] New post   21 Oct 2018, 08:22
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Question Stats:

89% (00:40) correct 10% (00:44) wrong based on 75 sessions

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\(x > 0\)

Quantity A
Quantity B
\(\frac{x+1}{x}\)
\(\frac{x}{x+1}\)


Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.
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pranab223
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Re: x>0 [#permalink] New post   21 Oct 2018, 19:40
Carcass wrote:
\(x > 0\)

Quantity A
Quantity B
\(\frac{x+1}{x}\)
\(\frac{x}{x+1}\)


Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
The relationship cannot be determined from the information given.


\(\frac{x+1}{x} = 1 + \frac{1}{x}\),

This equation must be greater than 1 always

where as,

\(\frac{x}{x+1}\) must be less than 1, because the denominator is greater compared to QTY A


therefore QTY A > QTY B
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Re: x>0 [#permalink] New post   25 Oct 2018, 21:39
it was a good one.
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nikitap
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Re: x>0 [#permalink] New post   01 Mar 2023, 10:23
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Multiple both quantities by x

QA -> (x+1/x) * x which is equal to x+1
QB -> (x/x+1) * x which is equal to x^2/x+1

Multiply both quantities by x+1

QA -> (x+1)^2
QB -> x^2

Now compare (x+1)^2 with (x)^2

Value of x+1 is always greater than x, therefore (x+1)^2 is greater than (x)^2

QA > QB
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Re: x>0 [#permalink]
01 Mar 2023, 10:23
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