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r > zero

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r > zero [#permalink] New post 04 Feb 2018, 15:07
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Re: r > zero [#permalink] New post 04 Feb 2018, 20:32
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The easiest way to solve this one is to cross-multiply and then subtract the common values. We're allowed to do this, btw, because we've been told r is positive. If we hadn't been told anything about whether r is positive or negative, we wouldn't be able to use this technique. See attached picture. Since the simplest version of this problem is 0 vs 56, we can see that Quantity B is bigger.

IMG_20180208_113654.jpg [ 39.25 KiB | Viewed 3931 times ]



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Last edited by SherpaPrep on 08 Feb 2018, 08:43, edited 2 times in total.
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Re: r > zero [#permalink] New post 05 Feb 2018, 02:37
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Carcass wrote:

\(r > 0\)

Quantity A
Quantity B

Here given r>0

So the value of r has to be positive. It can be either integer or a fraction.

Now we know if denominator > numerator then the value will be always less than 1.

Now compare QTY A and QTY b,

we can see that QTY A has denominator r+8, and the denominator of QTY B is r, so what ever the value of r (i.e integer or fraction) the value of QTY B > QTY A

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Re: r > zero [#permalink] New post 07 Feb 2018, 02:12
Answer: B
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Re: r > zero [#permalink] New post 20 May 2018, 01:55
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Moreover , we can see that the denominator of quantity A is getting added to a positive number where as the quantity B is just remaining as r.

hence quantity A's fraction is getting smaller as compared to quantity B.

Hence option B.
Re: r > zero   [#permalink] 20 May 2018, 01:55
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