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# Is j-k > 1/j - 1/k ?

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Is j-k > 1/j - 1/k ? [#permalink]

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13 Apr 2013, 11:37
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65% (hard)

Question Stats:

50% (02:24) correct 50% (01:05) wrong based on 20 sessions

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Is j-k > 1/j - 1/k ?

1. jk> -1

2. -1< k< j

can I prove it using algebra and not picking number ?
[Reveal] Spoiler: OA

Last edited by guerrero25 on 13 Apr 2013, 12:00, edited 2 times in total.
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Re: Is j-k > 1/j - 1/k ? [#permalink]

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13 Apr 2013, 12:07
1
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guerrero25 wrote:
Is j-k > 1/j - 1/k ?

1. jk> -1

2. -1< k< j

can I prove it using algebra and not picking number ?

is j-k > k-j / jk i.e. (j-k) - [(k-j)/jk] > 0 , is [jk(j-k) - (k-j)]/jk > 0 , i.e is[ jk(j-k) + (j-k)]/jk > 0 , i.e. is [(J-K)(JK+1)\ JK > 0

from 1

-1<jk<0 JK not = 0

i.e jk is -ve fraction, now to evaluate the question we ve ( jk+1) +ve and jk -ve , and thus we need to know j, k individually to evaluate (j-k).... no enough info ... insuff

from 2

still same like above .......insuff

both .......insuff E
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Re: Is j-k > 1/j - 1/k ? [#permalink]

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13 Apr 2013, 12:22
1
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guerrero25 wrote:
Is j-k > 1/j - 1/k ?

1. jk> -1

2. -1< k< j

can I prove it using algebra and not picking number ?

$$j-k > \frac{1}{j} - \frac{1}{k}$$ equals $$\frac{jk(j-k)-(k-j)}{jk}>0$$ don't divide by jk because it could be 0
this is true if N+ and D+ or N- and D-

1. jk> -1 D could be -ve or +ve, no info about N. Not sufficient
2. -1< k< j j-k>0 and k-j<0 Num= $$jk(+ve)-(-ve)$$
jk could be +ve or -ve , Not sufficient

1+2 The numerator is jk(j-k)-(k-j) and from 2 we know that j-k>0 and k-j<0 so Num= $$jk(+ve)-(-ve)=$$ and the Den is jk> -1.
Pick jk>0: Num>0 [(+ve)(+ve)-(-ve)>0] Den>0 so the answer is YES
pick -1<jk<0: Den<0 and there are no douts here, but what about the N? is N=(-ve)(+ve)-(-ve) a negative number? it depends, we cannot say.
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Re: Is j-k > 1/j - 1/k ? [#permalink]

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15 Apr 2013, 11:09
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guerrero25 wrote:
Is j-k > 1/j - 1/k ?

1. jk> -1

2. -1< k< j

Basically the question boils down to : Is (j-k)(1+jk)jk>0.

From F.S 1, we have (1+jk)>0. Thus, we have to answer whether (j-k)jk>0. Also for jk>-1, we could have both j,k>0; both j,k<0 or j,k of the opposite signs.Insufficient.

From F.S 2, we have (j-k)>0. Thus, we have to answer whether (1+jk)jk>0. This is possible only if jk>0 OR if jk<-1. Insufficient.

Taking both together, we know that both (1+jk) and (j-k) are positive. Thus, we have to answer whether jk>0. Just as above, Insufficient.

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Re: Is 100 j-k > 1/j - 1/k ? [#permalink]

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13 Apr 2013, 11:39
guerrero25 wrote:
Is 100 j-k > 1/j - 1/k ?

1. jk> -1

2. -1< k< j

can I prove it using algebra and not picking number ?

plz revisit question is it 100j-k or 100(j-k) or the 100 is not there to start with
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Re: Is j-k > 1/j - 1/k ? [#permalink]

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13 Apr 2013, 11:54
[/quote]

plz revisit question is it 100j-k or 100(j-k) or the 100 is not there to start with[/quote]

Edited . Sorry about the junk number .
Re: Is j-k > 1/j - 1/k ?   [#permalink] 13 Apr 2013, 11:54
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