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# Is rs = rx − 2 ? (1) r is an odd number (2) x = s + 2

Author Message
Manager
Joined: 17 Nov 2009
Posts: 235

Kudos [?]: 58 [0], given: 17

Is rs = rx − 2 ? (1) r is an odd number (2) x = s + 2 [#permalink]

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17 May 2010, 13:44
00:00

Difficulty:

(N/A)

Question Stats:

67% (00:24) correct 33% (00:47) wrong based on 6 sessions

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Is rs = rx − 2 ?

(1) r is an odd number
(2) x = s + 2

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/is-rs-rx-158366.html

Kudos [?]: 58 [0], given: 17

Senior Manager
Joined: 25 Jun 2009
Posts: 298

Kudos [?]: 145 [0], given: 6

Re: Is rs = rx − 2 ? (1) r is an odd number (2) x = s + 2 [#permalink]

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17 May 2010, 13:50
agnok wrote:
Is rs=rx-2?

1. r is odd
2. x=s+2

Is there some thing missing in this question ?

Anyway, Question is asking whether 2= rx-rs ?

1. Clearly insufficient

2. putting the value of x = > 2= rs + 2r -rs = > 2= 2r

which means the st 2 will be sufficient if we know that r =1 but else its not and hence answer should be E and not B

Kudos [?]: 145 [0], given: 6

Manager
Joined: 17 Nov 2009
Posts: 235

Kudos [?]: 58 [0], given: 17

Re: Is rs = rx − 2 ? (1) r is an odd number (2) x = s + 2 [#permalink]

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24 Jun 2010, 08:22
Thanks nitishmahajan! I see it

Kudos [?]: 58 [0], given: 17

Math Expert
Joined: 02 Sep 2009
Posts: 41699

Kudos [?]: 124810 [0], given: 12079

Re: Is rs = rx − 2 ? (1) r is an odd number (2) x = s + 2 [#permalink]

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11 Aug 2017, 03:09
Is rs = rx − 2 ?

Is $$rs = rx-2$$? --> is $$r(x-s)=2$$?

(1) r is an odd number. Not sufficient: consider r=1 and x-s=2 AND r=1 and x-s=3.

(2) x = s + 2 -->$$x-s=2$$ --> the question becomes: is $$r=1$$? We don't know that. Not sufficient.

(1)+(2) From (2) the question became: is $$r=1$$? and (1) says that r=odd, but we don't know whether it's 1 or any other odd number. Not sufficient.

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/is-rs-rx-158366.html
_________________

Kudos [?]: 124810 [0], given: 12079

Re: Is rs = rx − 2 ? (1) r is an odd number (2) x = s + 2   [#permalink] 11 Aug 2017, 03:09
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