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Fun with r!

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Fun with r!

by TryHarder » Fri May 29, 2009 6:24 pm
Is rs = rx - 2?
1. r is an odd number
2. x = s + 2

Note: For those of you who have seen this question in the official source, try not to copy the official explanation. Is is not very convincing, hence this query.
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by raleigh » Fri May 29, 2009 8:00 pm
(1) r is odd.
Choose r = 1, s = 2, x = 4. Then rs = 2, rx - 2 = 2.
Choose r = 1, s = 2, x = 2. Then rs = 2, rx - 2 = 0.

Insufficient.

(2) x = s + 2. Substitute this into rx - 2 = r (s + 2) - 2 = rs + 2r - 2.

Set this equal to rs and we see this will be valid for rs= rs + 2r - 2 which simplifies to r=1.

Choose r = 1, x = 1, s = -1. Then rs = -1, and rx - 2 = 1 -2 = -1.
Choose r = 3, x = 2, s = 0. Then rs = 0, and rx - 2 = 6 - 2 = 4.

Insufficient.

(1)+(2)
Notice that in (2) we chose r to be odd in both cases. So the argument for (2) applies to this case.

Insufficient.

The answer is E.
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by TryHarder » Fri May 29, 2009 8:29 pm
raleigh,
I came up with the same answer, but the official answer is B, even though is is true only when r=2.

Came an expert shed some light on this?
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by raleigh » Fri May 29, 2009 8:54 pm
My counter example for (2) is correct so it cannot be B.

The OA is incorrect unless you mistyped the problem. Which problem is this in the OG?
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by TryHarder » Fri May 29, 2009 8:59 pm
I think you misunderstood my point, I did imply that I too think the answer should be E for the very obvious reason that statement 2 is true only for r=2 and not always. Also, there is nothing wrong with the problem as stated.
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by raleigh » Fri May 29, 2009 9:05 pm
I understand your point. I'd like to see the actual OA; what book is this from and what is the number for this problem? It has to be wrong.
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by TryHarder » Fri May 29, 2009 9:32 pm
McGraw Hills GMAT 2009 Practice Test 3: Q27
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by greymatter » Sat May 30, 2009 2:57 am
yes E.
B would have been the answer if it also tells us that r =1; which is is not so.
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by riya.sharma.sweet1984 » Mon Jun 01, 2009 1:58 am
Answer E :D
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by aj5105 » Mon Jun 01, 2009 3:58 am
In statement (2), we get r = 1. In the part underlined, you chose r = 3

Is this right? Please clarify.

IMO, you can as well use the same values what you used in Statement (1)

raleigh wrote:(1) r is odd.
Choose r = 1, s = 2, x = 4. Then rs = 2, rx - 2 = 2.
Choose r = 1, s = 2, x = 2. Then rs = 2, rx - 2 = 0.

Insufficient.

(2) x = s + 2. Substitute this into rx - 2 = r (s + 2) - 2 = rs + 2r - 2.

Set this equal to rs and we see this will be valid for rs= rs + 2r - 2 which simplifies to r=1.

Choose r = 1, x = 1, s = -1. Then rs = -1, and rx - 2 = 1 -2 = -1.
Choose r = 3, x = 2, s = 0. Then rs = 0, and rx - 2 = 6 - 2 = 4.

Insufficient.

(1)+(2)
Notice that in (2) we chose r to be odd in both cases. So the argument for (2) applies to this case.

Insufficient.

The answer is E.
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by sacx » Mon Jun 01, 2009 5:21 am
r0 + s = r0 + x - 2
s = x -2
x = s +2

STMT I

Does not tell anything about s or x

Insuff

STMT II

this is what the question is asking

Insuff

STMT I & II

Insuff, as we do not know anything about s or x

E
SACX
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