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DS question

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 Post subject: DS question
 Post Posted: Tue Nov 09, 2010 5:23 pm 
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if p^2 + q^2 =1 , Is p-q=1 ?

stmt1: p+q=1
stmt2: p is a positive integer.

OA: B

I have know clue how B is the answer. I got A. Please help.
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tarun.mahan
 Post subject: Re: DS question
 Post Posted: Wed Nov 10, 2010 12:51 pm 
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Because the option B says clearly that p is a positive INTEGER. so the value of p can only be 1 and th value of q = 0. Therefore the question is solved p - q = 1.

Hope this helps.
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atul.prasad
 Post subject: Re: DS question
 Post Posted: Wed Nov 10, 2010 3:44 pm 
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How did you get A?

if P+Q = 1 then (P+Q)^2 = P^2+Q^2+2PQ = 1

or 2PQ = 0 since P^2 + Q^2 = 1
so either of P or Q can be 0/1
so you can't say for sure if P-Q = 1

What tarun said is right

From the given equation Q^2 = 1 - P^2
and p^2 >=1 , since P is a +ve integer
Q^2 has to be 0 (square can't be negative).
So P = 1 and Q is 0

So B is sufficient
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RonPurewal
 Post subject: Re: DS question
 Post Posted: Fri Nov 12, 2010 8:37 pm 
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tarun.mahan wrote:
Because the option B says clearly that p is a positive INTEGER. so the value of p can only be 1 and th value of q = 0. Therefore the question is solved p - q = 1.

Hope this helps.


yep -- this is good.
it also falls under the general rubric of “problems on which you can't use pure algebra”. for a bunch more of these, check out the NOVEMBER 4 recording here:
http://www.manhattangmat.com/thursdays-with-ron.cfm

as for the first statement, after a little bit of reflection (or a lot of algebra -- a little bit of reflection is probably better (: ) you can come up with two possibilities:
p = 0, q = 1
or
p = 1, q = 0
one of these possibilities yields a "yes" to the prompt question, while the other yields a "no". so that's insufficient.
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