DS with Integer

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DS with Integer

by anwarma » Thu Jun 12, 2008 3:05 pm
If N and P are integer, is P > 0?

1) N +1 > 0
2) NP > 0

=====================================

The question I have is , if you choose negative numbers for N and P because integers in this case could be positive and negative numbers, then in that case only statement 2 holds when N and P are negative integer numbers, the NP remains greater than 0.

Can someone just help me understand why the answer is C not B as I thought for this DS problem. How is statement 1 helps in C.
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Source: — Data Sufficiency |

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Re: DS with Integer

by Ian Stewart » Thu Jun 12, 2008 4:41 pm
anwarma wrote:If N and P are integer, is P > 0?

1) N +1 > 0
2) NP > 0

=====================================

The question I have is , if you choose negative numbers for N and P because integers in this case could be positive and negative numbers, then in that case only statement 2 holds when N and P are negative integer numbers, the NP remains greater than 0.
From your explanation, it sounds to me as though you're approaching Data Sufficiency 'backwards', which is very common for people to do before they gain some experience with this type of question. In Data Sufficiency, you don't ever test whether Statements 1) and 2) are true. They must be true- they are facts. Your task is, by using the facts in the Statements, to decide whether you can answer the question.

Statement 1) alone doesn't help, since it provides no information about P.

Statement 2) alone doesn't tell us whether P is positive. It may be that both P and N are negative.

Looking at Statement 1) and 2) together, we know, from Statement 2:

NP > 0

This means that N and P are either both negative or both positive. From Statement 1), we know

N + 1 > 0
N > -1

and since N is an integer, N >= 0. But we know from the above that N can't be zero, so N > 0, and from Statement 2, if N > 0, P must be positive as well. So we have enough information to answer the question if we use both statements.

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by Rodrigo » Wed May 13, 2009 11:25 pm
What if we only consider statement 2 in the following way:

np>0
p>0/n

since 0 devided by any number is 0, therefore: p>0

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Re: DS

by Ian Stewart » Thu May 14, 2009 3:00 am
Rodrigo wrote:What if we only consider statement 2 in the following way:

np>0
p>0/n
When you rewrite the inequality above, you've divided on both sides by n. If n is negative, you'd need to reverse the inequality when you do that. So unless you know that n is positive, what you've done above isn't correct.

In any case, if np > 0, that just means that the product of n and p is positive, or that n and p have the same sign: both are positive, or both are negative.
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DS

by Rodrigo » Thu May 14, 2009 4:36 am
Thanks Ian!

I missed out the possibility of n being negative.

Rodrigo