Inequality

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Source: — Data Sufficiency |

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IMO C

by sethids » Sat Aug 23, 2008 3:34 pm
C should be the answer.

Question stem indicates n and p are integers.

Stmt1: n+1 > 0. Says nothing about p. Insufficient. Left with B, C, or E.
Stmt2: np > 0. n and p could both be positive or negative. Insufficient. Left with C or E.

Combining both, from stmt1 we can deduce that n is positive as the sum of n and 1 is positive. (having n as a negative number < -1 will make the sum negative, if n = 0 or 1 then also the inequality would not hold true).

From stmt2 we know that product of n and p is positive. As n is positive (from stmt1) p has to be positive for the product to be positive.

Hence C.

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by II » Sat Aug 23, 2008 4:47 pm
Here's how I approached it ... and provide you with a view of my thought process.

Q-stem analysis:
Note: n and p are integers.
Is p>0 ? In other words is p positive.
(at this stage I am thinking that the question is going to test my knowledge of the positive/negative number properties)

Which of the 2 statements are easier to work with. Statement 1 is an easy. Lets start with that ... and draw out the AD/BCE answer grid.

(1) n+1 > 0 .... no information about p here. So we can cross out AD. Left with BCE.
Note: at this stage you may also want to quickly rearrange n+1>0 to n>-1

(2) np > 0
So based on our knowledge of positive and negatives ... we know that in order for the product of 2 integers to be positive, then the integers need to have the same sign. So either n and p are both negative, or they are both positive, in order for this inequality to be true.
So we cannot say whether p is positive or negative. Hence INSUFF. Cross out B. Left with choices C and E.

(1) and (2) together.

From (1) we know n>-1 ... so this means that n cannot be a negative integer. And this leads onto statement (2) If n cannot be a negative, then it has to be positive, which means that p has to be the same sign, and hence positive. SUFF. Choose C.

Hope this helps.

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by pepeprepa » Sun Aug 24, 2008 12:49 am
Thanks men, OA is C