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gmat6nplus1
Bunuel
If mn < np < 0, is n < 1?

(1) n is an integer
(2) m < p


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Reprhase mn<np<0 --> n(m-p)<0. Thus If n<0 m>p; If n>0 m<p.

Statement 1: n can be positive, negative, zero. Both scenarios are feasible. Not sufficient.

Statement 2. m<p thus n>0. n can be a proper fraction, an improper fraction, an irrational number, or a positive integer. We are not sure whether n < 1.

1+2) n is an integer --> eliminate proper fraction. Thus n is greater or equal to 1.

Answer C.


From what you said in Statement 1, I am pretty sure that n cannot equal 0 since mn < np < 0
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Bunuel
If mn < np < 0, is n < 1?

(1) n is an integer
(2) m < p


Kudos for a correct solution.

Reprhase mn<np<0 --> n(m-p)<0. Thus If n<0 m>p; If n>0 m<p.

Statement 1: n can be positive, negative, zero. Both scenarios are feasible. Not sufficient.

Statement 2. m<p thus n>0. n can be a proper fraction, an improper fraction, an irrational number, or a positive integer. We are not sure whether n < 1.

1+2) n is an integer --> eliminate proper fraction. Thus n is greater or equal to 1.

Answer C.


From what you said in Statement 1, I am pretty sure that n cannot equal 0 since mn < np < 0

Yes, n cannot be 0.

mn < np < 0
What does this imply? One of two cases:
Case 1: n is negative, m and p are positive. Also, m > p so that absolute value of m is greater than absolute value of p. This will make mn < np since n is negative.
Or
Case 2: n is positive, m and p are negative. In this case m < p

(1) n is an integer
We don't know whether n is positive or negative so not sufficient.

(2) m < p
If m < p, then n must be positive and m and p must be negative.
But n may lie between 0 and 1 or it may be 1 or greater. So we cannot say whether n is less than 1 or not.

Using both statements, we know that n cannot lie between 0 and 1 since it must be an integer. So n must be 1 or greater. We can answer the question with a definite 'no'.

Answer (C)
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gmat6nplus1
Bunuel
If mn < np < 0, is n < 1?

(1) n is an integer
(2) m < p


Kudos for a correct solution.

Reprhase mn<np<0 --> n(m-p)<0. Thus If n<0 m>p; If n>0 m<p.

Statement 1: n can be positive, negative, zero. Both scenarios are feasible. Not sufficient.

Statement 2. m<p thus n>0. n can be a proper fraction, an improper fraction, an irrational number, or a positive integer. We are not sure whether n < 1.

1+2) n is an integer --> eliminate proper fraction. Thus n is greater or equal to 1.

Answer C.


From what you said in Statement 1, I am pretty sure that n cannot equal 0 since mn < np < 0

Yes it cannot be zero, that was a typo. I am gonna edit it thanks
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Bunuel
If mn < np < 0, is n < 1?

(1) n is an integer
(2) m < p


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VERITAS PREP OFFICIAL SOLUTION:

Correct Answer: (C)

The very first part of this question – that mn<np – is key here. First of all, this provision makes it impossible that n = 0, because otherwise mn and np would equal 0, not less than 0. Next, observe that because of positive/negative number properties, if n is positive, both m and p have to be negative. In this case, since mn is less than np, m<p.

There is second possibility to consider: if n is negative, both m and p have to be positive. In this second case, m>p. After considering the implications of the question stem, we are ready to consider the statements. Statement (1) by itself is not sufficient because it gives us no information about whether n is positive or negative. Statement (2) gets us further, because it means that n must be positive. It is not fully sufficient though, since if n is positive, it could still be less than or greater than 1. At this point, we have eliminated answers (A), (B), and (D), and we have to decide between (C) and (E). Combining the statements gives us that n is a positive integer, which means that n could be 1 or any higher integer. None of these possibilities is less than 1. With that knowledge, we can answer the original yes/no question with a definitive no. So the correct answer is (C).
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If mn < np < 0, is n < 1?

mn-np <0 => n (m-p)<0

(1) n is an integer
n may take any values depending on values of m and p. Insufficient.
(2) m < p
(m-p)<0
Given n (m-p) <0 => n >0 but n can be 0<n<1 and n>1 so cannot answer. INSUFFICIENT

Combining 1 and 2, 0<n<1 is ruled out and only n>1 is left.
So can be answered. SUFFICIENT.
Hence C.
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