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(1) The positive diff erence between p and q is 2. (2) q - p < 1

If p and q are negative, is p / q > 1

Given: \(p<0\) and \(q<0\). Question: is \(\frac{p}{q}>1\) --> multiply both sides by \(q\) and as it's negative flip the sign: is \(p<q\)? or is \(p-q<0\)?

(1) The positive diff erence between p and q is 2 --> \(|p-q|=2\): either \(p-q>0\) (answer NO) and \(p-q=2\) or \(p-q<0\) (answer YES) and \(p-q=-2\). Not sufficient.

(2) q - p < 1 (\(p-q>-1\)) --> if \(q=-1\) and \(p=-1\) then the answer will be NO but if \(q=-1\) and \(p=-1.5\) then the answer will be YES. Not sufficient.

(1)+(2) As from (2) \(p-q>-1\) then from (1) \(p-q=2\) so \(p-q>0\) and we have the answer NO. Sufficient.

(1) The positive diff erence between p and q is 2. (2) q - p < 1

If p and q are negative, is p / q > 1

Given: \(p<0\) and \(q<0\). Question: is \(\frac{p}{q}>1\) --> multiply both sides by \(q\) and as it's negative flip the sign: is \(p<q\)? or is \(p-q<0\)?

(1) The positive diff erence between p and q is 2 --> \(|p-q|=2\): either \(p-q>0\) (answer NO) and \(p-q=2\) or \(p-q<0\) (answer YES) and \(p-q=-2\). Not sufficient.

I'm very confused. First why do you have absolute value? How did you derive p-q>0? p-q<0? p-q=-2? Any way to demonstrate? or explain the concepts? Thank you very much.

rxs0005 wrote:

(2) q - p < 1 (\(p-q>-1\)) --> if \(q=-1\) and \(p=-1\) then the answer will be NO but if \(q=-1\) and \(p=-1.5\) then the answer will be YES. Not sufficient.

(1) The positive diff erence between p and q is 2. (2) q - p < 1

If p and q are negative, is p / q > 1

Given: \(p<0\) and \(q<0\). Question: is \(\frac{p}{q}>1\) --> multiply both sides by \(q\) and as it's negative flip the sign: is \(p<q\)? or is \(p-q<0\)?

(1) The positive diff erence between p and q is 2 --> \(|p-q|=2\): either \(p-q>0\) (answer NO) and \(p-q=2\) or \(p-q<0\) (answer YES) and \(p-q=-2\). Not sufficient.

I'm very confused. First why do you have absolute value? How did you derive p-q>0? p-q<0? p-q=-2? Any way to demonstrate? or explain the concepts? Thank you very much.

rxs0005 wrote:

(2) q - p < 1 (\(p-q>-1\)) --> if \(q=-1\) and \(p=-1\) then the answer will be NO but if \(q=-1\) and \(p=-1.5\) then the answer will be YES. Not sufficient.

How can q and p both equal -1?

"The positive difference between p and q is 2" means that the distance between p and q is 2, which can be expressed as \(|p-q|=2\). For example positive difference between -5 and -3 is 2: |-5-(-3)|=2.

Next: Absolute value properties: When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\);

So, for \(|p-q|=2\): If \(p-q>0\) then \(|p-q|=p-q=2\) (example: \(p=-3\) and \(q=-5\)); If \(p-q<0\) then \(|p-q|=-(p-q)=q-p=2\) (example: \(p=-5\) and \(q=-3\));

(1) The positive diff erence between p and q is 2. (2) q - p < 1

It is a good question and you can solve it logically too.

Given p and q are negative so p/q must be positive (negative/negative). Whether p/q is greater than 1 depends on whether p < q. If p < q, then yes, p/q > 1 (if p is more negative, it has higher absolute value). Else p/q is not greater than 1.

So we have to find out whether p is less than q.

(1) The positive diff erence between p and q is 2.

This only tells us that the difference between them is 2. It doesn't tell us which one is greater so not sufficient.

(2) q - p < 1 This tells us that if q is greater than p, it is less than 1 greater than p. q can be equal to p or less than p but if it is greater than p, it is certainly less than 1 greater than p. This means (q = -1.2, p = -1.9), (q = -23, p = -23.4), (q = -3, p = -3), (q = -4, p = -2) are possible pairs (and many more). Again, we don't know whether p is greater or q so not sufficient.

Using both together, we know that the difference between p and q is 2 and if q is greater than p, it is less than 1 greater than p. Since the difference between them is 2, q cannot be greater than p so p must be greater than q. We can say that "No. p is not less than q." Hence sufficient. Answer (C)
_________________

(1) The positive diff erence between p and q is 2. (2) q - p < 1

st. (1) the +ve difference = |p-q| = 2 implies p-q > 0 or p-q < 0 if p-q > 0 then p>q then p/q> 1

But if p-q < 0 then p<q or p/q cannot be greater than 1 anyway st. (1) gives two options which leads "insufficient"

|p-q| = 2 gives you two cases: Either p-q = 2 or q-p = 2 We do not know whether p is smaller than q.

st. (2) q-p<1 (this could be p-q < -1 which mean p-q> -1 )

q-p < 1 is the same as p-q > -1 (when you multiply both sides by -1)

implies q-p = 0 or q-p is -ve if q-p=0then p/q > 1 is not possible

But if q-p is -ve then it gives different values of p and q which says both - p/q>1 or p/q<1 however st.(2) insufficient

If q-p<1, q could be greater or p could be greater. So we again cannot figure whether p is smaller than q

Combining together st. (1) and st. (2) p-q > -1 and p-q =2 implies p>q or we can say p/q>1 Sufficient

Combining, stmnt 1 tells us that either p-q = 2 or q-p = 2. Stmnt 2 tells us that q-p<1. Hence q-p cannot be 2. Therefore, p-q must be 2. p must be greater than q. We know that p is greater so p/q is not greater than 1 (since p and q are both negative) Answer (C).

Re: If p and q are negative, is p / q > 1 (1) The positive [#permalink]

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27 Nov 2012, 18:56

Luckily I went for the diagram and was able to do it in under 2 minutes.

Draw a line with 0 in the middle. p and q are both to the left of 0. We only know this much. We don't know their position with respect to 0, i. e we don't know whether q or p is closer to 0 or even whether p and q are the same number, both negative. Question asks whether p/q>1 meaning is p more negative than q? This suggests that the question is about the position of p and q with respect to each other and 0.

1. the positive difference between them is 2 suggests that the distance between p and q is 2 units. this only tells us the distance and not which one is more negative than the other. Not Sufficient. 2. q-p<1 suggests that the difference between the two is less than 1. but their degree of negativity is not clear. by itself, this statement points to the possibility that q and p may be the same number; -2-(-2)=0<1, or one of them could be slightly more negative and still have satisfy q-p<1. so Not Sufficient.

when you take 1 and 2 together, the possibility that the two are the same number is eliminated because 1 says that the two numbers are 2 units apart. so now, the number line will have p and q standing at 2 units apart and based on statement 2, q has to be more negative than p. Hence C.

vikram4689 wrote:

i was able to solve but within 3 minutes.... how to solve this question in less than 2 min.

(1) The positive diff erence between p and q is 2. (2) q - p < 1

If p and q are negative, is p / q > 1

Given: \(p<0\) and \(q<0\). Question: is \(\frac{p}{q}>1\) --> multiply both sides by \(q\) and as it's negative flip the sign: is \(p<q\)? or is \(p-q<0\)?

(1) The positive diff erence between p and q is 2 --> \(|p-q|=2\): either \(p-q>0\) (answer NO) and \(p-q=2\) or \(p-q<0\) (answer YES) and \(p-q=-2\). Not sufficient.

(2) q - p < 1 (\(p-q>-1\)) --> if \(q=-1\) and \(p=-1\) then the answer will be NO but if \(q=-1\) and \(p=-1.5\) then the answer will be YES. Not sufficient.

(1)+(2) As from (2) \(p-q>-1\) then from (1) \(p-q=2\) so \(p-q>0\) and we have the answer NO. Sufficient.

Hi Bunuel! Could you please explain to me, why you concluded from (1) + (2) that p-q>0? The question doesn't say that p and q are integers, so shouldn't the answer be E then? because p-q>-1 (from I) could mean that p-q<0 or p-q>0 ..

Answer: C.

Hi Bunuel! Could you please explain to me, why you concluded from (1) + (2) that p-q>0? The question doesn't say that p and q are integers, so shouldn't the answer be E then? because p-q>-1 (from I) could mean that p-q<0 or p-q>0 ..

(1) The positive diff erence between p and q is 2. (2) q - p < 1

If p and q are negative, is p / q > 1

Given: \(p<0\) and \(q<0\). Question: is \(\frac{p}{q}>1\) --> multiply both sides by \(q\) and as it's negative flip the sign: is \(p<q\)? or is \(p-q<0\)?

(1) The positive diff erence between p and q is 2 --> \(|p-q|=2\): either \(p-q>0\) (answer NO) and \(p-q=2\) or \(p-q<0\) (answer YES) and \(p-q=-2\). Not sufficient.

(2) q - p < 1 (\(p-q>-1\)) --> if \(q=-1\) and \(p=-1\) then the answer will be NO but if \(q=-1\) and \(p=-1.5\) then the answer will be YES. Not sufficient.

(1)+(2) As from (2) \(p-q>-1\) then from (1) \(p-q=2\) so \(p-q>0\) and we have the answer NO. Sufficient.

Hi Bunuel! Could you please explain to me, why you concluded from (1) + (2) that p-q>0? The question doesn't say that p and q are integers, so shouldn't the answer be E then? because p-q>-1 (from I) could mean that p-q<0 or p-q>0 ..

Answer: C.

Hi Bunuel! Could you please explain to me, why you concluded from (1) + (2) that p-q>0? The question doesn't say that p and q are integers, so shouldn't the answer be E then? because p-q>-1 (from I) could mean that p-q<0 or p-q>0 ..

As @grumpytesttaker said the best way to solve such problems is to use the number line.

The question stem says that p and q are negative so we can have 2 scenarios (p to the left of q or p to the right of q)

Now to definitively say whether p/q>1 we need to find if |p| > |q| as both are negative so there is no question of signs. Since both are negative nos |p| > |q| only if p is to the left of q on the Number Line. So we just need to find if p is to the left or right of q.

(1) |p - q| = 2, this means p and q have a separation of 2. But this is possible if p is to the left of q or p is to the right of q. So this statement doesn't help us. Not Sufficient.

(2) q - p < 1, Since both nos are negative we can rewrite this statement as |p| - |q| < 1. Now if p is to the left of q (|p| > |q|) then the separation between p and q have to be less than 1. But if p is to the right of q (|p| < |q|) then the separation can be anything. Since this statement doesn't say if p is to the left or right of q, it is Not Sufficient.

(1) & (2) Now if we combine the 2 statements we can see that p cannot be to the left of q because (1) -- |p| - |q| = 2 and (2) -- |p| - |q| < 1 together is not possible. So the only possibility is p is to the right of q, which answers the question, since |p| < |q| hence p/q < 1

So (1) & (2) put together answers the question. Sufficient. Answer C _________________

(1) The positive diff erence between p and q is 2. (2) q - p < 1

If p and q are negative, is p / q > 1

Given: \(p<0\) and \(q<0\). Question: is \(\frac{p}{q}>1\) --> multiply both sides by \(q\) and as it's negative flip the sign: is \(p<q\)? or is \(p-q<0\)?

(1) The positive difference between p and q is 2 --> \(|p-q|=2\): either \(p-q>0\) (answer NO) and \(p-q=2\) or \(p-q<0\) (answer YES) and \(p-q=-2\). Not sufficient.

(2) q - p < 1 (\(p-q>-1\)) --> if \(q=-1\) and \(p=-1\) then the answer will be NO but if \(q=-1\) and \(p=-1.5\) then the answer will be YES. Not sufficient.

(1)+(2) As from (2) \(p-q>-1\) then from (1) \(p-q=2\) so \(p-q>0\) and we have the answer NO. Sufficient.

Hi Bunuel! Could you please explain to me, why you concluded from (1) + (2) that p-q>0? The question doesn't say that p and q are integers, so shouldn't the answer be E then? because p-q>-1 (from I) could mean that p-q<0 or p-q>0 ..

Answer: C.

Hi Bunuel! Could you please explain to me, why you concluded from (1) + (2) that p-q>0? The question doesn't say that p and q are integers, so shouldn't the answer be E then? because p-q>-1 (from I) could mean that p-q<0 or p-q>0 ..

Sure. From (1) we have two possible cases: \(p-q=2\) or \(p-q=-2\). Since from (1) we have that \(p-q>-1\), then \(p-q\neq{-2}\), thus \(p-q=2>0\).

Re: If p and q are negative, is p/q > 1 [#permalink]

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27 May 2015, 11:28

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Re: If p and q are negative, is p/q > 1 [#permalink]

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30 Jun 2016, 02:59

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Re: If p and q are negative, is p/q > 1 [#permalink]

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14 Aug 2017, 10:39

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