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If p, q, r, and s are nonzero numbers, is pr/qs > r/q?
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Updated on: 29 Aug 2014, 03:49
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If p, q, r, and s are nonzero numbers, is pr/qs > r/q? (1) p > s (2) rq > 0 Dear Experts, kindly help me with the above problem. My pick was C, but it's incorrect. pr/qs>r/q =>pqr >rqs => qr(ps)>0. So i picked C. I guess i am missing something here. Request your help.
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Originally posted by Aurion on 28 Aug 2014, 23:11.
Last edited by Bunuel on 29 Aug 2014, 03:49, edited 1 time in total.
Renamed the topic and edited the question.



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Re: If p, q, r, and s are nonzero numbers, is pr/qs > r/q?
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28 Aug 2014, 23:25
Aurion wrote: If p, q, r & s are nonzero numbers, Is pr/qs > r/q? 1) p>s 2) rq>0 We have to find whether when we multiply \(\frac{r}{q}\) by \(\frac{p}{s}\) will be greater than \(\frac{r}{q}\) Statement 1: Give no valuable information to solve. So insufficient. Statement 2: we know that \(\frac{r}{q}\) is positive. But no info about \(\frac{p}{s}\). So insufficient. Combining: If both p and s positive, we have a bigger value. But if p or s is negative we have smaller value. for example, if p=3 and s=1 and if p=1 and s=3. So we have 2 possibilities. So even together we cannot solve this question. Hence the answer is E.
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Re: If p, q, r, and s are nonzero numbers, is pr/qs > r/q?
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29 Aug 2014, 02:50
we can cancel r from numerator and q from denominator of both the sides. so simplified form will be is p/s>1 i.e. p and s must have same sign and p must be greater than s.
1. p>s... not sufficient to answer whether p and s have same sign 2. rq>0.... not relevant
1+2... still we can't say whether p and s have same sign Therefore ans=E



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Re: If p, q, r, and s are nonzero numbers, is pr/qs > r/q?
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29 Aug 2014, 03:06
If p, q, r & s are nonzero numbers, Is pr/qs > r/q? 1) p>s 2) rq>0
From St. 1 p>s as not sufficient caus no information about signs of p&s and about r & q From rq>o r & q can be positive or negative if both positive pr/qs>r/q cancel r/q both sides and we get p>s (St. 1.) both negative than p<s useless Combining St. 1 and 2 we get p>s hence not relevant Option E best answer.



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Re: If p, q, r, and s are nonzero numbers, is pr/qs > r/q?
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29 Aug 2014, 04:08
Aurion wrote: If p, q, r, and s are nonzero numbers, is pr/qs > r/q? (1) p > s (2) rq > 0 Dear Experts, kindly help me with the above problem. My pick was C, but it's incorrect. pr/qs>r/q =>pqr >rqs => qr(ps)>0. So i picked C. I guess i am missing something here. Request your help. If p, q, r, and s are nonzero numbers, is pr/qs > r/q? Is \(\frac{r}{q}*\frac{p}{s} > \frac{r}{q}\)? Note that we can neither reduce this inequality by r/q, nor crossmultiply because we don't know signs of the variables, thus don't know whether we should flip the sign of the inequality (recall that we must flip the sign of an inequality when multiplying/reducing by a negative value). (1) p > s. Not sufficient: we know nothing about r and q. (2) rq > 0. This implies that r/q is also greater than 0, so we can reduce by it and the question becomes: is \(\frac{p}{s} >0\)? We don't know that. Not sufficient. (1)+(2) The question became "is \(\frac{p}{s} >0\)?" and (1) says that p > s, which is clearly insufficient to answer that. Not sufficient. Answer: E. Hope it's clear.
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Re: If p, q, r, and s are nonzero numbers, is pr/qs > r/q?
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01 Nov 2018, 02:19
IF we solve the inequality pr/qs >rq by taking r/q to LHS
we get two solutions either r/q>0 & P/s >1 or r/q<0 and P/s <1
Statement 1 says P/s can be greater than 1 or can be less than 1
statement two on solving says that r/q is definitely greater than 0
when we combines stmt 1 and 2 don't we get r/q>0 & P/s >1 or R/q >0 & P/s <1
out of these the first one is a definite solution for our ineuality in question
Pls tell me what am i doing wrong ?



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If p, q, r, and s are nonzero numbers, is pr/qs > r/q?
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10 Nov 2018, 04:42
Bunuel wrote: Aurion wrote: If p, q, r, and s are nonzero numbers, is pr/qs > r/q?
(1) p > s (2) rq > 0
(2) rq > 0. This implies that r/q is also greater than 0, so we can reduce by it and the question becomes: is \(\frac{p}{s} >0\)? We don't know that. Not sufficient.
(1)+(2) The question became "is \(\frac{p}{s} >0\)?" and (1) says that p > s, which is clearly insufficient to answer that. Not sufficient.
Answer: E.
Hope it's clear. This may be a ridiculous question, and I'm sure I'm overlooking something glaringly obvious, but: rq>0 tells us both r and q are positive or negative. So, p/s*r/q>r/q is divided by "r/q", should it not result in p/s>1? I'm unable to figure out how p/s>0.



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Re: If p, q, r, and s are nonzero numbers, is pr/qs > r/q?
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10 Nov 2018, 05:54
NCRanjan wrote: IF we solve the inequality pr/qs >rq by taking r/q to LHS
we get two solutions either r/q>0 & P/s >1 or r/q<0 and P/s <1
Statement 1 says P/s can be greater than 1 or can be less than 1
statement two on solving says that r/q is definitely greater than 0
when we combines stmt 1 and 2 don't we get r/q>0 & P/s >1 or R/q >0 & P/s <1
out of these the first one is a definite solution for our ineuality in question
Pls tell me what am i doing wrong ? Hello Few things which are probably not right here: First, you are assuming pr/qs > r/q to be already true, thats why you have already started solving this inequality in your solution. This is NOT to be assumed to be true, infact this is what has to be determined whether its true or not. Second, from (1) you are assuming p/s > 1. We are NOT given that p/s > 1, rather we are given that p > s. You have divided this given inequality on both sides by 's', without knowing the sign of s (positive or negative). We cannot do that, or we have to take both the cases (one case where s > 0 and another case where s < 0).



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Re: If p, q, r, and s are nonzero numbers, is pr/qs > r/q?
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10 Nov 2018, 05:58
shaarang wrote: Bunuel wrote: Aurion wrote: If p, q, r, and s are nonzero numbers, is pr/qs > r/q?
(1) p > s (2) rq > 0
(2) rq > 0. This implies that r/q is also greater than 0, so we can reduce by it and the question becomes: is \(\frac{p}{s} >0\)? We don't know that. Not sufficient.
(1)+(2) The question became "is \(\frac{p}{s} >0\)?" and (1) says that p > s, which is clearly insufficient to answer that. Not sufficient.
Answer: E.
Hope it's clear. This may be a ridiculous question, and I'm sure I'm overlooking something glaringly obvious, but: rq>0 tells us both r and q are positive or negative. So, p/s*r/q>r/q is divided by "r/q", should it not result in p/s>1? I'm unable to figure out how p/s>0. Hello What you have done to conclude that p/s > 1 is correct, mathematically. BUT  you have already started working with the inequality pr/sq > r/q; meaning you have already assumed it to be true. This is NOT given, this is something we have to determine whether its true or not, so we cannot start working with it the way you have done here. Or maybe I am missing something which you would want to explain. Also what is your confusion with p/s > 0, I dont understand that query of yours.



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If p, q, r, and s are nonzero numbers, is pr/qs > r/q?
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10 Nov 2018, 06:29
reduced the expression to (r/q)[(ps)/s] So is (r/q)[(ps)/s] > 0 ? 1. Nothing about r/q or s Insuff 2. Nothing about ps or s Insuff Combining 1 and 2 Still nothing about sign of s. Hence E
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Re: If p, q, r, and s are nonzero numbers, is pr/qs > r/q?
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10 Nov 2018, 08:40
I went with thinking as the weights of p q r s
1. p>s doesnt tell us any thing about the weights of r and q... maybe r is less than q, which gives different answers for pr>qs 2. rq>0 so either both r and q are positive or negative, again nothing on the weights of these numbers.
Even together, the weights of r and q are not known Hence, E.



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Re: If p, q, r, and s are nonzero numbers, is pr/qs > r/q?
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10 Nov 2018, 18:14
Quote: Hello
What you have done to conclude that p/s > 1 is correct, mathematically. BUT  you have already started working with the inequality pr/sq > r/q; meaning you have already assumed it to be true.
This is NOT given, this is something we have to determine whether its true or not, so we cannot start working with it the way you have done here.
Or maybe I am missing something which you would want to explain.
Also what is your confusion with p/s > 0, I dont understand that query of yours. Ah, shoot. I knew it was something stupid. Thanks for clearing that up!




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