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If p, q, r, and s are non-zero numbers, is pr/qs > r/q?

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Aurion
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If p, q, r, and s are non-zero numbers, is pr/qs > r/q? [#permalink] If p, q, r, and s are non-zero numbers, is pr/qs > r/q?   Updated on: 29 Aug 2014, 04:49
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If p, q, r, and s are non-zero numbers, is pr/qs > r/q?

(1) p > s
(2) rq > 0

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Dear Experts, kindly help me with the above problem. My pick was C, but it's incorrect.
pr/qs>r/q =>pqr >rqs => qr(p-s)>0. So i picked C. I guess i am missing something here. Request your help.
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E

Originally posted by Aurion on 29 Aug 2014, 00:11.
Last edited by Bunuel on 29 Aug 2014, 04:49, edited 1 time in total.
Renamed the topic and edited the question.
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Re: If p, q, r, and s are non-zero numbers, is pr/qs > r/q? [#permalink] If p, q, r, and s are non-zero numbers, is pr/qs > r/q?   29 Aug 2014, 00:25
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Aurion
If p, q, r & s are non-zero 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 non-zero numbers, is pr/qs > r/q? [#permalink] If p, q, r, and s are non-zero numbers, is pr/qs > r/q?   29 Aug 2014, 03:50
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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 non-zero numbers, is pr/qs > r/q? [#permalink] If p, q, r, and s are non-zero numbers, is pr/qs > r/q?   29 Aug 2014, 04:06
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If p, q, r & s are non-zero 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 non-zero numbers, is pr/qs > r/q? [#permalink] If p, q, r, and s are non-zero numbers, is pr/qs > r/q?   29 Aug 2014, 05:08
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Aurion
If p, q, r, and s are non-zero numbers, is pr/qs > r/q?

(1) p > s
(2) rq > 0

Show Spoiler
Dear Experts, kindly help me with the above problem. My pick was C, but it's incorrect.
pr/qs>r/q =>pqr >rqs => qr(p-s)>0. So i picked C. I guess i am missing something here. Request your help.

If p, q, r, and s are non-zero 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 cross-multiply 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 non-zero numbers, is pr/qs > r/q? [#permalink] If p, q, r, and s are non-zero numbers, is pr/qs > r/q?   01 Nov 2018, 03:19
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 non-zero numbers, is pr/qs > r/q? [#permalink] If p, q, r, and s are non-zero numbers, is pr/qs > r/q?   10 Nov 2018, 05:42
Bunuel
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If p, q, r, and s are non-zero 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 non-zero numbers, is pr/qs > r/q? [#permalink] If p, q, r, and s are non-zero numbers, is pr/qs > r/q?   10 Nov 2018, 06:54
NCRanjan
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 non-zero numbers, is pr/qs > r/q? [#permalink] If p, q, r, and s are non-zero numbers, is pr/qs > r/q?   10 Nov 2018, 06:58
shaarang
Bunuel
Aurion
If p, q, r, and s are non-zero 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 non-zero numbers, is pr/qs > r/q? [#permalink] If p, q, r, and s are non-zero numbers, is pr/qs > r/q?   10 Nov 2018, 07:29
reduced the expression to (r/q)[(p-s)/s]
So is (r/q)[(p-s)/s] > 0 ?
1. Nothing about r/q or s Insuff
2. Nothing about p-s 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 non-zero numbers, is pr/qs > r/q? [#permalink] If p, q, r, and s are non-zero numbers, is pr/qs > r/q?   10 Nov 2018, 09:40
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 non-zero numbers, is pr/qs > r/q? [#permalink] If p, q, r, and s are non-zero numbers, is pr/qs > r/q?   10 Nov 2018, 19:14
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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Re: If p, q, r, and s are non-zero numbers, is pr/qs > r/q? [#permalink] If p, q, r, and s are non-zero numbers, is pr/qs > r/q?   12 Aug 2022, 01:25
Aurion
If p, q, r, and s are non-zero numbers, is pr/qs > r/q?

(1) p > s
(2) rq > 0

Show Spoiler
Dear Experts, kindly help me with the above problem. My pick was C, but it's incorrect.
pr/qs>r/q =>pqr >rqs => qr(p-s)>0. So i picked C. I guess i am missing something here. Request your help.


Dear Expert/any kind soul,
My solution is as follows and I have not been able to see why E is the answer even after reading the existing posts. Please help to explain.

We need to find if the following is true-
pr/qs > r/q ?
r/q ( p/s - 1 ) > 0 ?
r/q > 0 ? AND p/s > 1 ?

When we combine, we get both the conditions above, so Op. C is the answer.
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Re: If p, q, r, and s are non-zero numbers, is pr/qs > r/q? [#permalink] If p, q, r, and s are non-zero numbers, is pr/qs > r/q?   06 May 2024, 12:22
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Re: If p, q, r, and s are non-zero numbers, is pr/qs > r/q? [#permalink]
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