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Is |qp + q^2| > qp?

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Is |qp + q^2| > qp?  [#permalink]

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New post 13 Jan 2017, 09:07
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Is |qp + q^2| > qp?  [#permalink]

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New post 13 Jan 2017, 09:55
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1
Bunuel wrote:
Is |qp + q^2| > qp?

(1) q > 0
(2) q/p > 1



Hi,

On the first look itself, the equation will hold true for every value except when left side is 0.
And the left side will be 0, only when q is 0..
So is |qp+q^2|>qp actually MEANS is q NOT EQUAL to 0

Let's see the statements..
1) q>0..
Ans is YES, q is not equal to 0..
Suff

2) q/P>1..
Again ans is YES..
If q was 0, q/P=0..
Suff

D
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: Is |qp + q^2| > qp?  [#permalink]

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New post 13 Jan 2017, 10:57
My answer is D.
Taking any value of q or p, absolute value will be greater than the product of the two value here.
Option 1: As q is positive, either value of p negative or positive will true for |qp + q^2| > qp
Option 2: As q>p, Either positive value of both, both negative value or q positive and p negative results in |qp + q^2| > qp.
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Is |qp + q^2| > qp?  [#permalink]

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New post Updated on: 13 Jan 2017, 22:24
Bunuel wrote:
Is \(|qp + q^{2}| > qp\) ?

(1) \(q > 0\)
(2) \(\frac{q}{p}> 1\)


Square both side \(|qp + q^{2}| > qp\)
\((qp+q^2)^{2}\) - \((qp)^{2}> 0\)
\((qp)^{2}+2q^{2}(qp) + (q^{2})^{2}-(qp)^{2}> 0\)
\(2q^{2}(qp) + (q^2)^{2}> 0\)
\(2(qp)+q^{2} > 0\)
\(q (2p+q) > 0\)

(1) \(q > 0\), q must be greater than zero in order for \(q (2p+q) > 0\) to hold. Sufficient

(2) \(\frac{2(qp)+q^{2}}{qp} > 0\), \(2 + \frac{q}{p} > 0\) , \(\frac{q}{p}> 1\) . Therefore, it is sufficient to solve the equation.

Answer : D
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Originally posted by hazelnut on 13 Jan 2017, 20:29.
Last edited by hazelnut on 13 Jan 2017, 22:24, edited 4 times in total.
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Re: Is |qp + q^2| > qp?  [#permalink]

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New post 13 Jan 2017, 21:52
ziyuenlau wrote:
Bunuel wrote:
Is \(|qp + q^{2}| > qp\) ?

(1) \(q > 0\)
(2) \(\frac{q}{p}> 1\)


Square both side \(|qp + q^{2}| > qp\)
\((qp+q^2)^{2}\) - \((qp)^{2}> 0\)
\((qp)^{2}+2q^{2}(qp) + (q^{2})^{2}-(qp)^{2}> 0\)
\(2q^{2}(qp) + (q^2)^{2}> 0\)
\(2(qp)+q^{2} > 0\)

(1) \(q > 0\), q must be greater than zero in order for \(2(qp)+q^{2}> 0\) to hold. Sufficient

(2) \(\frac{2(qp)+q^{2}}{qp} > 0\), \(2 + \frac{q}{p} > 0\) . \(\frac{q}{p}> 1\) . Therefore, it is sufficient to solve the equation.

Answer : D


Hi
Relook into the coloured portion above..
Say q is 2 and p is -8..
2pq+q^2=2*2*-8+2^2= -32+4=-28, this is not GREATER than 0..
On other hand if both q and p are negative, 2pq+q^2>0..
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: Is |qp + q^2| > qp?  [#permalink]

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New post 13 Jan 2017, 21:54
2
Bunuel wrote:
Is |qp + q^2| > qp?

(1) q > 0
(2) q/p > 1


Solution: Answer is D..!!

Take any value of q&p except zero , the situation would hold true..
Irrespective of whether q>p or p>q..Anything would hold true..
Example Let q=2 And p=3
|6+4|=6
Anything would hold true except the condition which i mentioned in the BOLD....

So answer is D...
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Is |qp + q^2| > qp?  [#permalink]

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New post 13 Jan 2017, 22:27
chetan2u wrote:
ziyuenlau wrote:
Bunuel wrote:
Is \(|qp + q^{2}| > qp\) ?

(1) \(q > 0\)
(2) \(\frac{q}{p}> 1\)


Square both side \(|qp + q^{2}| > qp\)
\((qp+q^2)^{2}\) - \((qp)^{2}> 0\)
\((qp)^{2}+2q^{2}(qp) + (q^{2})^{2}-(qp)^{2}> 0\)
\(2q^{2}(qp) + (q^2)^{2}> 0\)
\(2(qp)+q^{2} > 0\)

(1) \(q > 0\), q must be greater than zero in order for \(2(qp)+q^{2}> 0\) to hold. Sufficient

(2) \(\frac{2(qp)+q^{2}}{qp} > 0\), \(2 + \frac{q}{p} > 0\) . \(\frac{q}{p}> 1\) . Therefore, it is sufficient to solve the equation.

Answer : D


Hi
Relook into the coloured portion above..
Say q is 2 and p is -8..
2pq+q^2=2*2*-8+2^2= -32+4=-28, this is not GREATER than 0..
On other hand if both q and p are negative, 2pq+q^2>0..


chetan2u, I have factorised the equation. \(2 (qp)+q^{2} > 0\) , \(q (2p+q) > 0\) . Therefore, \(q > 0\) and \(2p + q > 0\) .
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Is |qp + q^2| > qp?  [#permalink]

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New post 16 Mar 2018, 08:01
chetan2u wrote:
Bunuel wrote:
Is |qp + q^2| > qp?

(1) q > 0
(2) q/p > 1



Hi,

On the first look itself, the equation will hold true for every value except when left side is 0.
And the left side will be 0, only when q is 0..
So is |qp+q^2|>qp actually MEANS is q NOT EQUAL to 0

Let's see the statements..
1) q>0..
Ans is YES, q is not equal to 0..
Suff

2) q/P>1..
Again ans is YES..
If q was 0, q/P=0..
Suff

D


Hi Chetan,

Have a doubt, if we solve the inequality after squaring just to remove the modulus in any case,
we end with,

q^3(p+2q) > 0; this means q > 0 and p+2q >0 || p+2q <0 and q <0; if this is true, then how can q > 0 sufficient ? What am I missing here?
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Re: Is |qp + q^2| > qp?  [#permalink]

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New post 21 Mar 2018, 20:12
srinjoy1990 wrote:
chetan2u wrote:
Bunuel wrote:
Is |qp + q^2| > qp?

(1) q > 0
(2) q/p > 1



Hi,

On the first look itself, the equation will hold true for every value except when left side is 0.
And the left side will be 0, only when q is 0..
So is |qp+q^2|>qp actually MEANS is q NOT EQUAL to 0

Let's see the statements..
1) q>0..
Ans is YES, q is not equal to 0..
Suff

2) q/P>1..
Again ans is YES..
If q was 0, q/P=0..
Suff

D


Hi Chetan,

Have a doubt, if we solve the inequality after squaring just to remove the modulus in any case,
we end with,

q^3(p+2q) > 0; this means q > 0 and p+2q >0 || p+2q <0 and q <0; if this is true, then how can q > 0 sufficient ? What am I missing here?



Hi....
You have gone wrong by squaring both sides..
You can square an INEQUALITY, only if both sides are POSITIVE..
Here you do not know if pq>0
Example..
|-3+2|>-3....1>-3
Square both sides..
1^2>(-3)^2......1>9?????

So be sure both sides are POSITIVE
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: Is |qp + q^2| > qp?  [#permalink]

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New post 22 Mar 2018, 08:59
1
chetan2u wrote:
Bunuel wrote:
Is |qp + q^2| > qp?

(1) q > 0
(2) q/p > 1



Hi,

On the first look itself, the equation will hold true for every value except when left side is 0.
And the left side will be 0, only when q is 0..
So is |qp+q^2|>qp actually MEANS is q NOT EQUAL to 0

Let's see the statements..
1) q>0..
Ans is YES, q is not equal to 0..
Suff

2) q/P>1..
Again ans is YES..
If q was 0, q/P=0..
Suff

D


Hello chetan2u,

Please assist me with this question.

My approach:

Is |qp + q^2| > qp?

if qp + q^2 >0 then
=> qp + q^2 > qp
=> q^2>0
=> |q|>0; basically q not = 0

if qp + q^2 <0 then
=> qp+q^2<-qp
=> 2qp+q^2<0
=> q(2p+q)<0

so we need to find is "q not = 0" and "q(2p+q)<0"

stmt 1: q>0 -->
then is 2p+q<0
=>is q<-2p
but we can't say that since we don't know sign of p; insufficient

stmt 2: q/p > 1 --> clearly insufficient since we don't know signs of p and q

stmt 1 and 2: q>0 therefore p must be +ve
from stmt 1: is q<-2p ----> since p and q are +ve so "-2p" would be -ve
Hence, the above inequality fails ----------> sufficient
My answer=C

Please correct the fault in my analysis

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Re: Is |qp + q^2| > qp?  [#permalink]

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New post 22 Mar 2018, 09:16
1
gmatexam439 wrote:
chetan2u wrote:
Bunuel wrote:
Is |qp + q^2| > qp?

(1) q > 0
(2) q/p > 1



Hi,

On the first look itself, the equation will hold true for every value except when left side is 0.
And the left side will be 0, only when q is 0..
So is |qp+q^2|>qp actually MEANS is q NOT EQUAL to 0

Let's see the statements..
1) q>0..
Ans is YES, q is not equal to 0..
Suff

2) q/P>1..
Again ans is YES..
If q was 0, q/P=0..
Suff

D


Hello chetan2u,

Please assist me with this question.

My approach:

Is |qp + q^2| > qp?

if qp + q^2 >0 then
=> qp + q^2 > qp
=> q^2>0
=> |q|>0; basically q not = 0

if qp + q^2 <0 then
=> qp+q^2<-qp
=> 2qp+q^2<0
=> q(2p+q)<0

so we need to find is "q not = 0" and "q(2p+q)<0"


stmt 1: q>0 -->
then is 2p+q<0
=>is q<-2p
but we can't say that since we don't know sign of p; insufficient

stmt 2: q/p > 1 --> clearly insufficient since we don't know signs of p and q

stmt 1 and 2: q>0 therefore p must be +ve
from stmt 1: is q<-2p ----> since p and q are +ve so "-2p" would be -ve
Hence, the above inequality fails ----------> sufficient
My answer=C

Please correct the fault in my analysis

Regards



Hi..

On your approach..
1) you have complicated the equation.
2) the coloured portion is incomplete..q(2p+q)<0
So if Q>0, 2p+Q<0..2p<-q.... Possible p is some NEGATIVE term
If q<0, 2p+q>0.....2p>-q..... P is some POSITIVE term possible
But the equation is NOT possible when q=0.. so here too, we have to find if Q=0, otherwise q can take any value..

From both cases ONLY one thing is to be checked-" if q=0?"
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2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: Is |qp + q^2| > qp?  [#permalink]

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New post 21 Apr 2018, 05:59
i have used the following approach-
but i think i am missing something here----
|qp+q^2| > qp
|q(p+q)|>qp

assume q is positive
|p+q|>p
as q is positive , hence p will be positive

so for the question - it will negate if any answer choice proves p is negative.

assume q is negative
|p+q|<p
as q is negative, p is positive.

stmt 1..
q>0, 1 st case....

stmt 2...
q/p>1
hence both q and p have same signs...
as p is always +ve, q will be +ve,
hence case 1 again...
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Is |qp + q^2| > qp?  [#permalink]

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New post 21 Apr 2018, 07:54
chetan2u wrote:
Bunuel wrote:
Is |qp + q^2| > qp?

(1) q > 0
(2) q/p > 1



Hi,

On the first look itself, the equation will hold true for every value except when left side is 0.
And the left side will be 0, only when q is 0..
So is |qp+q^2|>qp actually MEANS is q NOT EQUAL to 0

Let's see the statements..
1) q>0..
Ans is YES, q is not equal to 0..
Suff

2) q/P>1..
Again ans is YES..
If q was 0, q/P=0..
Suff

D



Hi chetan2u,

I think the approach that gmatexam439 used (above in the chain) is correct and necessary. I am saying this by the following general understanding in modulus that:
If |x|>a,
then either x>a, if x>0
or x<-a, if x<0

Now, the question is, is |qp + q^2| > qp?
This transcends into asking:
QI: Is qp + q^2 > qp, if (qp + q) >0?
OR
QII: Is qp + q^2 < -qp, if (qp + q) <0?

However, as my logic goes, if the stmt1 or stmt2 manages to answer "either of the 2 questions", then we are done. This is essentially because, there is a "OR" between the subordinate questions Q1 and Q2.
Here, both statements easily answer QI. So, we need not proceed to QII, which I agree complicates the question and hence the answer choice is D.

But, if the main question would have been in the format:
Is |x|<a?, which would mean asking:
Is x<a ? if x>0
AND
Is x>-a ? if x<0
In such a case, I think we will have to consider both statements equally and make sure that we are able to arrive at an answer for both the subordinate questions using the 2 statements provided with that question.

Please let me know if my analysis is correct.
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