Is |qp + q^2| qp?

Question

Is |qp + q^2| > qp?

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

chetan2u · Accepted Answer

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

prashantrchawla · Answer

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.

BillyZ · Answer

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

chetan2u · Answer

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..

tanmeet · Answer

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...

BillyZ · Answer

chetan2u , I have factorised the equation.  2 (qp)+q^{2} > 0  ,  q (2p+q) > 0  . Therefore,  q > 0  and  2p + q > 0  .

srinjoy1990 · Answer

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?

chetan2u · Answer

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

gmatexam439 · Answer

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

chetan2u · Answer

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?"

dewdrops909 · Answer

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...

aviejay · Answer

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|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.

chetan2u · Answer

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
eq{0}

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

2) \frac{q}{p}>1 ..
Again ans is YES..
If q was 0, q/p would be 0..
Suff

D

GyMrAT · Answer

Given 
is |qp + q^2| > qp ?

can be written as, |q(p + q)| > qp

Statement 1: q > 0

Plug in numbers, 
q = 1, p = -1, we get LHS = 0 & RHS = -1....we get YES 
q = 1, p = 0, we get LHS = 1, RHS = 0...we get YES
q = 1, p = 1, we get LHS = 2, RHS = 1...we get YES

Statement 1 is Sufficient.

Statement 2: q/p > 1, hence q & p have the same sign.

Plug in numbers
q = 2, p = 1, we get LHS = 6, RHS = 2...we get YES
q = -2, p = -1, we get LHS = 6, RHS = 2, we get YES

Statement 2 is Sufficient.

Answer D.

Thanks,
GyM

durgamadhab · Answer

Left side is always positive. q^2 is making left side greater than RHS when p,q>0. when p<0, RHS is negative and LHS positive.

So, D

PKGMAT · Answer

Hi Chetan,

Could you please explain why we are not considering value of q=0;

It is not given anywhere in the original question

chetan2u · Answer

I think you are referring to statement II
 (2) q/p > 1
Now if we consider q=0.. \frac{q}{p}=\frac{0}{p}=0 
But it is given that  \frac{q}{p}>1 , so how can 0>1
this means  q
eq{0}

the equation q/p>1 itself means that q is not 0

hope it helps

rahulkashyap · Answer

can you explain how this can be done by opening the absolute sign?

one way is -- qp+q^2-qp>o , i.e. is q not equal to 0

second way is -qp-q^2> qp
simplified to 2qp + q^2<0

with this second way, how can we say with statement 1 we can prove sufficient

Bunuel   chetan2u

chetan2u · Answer

Let's take the second statement..
2pq+p^2<0
p^2 is surely positive so 2pq<0
This means - are p AND q of different sign
Statement II tells us both are of same sign.. 
Sufficient

Is |qp + q^2| > qp?

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Is |qp + q^2| > qp?

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