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Is qp + q^2 > qp? [#permalink]
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13 Jan 2017, 09:07
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Is qp + q^2 > qp? [#permalink]
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13 Jan 2017, 09:55
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 0Let'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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Re: Is qp + q^2 > qp? [#permalink]
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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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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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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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Re: Is qp + q^2 > qp? [#permalink]
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13 Jan 2017, 21:54
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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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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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 0Let'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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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 0Let'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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Re: Is qp + q^2 > qp? [#permalink]
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22 Mar 2018, 08:59
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 0Let'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
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Re: Is qp + q^2 > qp? [#permalink]
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22 Mar 2018, 09:16
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 0Let'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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Re: Is qp + q^2 > qp? [#permalink]
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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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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 0Let'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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