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Is |qp + q^2| > qp? (1) q > 0 (2) q/p > 1

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Is |qp + q^2| > qp? (1) q > 0 (2) q/p > 1 [#permalink]

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New post 10 Jul 2018, 20:50
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Difficulty:

  85% (hard)

Question Stats:

33% (01:49) correct 68% (01:39) wrong based on 40 sessions

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Re: Is |qp + q^2| > qp? (1) q > 0 (2) q/p > 1 [#permalink]

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New post 10 Jul 2018, 20:56
1
Bunuel wrote:
Is |qp + q^2| > qp?

(1) q > 0

(2) q/p > 1





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 \(q\neq{0}\)

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
_________________

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? (1) q > 0 (2) q/p > 1 [#permalink]

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New post 10 Jul 2018, 21:39
Bunuel wrote:
Is |qp + q^2| > qp?

(1) q > 0

(2) q/p > 1


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,
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Re: Is |qp + q^2| > qp? (1) q > 0 (2) q/p > 1 [#permalink]

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New post 10 Jul 2018, 23:12
Bunuel wrote:
Is |qp + q^2| > qp?

(1) q > 0

(2) q/p > 1


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
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Re: Is |qp + q^2| > qp? (1) q > 0 (2) q/p > 1 [#permalink]

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New post 12 Jul 2018, 11:16
chetan2u wrote:
Bunuel wrote:
Is |qp + q^2| > qp?

(1) q > 0

(2) q/p > 1





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 \(q\neq{0}\)

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


Hi Chetan,

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

It is not given anywhere in the original question
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Re: Is |qp + q^2| > qp? (1) q > 0 (2) q/p > 1 [#permalink]

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New post 14 Jul 2018, 01:08
PKGMAT wrote:
chetan2u wrote:
Bunuel wrote:
Is |qp + q^2| > qp?

(1) q > 0

(2) q/p > 1





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 \(q\neq{0}\)

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


Hi Chetan,

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

It is not given anywhere in the original question


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 \(q\neq{0}\)

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

hope it helps
_________________

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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Is |qp + q^2| > qp? (1) q > 0 (2) q/p > 1 [#permalink]

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New post 14 Jul 2018, 01:46
chetan2u wrote:
Bunuel wrote:
Is |qp + q^2| > qp?

(1) q > 0

(2) q/p > 1





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 \(q\neq{0}\)

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


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
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Re: Is |qp + q^2| > qp? (1) q > 0 (2) q/p > 1 [#permalink]

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New post 14 Jul 2018, 02:34
rahulkashyap wrote:
chetan2u wrote:
Bunuel wrote:
Is |qp + q^2| > qp?

(1) q > 0

(2) q/p > 1





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 \(q\neq{0}\)

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


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-p^2> qp
simplified to 2qp + p^2<0

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

Bunuel chetan2u


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
_________________

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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Is |qp + q^2| > qp? (1) q > 0 (2) q/p > 1 [#permalink]

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New post 14 Jul 2018, 02:37
Hi chetan2u

The question can be simplified into either (1) or (2)

In 2, on simplification, we get 2pq+p² <0
Using statement 1 of the question, we can not prove this as sufficient

That is my question

Also, as stated by you "2pq+p^2<0
p^2 is surely positive so 2pq<0
This means - are p AND q of different sign"

this is simplified into 2pq<-p^2 and not otherwise as stated by you.

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Is |qp + q^2| > qp? (1) q > 0 (2) q/p > 1   [#permalink] 14 Jul 2018, 02:37
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