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Is pq = 1? (1) pqp = p (2) qpq = q

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Is pq = 1? (1) pqp = p (2) qpq = q  [#permalink]

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New post Updated on: 01 Oct 2013, 03:21
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Is pq = 1?

(1) pqp = p

(2) qpq = q

Originally posted by DFG5150 on 16 May 2009, 21:17.
Last edited by Bunuel on 01 Oct 2013, 03:21, edited 2 times in total.
Added the OA.
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Re: Is pq = 1?  [#permalink]

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New post 01 Oct 2013, 04:37
9
3
warabull wrote:
fameatop wrote:
abilash10 wrote:
yes but if (1) gives us PQ= 1 as a possible conclusion, and so does (2), then cant we conclude that PQ=1 ?


That's not correct always and for the same reason i combined both the statements.


Hey can you please explain when to combine the statements and when not to ?
I was under the impression that for such questions, one needs to look at the common solution from both the qquestions to arrive at the answer.
If you could explain when to use these two approaches, it would help.

Thanks !


Notice that p=0 from the first statement does not exclude q being 0 too and similarly q=0 from the second statement does not exclude p being 0 too.

Is pq = 1?

(1) pqp = p --> p(pq-1)=0 --> either p=0 and q=(any number), including 0 so in this case pq=0 not 1 or pq=1. Not sufficient.

(2) qpq = q --> q(pq-1)=0 --> either q=0 and p=(any number), including 0 so in this case pq=0 not 1 or pq=1. Not sufficient.

(1)+(2) When combined we have that either p=q=0 or pq=1. Not sufficient.

Answer: E.

When we consider two statements together we should take the values which satisfy both statements. For this question \(pq=1\) satisfies both statement, but \(p=q=0\) also satisfies both statements. So what you call "common solution" for this question is: \(pq=1\) OR \(pq=0\neq{1}\).

Hope it's clear.
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Re: Is pq = 1?  [#permalink]

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New post 28 Jul 2013, 20:50
8
1
abilash10 wrote:
Is pq = 1?
(1) pqp = p
(2) qpq = q


Statement 1-
pqp = p
p (pq-1)=0
It means p=0 or pq=1
Not sufficient

Statement 2-
qpq = q
q (pq-1)=0
It means q=0 or pq=1
Not sufficient

Statement 1 & 2-
Combining both statements we get \(p^3q^2\) - p = 0
p(\(p^2q^2\) - 1) = 0
p (pq-1)(pq+1) = 0
p=0 or pq=1 or pq=-1
Not sufficient
Answer E
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Re: A One product  [#permalink]

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New post 17 May 2009, 00:29
Statement 1

pqp = p
Dividing both sides in p
pq = 1

sufficient

Statement 2

qpq = q
Dividing both sides in q
qp = 1

sufficient

So the answer is (D)
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Re: A One product  [#permalink]

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New post 17 May 2009, 19:05
4
DFG5150 wrote:
Is pq = 1?

(1) pqp = p

(2) qpq = q


It's certainly possible, using either or both statements, that p = q = pq = 1. It's also possible that p = q = pq = 0. So the answer is E.
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Re: A One product  [#permalink]

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New post 17 May 2009, 19:32
1
Greenberg wrote:
Statement 1

pqp = p
Dividing both sides in p
pq = 1

sufficient

Statement 2

qpq = q
Dividing both sides in q
qp = 1

sufficient

So the answer is (D)


Since we don't know whether p=0 or q=0, we can't divide the ecuations by these variables.
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Re: A One product  [#permalink]

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New post 26 May 2009, 00:17
1
Greenberg wrote:
Statement 1

pqp = p
Dividing both sides in p
pq = 1

sufficient

Statement 2

qpq = q
Dividing both sides in q
qp = 1

sufficient

So the answer is (D)


You forgot to consider whether p or q are 0, in which case you'd be dividing by 0 (which is undefined). If it's not stated that the variable can't be 0, move the variables to one side and factor.
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Re: Is pq = 1?  [#permalink]

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New post 28 Jul 2013, 21:21
yes but if (1) gives us PQ= 1 as a possible conclusion, and so does (2), then cant we conclude that PQ=1 ?
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Re: Is pq = 1?  [#permalink]

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New post 28 Jul 2013, 21:29
abilash10 wrote:
yes but if (1) gives us PQ= 1 as a possible conclusion, and so does (2), then cant we conclude that PQ=1 ?


That's not correct always and for the same reason i combined both the statements.
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Re: Is pq = 1?  [#permalink]

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New post 01 Oct 2013, 03:05
fameatop wrote:
abilash10 wrote:
yes but if (1) gives us PQ= 1 as a possible conclusion, and so does (2), then cant we conclude that PQ=1 ?


That's not correct always and for the same reason i combined both the statements.


Hey can you please explain when to combine the statements and when not to ?
I was under the impression that for such questions, one needs to look at the common solution from both the qquestions to arrive at the answer.
If you could explain when to use these two approaches, it would help.

Thanks !
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Re: Is pq = 1? (1) pqp = p (2) qpq = q  [#permalink]

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New post 03 Oct 2013, 10:53
DFG5150 wrote:
Is pq = 1?

(1) pqp = p

(2) qpq = q


This is how I approached the question; if anyone can kindly point out my flaw.
The question is asking pq=1

or p=1/q or q=1/p

Meaning are p and q reciprocal?. Yes/No

Statement 1
pqp=p
p^2q=p
q=p/p^2
q=1/p so Sufficient

Statement 2

qpq=q
q^2p=q
p=q/q^2
p=1/q so Sufficient

Therefore the answer is C
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Re: Is pq = 1? (1) pqp = p (2) qpq = q  [#permalink]

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New post 04 Oct 2013, 01:18
suk1234 wrote:
DFG5150 wrote:
Is pq = 1?

(1) pqp = p

(2) qpq = q


This is how I approached the question; if anyone can kindly point out my flaw.
The question is asking pq=1

or p=1/q or q=1/p

Meaning are p and q reciprocal?. Yes/No

Statement 1
pqp=p
p^2q=p
q=p/p^2
q=1/p so Sufficient

Statement 2

qpq=q
q^2p=q
p=q/q^2
p=1/q so Sufficient

Therefore the answer is C


If you say that each statement is sufficient, then you mean that the answer should be D, not C, right?

Now, the problem in your solution is that you cannot reduce pqp=p by p to get pq=1. This is because p can be zero and division be zero is not allowed.

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

So when you divide by p you assume, with no ground for it, that p does not equal to zero thus exclude a possible solution. Notice that p=0 and pq=1 both satisfy pqp=p.

The same logic applies to the second statement.

Hope it helps.
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Re: Is pq = 1? (1) pqp = p (2) qpq = q  [#permalink]

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New post 04 Oct 2013, 01:36
Bunuel wrote:
suk1234 wrote:
DFG5150 wrote:
Is pq = 1?

(1) pqp = p

(2) qpq = q


This is how I approached the question; if anyone can kindly point out my flaw.
The question is asking pq=1

or p=1/q or q=1/p

Meaning are p and q reciprocal?. Yes/No

Statement 1
pqp=p
p^2q=p
q=p/p^2
q=1/p so Sufficient

Statement 2

qpq=q
q^2p=q
p=q/q^2
p=1/q so Sufficient

Therefore the answer is C


If you say that each statement is sufficient, then you mean that the answer should be D, not C, right?

Now, the problem in your solution is that you cannot reduce pqp=p by p to get pq=1. This is because p can be zero and division be zero is not allowed.

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

So when you divide by p you assume, with no ground for it, that p does not equal to zero thus exclude a possible solution. Notice that p=0 and pq=1 both satisfy pqp=p.

The same logic applies to the second statement.

Hope it helps.


Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.


Yeaps! got it now. Thanks for the reply, it's very helpful. And thanks for that important bit of information. :thumbup:
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Re: Is pq = 1? (1) pqp = p (2) qpq = q  [#permalink]

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New post 04 Oct 2013, 02:02
Bunuel wrote:

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.



This is a very important rule. Thanks bunnel.
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Re: Is pq = 1? (1) pqp = p (2) qpq = q  [#permalink]

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New post 30 Sep 2018, 15:00
DFG5150 wrote:
Is pq = 1?

(1) pqp = p

(2) qpq = q

\(pq\,\,\mathop = \limits^? \,\,1\)

Let´s go straight to bifurcate (1+2), to guarantee the right answer is (E), indeed.

\(\left( {1 + 2} \right)\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {p,q} \right) = \left( {1,1} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {p,q} \right) = \left( {0,0} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.\)


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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