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Is p + pz = p? (1) p = 0 (2) z = 0

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Is p + pz = p? (1) p = 0 (2) z = 0  [#permalink]

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New post 04 Jun 2017, 06:11
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A
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D
E

Difficulty:

  15% (low)

Question Stats:

75% (00:37) correct 25% (00:34) wrong based on 168 sessions

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Re: Is p + pz = p? (1) p = 0 (2) z = 0  [#permalink]

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New post 04 Jun 2017, 06:33
Bunuel wrote:
Is p + pz = p?

(1) p = 0

(2) z = 0


We can rewrite the statement

P + pz =p

Is pz = 0?
Statement 1) If P = 0, pz = 0, SUFFICIENT
Statement 2) If z = 0, pz = 0, SUFFICIENT

Either Statement is Sufficient

Answer D
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Re: Is p + pz = p? (1) p = 0 (2) z = 0  [#permalink]

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New post 05 Jun 2017, 06:20
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Bunuel wrote:
Is p + pz = p?

(1) p = 0

(2) z = 0


Target question: Is p + pz = p?

This is a great candidate for rephrasing the target question.

Take Is p + pz = p? and subtract p from both sides to get: Is pz = 0?

In order for pz to equal 0, it must be the case that either p = 0 or z = 0 (or they both equal 0)

REPHRASED target question: Does either p or z equal 0?

Statement 1: p = 0
This answers our REPHRASED target question perfectly.
Since we can answer the REPHRASED target question with certainty, statement 1 is SUFFICIENT

Statement 2: z = 0
This answers our REPHRASED target question perfectly.
Since we can answer the REPHRASED target question with certainty, statement 2 is SUFFICIENT

Answer:

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Re: Is p + pz = p? (1) p = 0 (2) z = 0  [#permalink]

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New post 06 Jun 2017, 16:49
Bunuel wrote:
Is p + pz = p?

(1) p = 0

(2) z = 0


Let’s simplify the question:

Is p + pz = p?

Is pz = 0?

Statement One Alone:

p = 0

Since p = 0, pz = 0. Statement one is sufficient to answer the question.

Statement Two Alone:

z = 0

Since z = 0, pz = 0. Statement two is sufficient to answer the question.

Answer: D
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Re: Is p + pz = p? (1) p = 0 (2) z = 0  [#permalink]

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New post 23 Oct 2017, 04:53
1
Bunuel wrote:
Is p + pz = p?

(1) p = 0

(2) z = 0



i took

First simplification as

p+pz = p
p(1+z)=p
1+z=p/p
1+z=1

then i got stuck with statement 1 ..

please help
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Re: Is p + pz = p? (1) p = 0 (2) z = 0  [#permalink]

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New post 12 Sep 2018, 15:01
Bunuel wrote:
Is p + pz = p?

(1) p = 0

(2) z = 0


\(?\,\,\,\,:\,\,\,\,\,p\left( {1 + z} \right)\,\,\mathop = \limits^? \,\,p\)

\(\left( 1 \right)\,\,\,p = 0\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle\)

\(\left( 2 \right)\,\,\,z = 0\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle\)


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

Regards,
Fabio.
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Is p + pz = p? (1) p = 0 (2) z = 0  [#permalink]

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New post 12 Sep 2018, 15:09
Harikrushna Kotadiya wrote:
Bunuel wrote:
Is p + pz = p?

(1) p = 0

(2) z = 0



i took

First simplification as

p+pz = p
p(1+z)=p
1+z=p/p
1+z=1

then i got stuck with statement 1 ..

please help


Hi Harikrushna Kotadiya!

p+pz = p is true if, and only if, p(1+z)=p is true

This is perfect (and marvellous, see my post above)!

But the following stage: p(1+z) = p implies 1+z=p/p is wrong, because p IS zero, and you cannot divide both sides of the equation in bold by p because of that!

In fact, to divide an equation by ANY unknown value (a "letter", a "variable"), you must be sure it cannot assume the zero value... got it?

Example:

\(abc = {b^2} + \frac{{4bc}}{d}\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,ac = b + \frac{{4c}}{d}\,\,\,\) only if b is different from zero!

(Important: we IMPLICITLY assume d is not zero, otherwise one of the parcels on the left equation - therefore the whole equation - would not have meaning...)

Regards,
Fabio.
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