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M01-12

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M01-12  [#permalink]

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New post 15 Sep 2014, 23:15
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

73% (00:32) correct 27% (00:39) wrong based on 165 sessions

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Re M01-12  [#permalink]

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New post 15 Sep 2014, 23:15
2
Official Solution:


(1) \(p-q=0\). Rearrange: \(p=q\), so \(p^3-q^3=p^3-p^3=0\). Sufficient.

(2) \(p+q=0\). Rearrange: \(q=-p\), so \(p^3-q^3=p^3-(-p^3)=2p^3\), so the value we are looking for depends on the value of \(p\). Not sufficient.


Answer: A
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Re: M01-12  [#permalink]

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New post 05 Feb 2015, 05:27
Hi Bunuel,
I solved the question in the following way.
1)p-q=0;
(p-q)^3=0;
p^3-q^3-3p^2q-3pq^2=0;
p^3-q^3-3pq(p-q)=0;
//But we know that p-q=0
p^3-q^3=0//Sufficient

2)p+q=0;
From this we will get p^3+q^3 which is not sufficient.

Is this approach correct?

Thanks in advance.
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Re: M01-12  [#permalink]

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New post 02 May 2017, 07:13
I, on the other hand, changed the question stem to "is p^3=q^3?"

I believe it's easier to tackle the question like this.
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M01-12  [#permalink]

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New post 20 Jul 2017, 04:53
Bunuel wrote:
Official Solution:


(1) \(p-q=0\). Rearrange: \(p=q\), so \(p^3-q^3=p^3-p^3=0\). Sufficient.

(2) \(p+q=0\). Rearrange: \(q=-p\), so \(p^3-q^3=p^3-(-p^3)=2p^3\), so the value we are looking for depends on the value of \(p\). Not sufficient.


Answer: A


Hi,

Is expanding the equation \(p^3 - q^3\) as\((p-q)\) \(p^2 - q^2\)wrong? From here we can use the \(a^2 - b^2\)exapnsion and solve.

Kindly explain!
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Re: M01-12  [#permalink]

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New post 20 Jul 2017, 05:01
ashikaverma13 wrote:
Bunuel wrote:
Official Solution:


(1) \(p-q=0\). Rearrange: \(p=q\), so \(p^3-q^3=p^3-p^3=0\). Sufficient.

(2) \(p+q=0\). Rearrange: \(q=-p\), so \(p^3-q^3=p^3-(-p^3)=2p^3\), so the value we are looking for depends on the value of \(p\). Not sufficient.


Answer: A


Hi,

Is expanding the equation \(p^3 - q^3\) as\((p-q)\) \(p^2 - q^2\)wrong? From here we can use the \(a^2 - b^2\)exapnsion and solve.

Kindly explain!


\(p^3 - q^3=(p - q) (p^2 + p q + q^2)\), not \((p-q)(p^2 - q^2)\).
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New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M01-12  [#permalink]

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New post 03 Apr 2018, 14:34
To make my life easier I used Picking Numbers for P^3 - Q^3 = ?


Statement 1:
P - Q = 0

P = Q

If P = 1 Q = 1 then 1^3 - 1^3 = 1 - 1 = 0

If P = 2 Q = 2 then 2^3 - 2^3 = 8 - 8 = 0

Sufficient.


Statement 2:
P + Q = 0

P = - Q

If P = -1 Q = 1 then -1^3 - 1^3 = - 1 - 1 = -2

If P = -2 Q = 2 then -2^3 - 2^3 = -8 - 8 = -16

2 different answers. Insufficient.

Answer A

Thanks! Ale :-)
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Re: M01-12  [#permalink]

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New post 01 Sep 2018, 10:45
Trick :-
    (p-q) is a factor of the given statement. Since p & q both have odd number of powers.
      (p+q) will be factor only if p & q are raised to even numbers of powers. for e.g. \(p2−q2\)
      GMAT Club Bot
      Re: M01-12 &nbs [#permalink] 01 Sep 2018, 10:45
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      M01-12

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