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a, b, c, and d are integers; abcd≠0

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a, b, c, and d are integers; abcd≠0 [#permalink]

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16 Jan 2012, 00:43
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a, b, c, and d are integers; abcd >< 0; what is the value of cd?
1) c/b = 2/d
2) b^3*a^4*c = 27*a^4*c

solution here
[Reveal] Spoiler:
http://www.platinumgmat.com/gmat-practice-test/?state=qc-cn#show-explanation

brief expl from me
1 - clearly not suff, and we got that cd=2b
2 - clearly not suff, and we got that |b^3|=27 cause what if c negative? so we can't say for sure that b=3 like stated in OA. Am I right?
[Reveal] Spoiler: OA

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Re: a, b, c, and d are integers; abcd≠0 [#permalink]

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16 Jan 2012, 02:09
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a, b, c, and d are integers; abcd >< 0; what is the value of cd?
1) c/b = 2/d
2) b^3*a^4*c = 27*a^4*c

SOLUTION:

statement 1: c/b = 2/d
cd = 2b, we don't know the value of b. so. we can't find the value of cd.
NOT SUFFICIENT

statement 2 : b^3*a^4*c = 27*a^4*c
==> a^4 * c (b^3-27) = 0
it means, a^4 =0 or c =0 or b^3 =27 so, b = 3
so, here we can get different values of cd.
NOT SUFFICIENT

after combining both statement , we can get value of cd = 2b =6

Hence the ans is C.

I HOPE IT WILL BE HELPFUL.
PS: EDITED after bunuel explanation
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Last edited by 321kumarsushant on 16 Jan 2012, 03:30, edited 2 times in total.

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a, b, c, and d are integers; abcd≠0 [#permalink]

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16 Jan 2012, 03:11
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Expert's post
a, b, c, and d are integers; abcd≠0; what is the value of cd?

(1) c/b = 2/d --> $$cd=2b$$, we don't know the value of $$b$$ to get the single numerical value of $$cd$$. Not sufficient.

(2) b^3*a^4*c = 27*a^4*c --> as $$a$$ and $$c$$ does not equal to zero we can safely reduce both parts by $$a^4*c$$ --> $$b^3=27$$ --> $$b=3$$. Not sufficient.

(1)+(2) As from (1) $$cd=2b$$ and from (2) $$b=3$$ then $$cd=2b=6$$. Sufficient.

Runner2 wrote:
2 - clearly not suff, and we got that |b^3|=27 cause what if c negative? so we can't say for sure that b=3 like stated in OA. Am I right?

Odd roots have the same sign as the base of the root. For example, $$\sqrt[3]{125} =5$$ and $$\sqrt[3]{-64} =-4$$.

So $$\sqrt[3]{27}=3$$ and not $$-3$$ --> $$3^3=27$$ and $$(-3)^3=-27$$.

Hope its' clear.
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Re: a, b, c, and d are integers; abcd≠0 [#permalink]

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17 May 2012, 09:22
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Bunuel wrote:
a, b, c, and d are integers; abcd≠0; what is the value of cd?

(1) c/b = 2/d --> $$cd=2b$$, we don't know the value of $$b$$ to get the single numerical value of $$cd$$. Sufficient.
.

Bunuel, i think what you meant here is Not Sufficient. Correct?

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Re: a, b, c, and d are integers; abcd≠0 [#permalink]

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13 Sep 2012, 08:12
1
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Bunuel wrote:
a, b, c, and d are integers; abcd≠0; what is the value of cd?

(1) c/b = 2/d --> $$cd=2b$$, we don't know the value of $$b$$ to get the single numerical value of $$cd$$. Sufficient.

(2) b^3*a^4*c = 27*a^4*c --> as $$a$$ and $$c$$ does not equal to zero we can safely reduce both parts by $$a^4*c$$ --> $$b^3=27$$ --> $$b=3$$. Not sufficient.

(1)+(2) As from (1) $$cd=2b$$ and from (2) $$b=3$$ then $$cd=2b=6$$. Not sufficient.

Runner2 wrote:
2 - clearly not suff, and we got that |b^3|=27 cause what if c negative? so we can't say for sure that b=3 like stated in OA. Am I right?

Odd roots have the same sign as the base of the root. For example, $$\sqrt[3]{125} =5$$ and $$\sqrt[3]{-64} =-4$$.

So $$\sqrt[3]{27}=3$$ and not $$-3$$ --> $$3^3=27$$ and $$(-3)^3=-27$$.

Hope its' clear.

Hi Bunuel,

There is a slight typing error in the explanation.
Statement "(1) c/b = 2/d --> $$cd=2b$$, we don't know the value of $$b$$ to get the single numerical value of $$cd$$. Sufficient."
"(1) c/b = 2/d --> $$cd=2b$$, we don't know the value of $$b$$ to get the single numerical value of $$cd$$. Insufficient."

Correct me if i am wrong.
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Re: a, b, c, and d are integers; abcd≠0 [#permalink]

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10 Nov 2014, 11:46
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Bunuel, I think there is a typo- the option (c) i.e, (1)+(2) is 'sufficient', right?

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Re: a, b, c, and d are integers; abcd≠0 [#permalink]

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10 Nov 2014, 11:53
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Expert's post
deeuk wrote:
Bunuel, I think there is a typo- the option (c) i.e, (1)+(2) is 'sufficient', right?

Right. Edited. Thank you.
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Re: a, b, c, and d are integers; abcd≠0 [#permalink]

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16 Jan 2012, 03:33
@bunuel
thanks for explanation. it looks that my mind was somewhere else while solving the question. many times i misses an obvious point , main reason never to the 51 in Quant. i will have to focus more.

anyway, i have edited my explanation.
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Re: a, b, c, and d are integers; abcd≠0 [#permalink]

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16 Jan 2012, 04:19
thanks for explanation, you should agree very stupid and easy question, I should sleep more not to make such mistakes....

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Re: a, b, c, and d are integers; abcd≠0 [#permalink]

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21 Jan 2012, 14:45
a, b, c, and d are integers; abcd >< 0; what is the value of cd?

1) c/b = 2/d

c = (2*b)/(d)
not sufficient

2) b^3*a^4*c = 27*a^4*c

b^3 = 27
b = 3

not sufficient.

1 + 2

c = (2*3)/d
c = (6)/d
cd = 6

sufficient.

sufficient.

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Re: a, b, c, and d are integers; abcd≠0 [#permalink]

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13 Sep 2012, 08:18
fameatop wrote:
Bunuel wrote:
a, b, c, and d are integers; abcd≠0; what is the value of cd?

(1) c/b = 2/d --> $$cd=2b$$, we don't know the value of $$b$$ to get the single numerical value of $$cd$$. Sufficient.

(2) b^3*a^4*c = 27*a^4*c --> as $$a$$ and $$c$$ does not equal to zero we can safely reduce both parts by $$a^4*c$$ --> $$b^3=27$$ --> $$b=3$$. Not sufficient.

(1)+(2) As from (1) $$cd=2b$$ and from (2) $$b=3$$ then $$cd=2b=6$$. Not sufficient.

Runner2 wrote:
2 - clearly not suff, and we got that |b^3|=27 cause what if c negative? so we can't say for sure that b=3 like stated in OA. Am I right?

Odd roots have the same sign as the base of the root. For example, $$\sqrt[3]{125} =5$$ and $$\sqrt[3]{-64} =-4$$.

So $$\sqrt[3]{27}=3$$ and not $$-3$$ --> $$3^3=27$$ and $$(-3)^3=-27$$.

Hope its' clear.

Hi Bunuel,

There is a slight typing error in the explanation.
Statement "(1) c/b = 2/d --> $$cd=2b$$, we don't know the value of $$b$$ to get the single numerical value of $$cd$$. Sufficient."
"(1) c/b = 2/d --> $$cd=2b$$, we don't know the value of $$b$$ to get the single numerical value of $$cd$$. Insufficient."

Correct me if i am wrong.

Thank you. Typo edited.
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Re: a, b, c, and d are integers; abcd≠0 [#permalink]

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02 Feb 2016, 04:49
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Re: a, b, c, and d are integers; abcd≠0 [#permalink]

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16 Sep 2016, 17:37
D is correct. Here's why:

(1) x^(2) yz = 12xy --> xz = 12

SUFFICIENT

(2) (z/4) - (3/x) --> xz = 12

SUFFICIENT

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Re: a, b, c, and d are integers; abcd≠0   [#permalink] 16 Sep 2016, 17:37
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