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a, b, c, and d are integers; abcd ≠ 0; what is the value of cd?
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15 Jan 2012, 23:43
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a, b, c, and d are integers; abcd ≠ 0; what is the value of cd? (1) \(\frac{c}{b} = \frac{2}{d}\) (2) \(b^3*a^4*c = 27*a^4*c\)
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Re: a, b, c, and d are integers; abcd ≠ 0; what is the value of cd?
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Updated on: 16 Jan 2012, 02:30
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^327) = 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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Re: a, b, c, and d are integers; abcd ≠ 0; what is the value of cd?
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16 Jan 2012, 02:11
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. Answer:C. As for your question: 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; what is the value of cd?
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16 Jan 2012, 02:33
@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; what is the value of cd?
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16 Jan 2012, 03:19
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; what is the value of cd?
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21 Jan 2012, 13:45
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; what is the value of cd?
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17 May 2012, 08:22
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; what is the value of cd?
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13 Sep 2012, 07:12
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. Answer:C. As for your question: 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." should read "(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; what is the value of cd?
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13 Sep 2012, 07:18
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. Answer:C. As for your question: 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." should read "(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; what is the value of cd?
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10 Nov 2014, 10:46
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; what is the value of cd?
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10 Nov 2014, 10:53



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Re: a, b, c, and d are integers; abcd ≠ 0; what is the value of cd?
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16 Sep 2016, 16:37
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; what is the value of cd?
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12 Jan 2019, 09:19
law258, that looks like a different problem from the one in this post.
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Re: a, b, c, and d are integers; abcd ≠ 0; what is the value of cd? &nbs
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12 Jan 2019, 09:19






