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Official Solution:If \(A = \frac{B}{C}\), is \(A \gt B\) ?Statement (1) by itself is insufficient. Consider \(A = 2\), \(B = 1\), \(C = 0.5\) (the answer is "yes") and \(A = 2\), \(B = 1\), \(C = 0.5\) (the answer is "no"). Statement (2) by itself is insufficient. Consider \(A = 1\), \(B = C = 1\) (the answer is "yes") and \(A = B=C=1\), (the answer is "no"). Statements (1) and (2) combined are sufficient. \(B = AC \lt A\). Answer: C
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Re: M2334
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28 Oct 2014, 16:47
I think the explanation for stmt 2 may have an error. In both examples given to disprove stmt 2, A actually is greater then B (1>1, and 1>0.5).
I think one of the examples was supposed to be A=B=C=1. In this case A isn't greater than B.
Please correct me if wrong.
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Re: M2334
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29 Oct 2014, 03:41



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Re: M2334
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25 Dec 2014, 08:23
Bunuel wrote: Official Solution:
Statement (1) by itself is insufficient. Consider \(A = 2\), \(B = 1\), \(C = 0.5\) (the answer is "yes") and \(A = 2\), \(B = 1\), \(C = 0.5\) (the answer is "no"). Statement (2) by itself is insufficient. Consider \(A = 1\), \(B = C = 1\) (the answer is "yes") and \(A = B=C=1\), (the answer is "no"). Statements (1) and (2) combined are sufficient.\(B = AC \lt A\).
Answer: C Hi Can somebody pls explain Statements (1) and (2) combined are sufficient.\(B = AC \lt A\).I understand B = AC, but how did you conclude that AC < A ? Thanks



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25 Dec 2014, 09:04
buddyisraelgmat wrote: Bunuel wrote: Official Solution:
Statement (1) by itself is insufficient. Consider \(A = 2\), \(B = 1\), \(C = 0.5\) (the answer is "yes") and \(A = 2\), \(B = 1\), \(C = 0.5\) (the answer is "no"). Statement (2) by itself is insufficient. Consider \(A = 1\), \(B = C = 1\) (the answer is "yes") and \(A = B=C=1\), (the answer is "no"). Statements (1) and (2) combined are sufficient.\(B = AC \lt A\).
Answer: C Hi Can somebody pls explain Statements (1) and (2) combined are sufficient.\(B = AC \lt A\).I understand B = AC, but how did you conclude that AC < A ? Thanks 0 < C < 1 > 0 < B/A < 1 > multiply by A (which we know is positive): 0 < B < A.
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Re: M2334
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11 Aug 2016, 04:02
We know that \(A=\frac{B}{C}\)
And we are asked whether \(A>B\) or not
We can change the question to : Is \(\frac{AC}{C}>\frac{B}{C}\) ?
Statement 1 : \(0<C<1\)
It gives us two information
a/ \(C>0\)
The answer to the question " is \(A>B\) ?" will be the answer to the question "is \(AC>B\) ?"
b/ \(C<1\)
Therefore \(AC<A\) if A is positive \(AC>A\) if A is negative
At this moment the answer to the question depends on whether A is positive or not
Statement 2 gives us this information
Hence the answer is C



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Re: M2334
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25 Oct 2016, 00:31
how come we are supposed to get the exact substituted values in such a little span of time?



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Re: M2334
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08 Nov 2016, 00:49
Bunuel wrote: Official Solution:
If \(A = \frac{B}{C}\), is \(A \gt B\) ?
Statement (1) by itself is insufficient. Consider \(A = 2\), \(B = 1\), \(C = 0.5\) (the answer is "yes") and \(A = 2\), \(B = 1\), \(C = 0.5\) (the answer is "no"). Statement (2) by itself is insufficient. Consider \(A = 1\), \(B = C = 1\) (the answer is "yes") and \(A = B=C=1\), (the answer is "no"). Statements (1) and (2) combined are sufficient. \(B = AC \lt A\).
Answer: C Can we not do it this way? If \(A = \frac{B}{C}\); Therefore is \(\frac{B}{C}>B\)? Thus is \(C<1\)? And Statement 1 tells us \(C<1\)



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Re: M2334
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08 Nov 2016, 01:46
yeshu_a wrote: Bunuel wrote: Official Solution:
If \(A = \frac{B}{C}\), is \(A \gt B\) ?
Statement (1) by itself is insufficient. Consider \(A = 2\), \(B = 1\), \(C = 0.5\) (the answer is "yes") and \(A = 2\), \(B = 1\), \(C = 0.5\) (the answer is "no"). Statement (2) by itself is insufficient. Consider \(A = 1\), \(B = C = 1\) (the answer is "yes") and \(A = B=C=1\), (the answer is "no"). Statements (1) and (2) combined are sufficient. \(B = AC \lt A\).
Answer: C Can we not do it this way? If \(A = \frac{B}{C}\); Therefore is \(\frac{B}{C}>B\)? Thus is \(C<1\)? And Statement 1 tells us \(C<1\) The highlighted part is not correct. B/C > B is true if: B > 0 and 0 < C < 1 B < 0 and C > 1 B < 0 and C < 0
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Re: M2334
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13 Jan 2017, 21:21
Hi Bunuel, Can I also do it this way?
From the question,
C=[B][/A]
I. 0<C<1 then 0<[B][/A]<1 which can be in these two ways: 0<B<A or 0>B>A so Not Sufficient
II. A>0. Obviously Not Sufficient since nothing about B and C was mentioned.
Combining the I and II will eliminate the second solution (0>B>A) so the actually solution is 0<B<A and in this case A>B. Sufficient. C



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14 Jan 2017, 04:35



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Re: M2334
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02 Apr 2018, 02:22
+1 for option C
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Re: M2334
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23 Aug 2018, 12:12
alados14 wrote: Hi Bunuel, Can I also do it this way?
From the question,
C=[B][/A]
I. 0<C<1 then 0<[B][/A]<1 which can be in these two ways: 0<B<A or 0>B>A so Not Sufficient
II. A>0. Obviously Not Sufficient since nothing about B and C was mentioned.
Combining the I and II will eliminate the second solution (0>B>A) so the actually solution is 0<B<A and in this case A>B. Sufficient. C How is 0>B>A?










