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M23-34

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M23-34  [#permalink]

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New post 16 Sep 2014, 01:20
2
1
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

55% (01:03) correct 45% (00:47) wrong based on 183 sessions

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M23-34  [#permalink]

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New post 16 Sep 2014, 01:20
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: M23-34  [#permalink]

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New post 28 Oct 2014, 17:47
1
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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New post 29 Oct 2014, 04:41
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Re: M23-34  [#permalink]

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New post 25 Dec 2014, 09:23
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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New post 25 Dec 2014, 10:04
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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: M23-34  [#permalink]

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New post 11 Aug 2016, 05:02
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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New post 25 Oct 2016, 01:31
how come we are supposed to get the exact substituted values in such a little span of time?
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Re: M23-34  [#permalink]

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New post 08 Nov 2016, 01:49
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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New post 08 Nov 2016, 02:46
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: M23-34  [#permalink]

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New post 13 Jan 2017, 22:21
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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New post 14 Jan 2017, 05:35
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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

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Yes, that's correct.
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Re: M23-34  [#permalink]

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New post 02 Apr 2018, 03:22
+1 for option C
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Re: M23-34  [#permalink]

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New post 23 Aug 2018, 13:12
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?
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Re: M23-34   [#permalink] 23 Aug 2018, 13:12
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