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# M23-34

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Math Expert
Joined: 02 Sep 2009
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16 Sep 2014, 01:20
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Difficulty:

45% (medium)

Question Stats:

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

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If $$A = \frac{B}{C}$$, is $$A \gt B$$ ?

(1) $$0 \lt C \lt 1$$

(2) $$A \gt 0$$

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Joined: 02 Sep 2009
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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. $$B = AC \lt A$$.

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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.

Best
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29 Oct 2014, 04:41
JackSparr0w wrote:
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.

Best

Edited. Thank you.
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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.$$B = AC \lt A$$.

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, 10:04
1
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.$$B = AC \lt A$$.

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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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

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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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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. $$B = AC \lt A$$.

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 $$C<1$$
Math Expert
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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. $$B = AC \lt A$$.

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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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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14 Jan 2017, 05:35
1
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

_______________
Yes, that's correct.
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02 Apr 2018, 03:22
+1 for option C
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23 Aug 2018, 13:12
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?
Re: M23-34   [#permalink] 23 Aug 2018, 13:12
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# M23-34

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