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# If d >0 and (1-c/d) is greater than or equal to zero but

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If d >0 and (1-c/d) is greater than or equal to zero but [#permalink]

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28 Nov 2003, 11:17
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If d >0 and (1-c/d) is greater than or equal to zero but less than 1, which of the following must be true
I) c>0
II) c/d<1
III) c^2 +d^2>1( just to be clear, ^ is the exponent)
A) I only
B) II only
C) I and II
D) II and III
E) I , II and III
I chose A, but the official answer is different
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28 Nov 2003, 18:39
I agree with A

c/d<1 - it can be 1 if c=d, if c/d =1 then 0 <= will fail, so II cannot be.

c^2 + d^2 > 1 is false, because
say c=.25 & d=0.5, c/d= .5 and is true for the inequality.
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28 Nov 2003, 22:26
"I agree with A c/d<1 - it can be 1 if c=d, if c/d =1 then 0 <= will fail, so II cannot be.

c^2 + d^2 > 1 is false, because
say c=.25 & d=0.5, c/d= .5 and is true for the inequality."

Last edited by Titleist on 28 Nov 2003, 22:42, edited 1 time in total.
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28 Nov 2003, 22:39
(1-c/d) is greater than or equal to zero

The only way 1 - c/d can be equal to 0 is if c=d, hence II is not a possibility.
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28 Nov 2003, 22:41
Titleist wrote:
"I agree with A c/d<1 - it can be 1 if c=d, if c/d =1 then 0 <= will fail, so II cannot be.

c^2 + d^2 > 1 is false, because
say c=.25 & d=0.5, c/d= .5 and is true for the inequality."

Read the statement carefully. I don't think that you can assume that C can equal D given the stated rule that 0<(1-c/d)<0. This implies that C must be another value besides D. In addition, c must also be < d. Also implicit in the statement given above, D is a positive number therefore C must be a positive number less than D - which makes c/d a fraction - a fraction is less than 1. Therefore, II must be true.

I don't think III MUST BE TRUE. C and D do not have to be integers - just greater than 0. If C = 1/3 and D= 1/2 then c/d = 2/3. And C^2 = 1/9 and D^2 = 1/4 combine the two fractions and of course the sum is less than 1.

read carefully. The stimulus states that 1 - c/d is greater then OR EQUAL to zero.
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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28 Nov 2003, 22:43
Yes, I had some cider before i turned on the computer - I will never do so again!
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