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If ab/cd < 0 , is bc > ad (1) cd > ab (2) b < c < 0 [#permalink]
 16 Aug 2009, 00:46
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If ab/cd < 0 , is bc > ad (1) cd > ab (2) b < c < 0
Last edited by Bunuel on 30 Jan 2012, 16:00, edited 1 time in total.
Added the OA
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IMO B
given ab/cd < 0 , so two cases
1) ab < 0, cd >0 or
2) ab >0 , cd <0
stmt 1 cd > ab
so case 1 satisfy, qstn is bc > ad? not sufficnt
stmt 2
b<c<0
applying in above two cases
1) ab<0 , cd> 0
b< 0 , so for ab<0, a must be >0
c<0 , so for cd>0, d must be <0
so a>0, b<0, c<0, d<0 , apply to check bc > ad?
defintely bc>ad
2) ab >0 , cd < 0
b<0 , so for ab>0 , a must be <0
c<0 , so for cd<0 , d must be >0
so a<0, b<0, c<0, d>0 , apply to check bc>ad?
defintely bc>ad.
so stmt2 suffint
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given ab/cd<0 case 1 a,b are positive and either c or d is negative case 2 c,d are positive and either a or b is negative case 3 a,b are negative and either c or d negative case 4 c,d are negative and either a or b negative st1 - cd>ab we have to consider case 2 and case 4 i.e c,d both are postive or negative and either a or b postive case 2: consider cd postive and a negative,b positive so will get bc>ad consider cd postive and b negative,a positive so will get bc<ad Not sufficient st2-b<c<0 we have to consider case 3 and 4 case 3 consider a,b negative and c negative,d positive bc>ad case 4 consider c,d negative and b negative,a positive bc>ad sufficient. My answer is B
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Good question.
1)CD>AB;
In this case CD should pe positive and AB negative in order to satisfy the condition that ab/cd<0; So Eiher two numbers from CD should be positive or negative and one number from AB should be positive and another should be negative or vice versa. If we put different signs of numbers in equation bc>ad we will have different results, so insuff.
2) B<C<0
ab<0 and cd>0
so a - positive, and d - negative, let's put them into equation
bc>ad; on the left side we have positive number, on the right side we have negative number, Suff.
ab>0 and cd<0
so a - negative, and d - positive, and again
bc>ad; on the left side we have positive number, on the right side we have negative number, Suff.
So I choose B.
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Re: 7)if ab/cd <0, is bc>ad? st1 cd>ab st2 b<c<0 [#permalink]
30 Jan 2012, 14:39
Quote: 7)if ab/cd <0, is bc>ad?
st1 cd>ab
st2 b<c<0
7)if ab/cd <0, is bc>ad?
st1 cd>ab
st2 b<c<0 +1 B, good question If \frac{ab}{cd} <0, ab > 0 and cd < 0, OR ab<0 and cd > 0. Statement (1): cd > ab So, cd > 0 and ab < 0 This could mean that c,d > 0 OR c,d<0 AND a>0, b<0, OR a<0, b>0 If c,d >0 , a>0, and b<0, the result is bc<0 and ad>0, then answer is NO If c,d <0 , a>0, and b<0, the result is bc>0 and ad<0, then answer is YES INSUFF Statement (2): b < c < 0, this means that bc >0 Then, if a<0, d>0, so bc >0 AND ad<0, the answer is YES. If a>0, d<0, so bc>0 AND ad<0, the answer is also YES. SUFFICIENT ANSWER : B If someone has a faster method, that would be a great help!
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Re: 7)if ab/cd <0, is bc>ad? st1 cd>ab st2 b<c<0 [#permalink]
30 Jan 2012, 15:59
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Re: If ab/cd < 0 , is bc > ad (1) cd > ab (2) b < c < 0 [#permalink]
02 Nov 2012, 16:37
what is the difficulty level of this question?
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Re: If ab/cd < 0 , is bc > ad (1) cd > ab (2) b < c < 0
[#permalink]
02 Nov 2012, 16:37
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close
