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If a < b < c < d, is abcd > 0? (1) abc > 0 (2) bcd < 0

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If a < b < c < d, is abcd > 0? (1) abc > 0 (2) bcd < 0  [#permalink]

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New post 29 Dec 2016, 10:56
00:00
A
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Difficulty:

  85% (hard)

Question Stats:

52% (01:42) correct 48% (02:19) wrong based on 93 sessions

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Re: If a < b < c < d, is abcd > 0? (1) abc > 0 (2) bcd < 0  [#permalink]

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New post 29 Dec 2016, 11:21
as it is clear that none of these(abcd)can't be zero.so,
st.1 we may have two cases
1. -2<-1<1, always abcd>0
2. 2<3<4, always abcd>0 hence suff.
st.2 whatever values we take for bcd one has to be -ve, say b is -ve then a must has to be -ve and product of abcd>0 always. or we can take values of bcd -ve then a must has to be -ve , hence abcd>0 so suff.

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Re: If a < b < c < d, is abcd > 0? (1) abc > 0 (2) bcd < 0  [#permalink]

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New post 30 Dec 2016, 00:24
Statement 1, abc>0, the possible cases are
i) a, b <0 and c>0, which means d>0.
ii) a, b, c >0, which means d>0.
And in both cases, abcd>0.

Statement 2, bcd<0, the possible cases are
i) b <0 and c, d>0, which means a<0.
ii) b, c, d <0, which means a<0.
And in both cases, abcd>0.

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Re: If a < b < c < d, is abcd > 0? (1) abc > 0 (2) bcd < 0  [#permalink]

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New post 30 Dec 2016, 05:33
1
Bunuel wrote:
If a < b < c < d, is abcd > 0?

(1) abc > 0
(2) bcd < 0


Question: If a < b < c < d, is a*b*c*d > 0?

Answer to the question is YES if all variables are +ve or all variables are -ve or only two variables are +ve or -ve.

As per statement-1

a*b*c > 0, this can happen in two cases when all the variables are +ve or two variables are -ve.

So if a = 1, b =2 and c = 3 then d has to be greater than 3 as per question

or if a = -2 , b = -1 and c = 1 then d has to be greater than 2 (no variable in our case can be equal to "0").

In both the cases answer to our question is YES a*b*c*d > 0

So sufficient

As per statement-2

b*c*d < 0 , this can happen in two cases when all variables are -ve or only one variable is -ve.

If b = -1, c = 1, d = 2 then a has to be less than -1, So answer to the question is YES a*b*c*d > 0

If b = -3, c = -2, d = -1 then a has to be less than -3, Again answer to the question is YES a*b*c*d > 0

So sufficient.

(D)
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Re: If a < b < c < d, is abcd > 0? (1) abc > 0 (2) bcd < 0  [#permalink]

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New post 30 Dec 2016, 14:23
Bunuel wrote:
If a < b < c < d, is abcd > 0?

(1) abc > 0
(2) bcd < 0

1) abc>0 .. we already know that d is the greatest. Yes. So sufficient.
2) bcd<0 .. we already know that a is the least. No. So sufficient.

D.

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Re: If a < b < c < d, is abcd > 0? (1) abc > 0 (2) bcd < 0  [#permalink]

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New post 30 Dec 2016, 20:28
IMO D.

abc> 0 implies, either all of them are greater than 0 or only two of them are < 0 which would be a and b. Thus d would be positive. hence abcd>0. Suff
bcd<0 implies either all of them are less than 0 that also means a<0 -- abcd>0 or only one of them is <0 which would be b. and so a also should be less than 0. -- abcd>0. Suff.
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Re: If a < b < c < d, is abcd > 0? (1) abc > 0 (2) bcd < 0  [#permalink]

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New post 27 Feb 2018, 08:55
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Re: If a < b < c < d, is abcd > 0? (1) abc > 0 (2) bcd < 0 &nbs [#permalink] 27 Feb 2018, 08:55
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