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Bunuel
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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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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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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