If abc 0, which of the following is definitely negative

Question

If abc < 0 and bcd > 0, which of the following is definitely negative ?

A.  a^2b^3c^3d

B.  a^2b^4c^4d^2

C.  a^3b^4c^5d

D.  a^3b^4c^4d

E.  a^3b^5c^4d^2

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Adit_ · Answer

EDIT: Made a big mistake initially. abc<0 and bcd>0 Divide both equations by bc then: a<0 and d>0 If we divide both equations by -bc then: a>0 and d<0. So it shows a,d have opposing signs, so their product if both raised to odd powers will always be negative. Option D perfectly portrays this.

SteepOat · Answer

a^3*b^4*c^4*d = (a^3*b^3*c^3)*(bcd) = (abc)^3*(bcd) = (-)^3*(+) = (-)

Answer is D

BongBideshini · Answer

abc < 0 
 Either all are negative or exactly one negative among a, b, c

bcd > 0 
Either all positive or exactly two negatives among b, c, d

Case 
 a 
 b 
 c 
 d 
 abc 
 bcd 
 
  1 
 − 
 + 
 + 
 + 
 − 
 + 
 
  2 
 − 
 − 
 − 
 + 
 − 
 + 
 
  3 
 + 
 + 
 − 
 − 
 − 
 + 
 
  4 
 + 
 − 
 + 
 − 
 − 
 +

From this table we can see that a & d are always opposite signs,

A. 
Sign (a^2 b^3 c^3 d)=sign(bc) 
-> Depends

B. 
a^2 b^4 c^4 d^2 
all even powers → positive

C.
Sign( a^3 b^4 c^5 d) =sign(acd)
->depends on c, not fixed

D. 
sign(a^3 b^4 c^4 d)
->sign(ad)
 always negative

E. 
sign(a^3 b^5 c^4 d^2)=sign(ab)
->depends

Ans D

ExpertsGlobal5 · Answer

D is the correct answer choice.

Video explanation:

https://www.youtube.com/watch?v=MT8LxvYds_g

If abc < 0 and bcd > 0, which of the following is definitely negative

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