12 Days of Christmas GMAT Competition - Day 11: If abc < 0 and bcd > 0

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

A. \(a^{2}b^{3}c^{3}d\)
B. \(a^{2}b^{4}c^{4}d^{2}\)
C. \(a^{3}b^{4}c^{5}d\)
D. \(a^{3}b^{4}c^{4}d\)
E. \(a^{3}b^{5}c^{4}d^{2}\)

Case 1: \(a<0, b>0, c>0, d>0\)
A, B are positive, Eliminate
Case 2: \(a<0, b<0, c<0, d>0\)
C, E are positive, Eliminate
Case 3: \(a>0, b>0, c<0, d<0\)
Case 4: \(a>0, b<0, c>0, d<0\)

D is negetive for all the cases

Ans : D

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

option D is a^3 b^4 c^4 d
= (abc)^3 (bcd)
= (-ive)^3 (+ive)
= -ive value
Hence Option D is correct

abc<0 and bcd>0
the following cases are possible:

a<0, b>0, c>0, d>0 This gives option A and B positive
a>0, b<0, c>0, d<0
a>0, b>0, c<0, d<0 This gives option C and E positive
a<0, b<0, c<0, d>0

The only choice remains is option D which is not positive for any of the cases mentioned above.
Option D

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

Let's consider a<0 b<0 c<0 and d>0 to match the given inequality and substitute these values to the answers

A.\( a^2b^3c^3d\) Positive, since the product of 2 negatives, here b and c, is positive
B. \(a^2b^4c^4d^2\) Positive, because of the even powers
C. \(a^3b^4c^5d\) Positive, since the product of 2 negatives, here a and c, is positive
D. \(a^3b^4c^4d \) Negative because only \( a^3\) is negative
E. \(a^3b^5c^4d^2 \) Positive, since the product of 2 negatives, here a and b, is positive

Answer D

OA) D
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. \(a^3b^5c^4d^2\)

now based on the question abc<0 & bcd>0

case 1) a<0,b<0,c<0,d>0
case 2) a>0,b>0,c<0,d<0
case 3) a>0,b>0,c>0,d<0
case 4) a<0,b>0,c>0,d>0

option A) \(a^2b^3c^3d\) sign depends on \(b^3c^3d\) ,
case 1) a<0,b<0,c<0,d>0 then \(b^3c^3d\)>0
options A cannot be the answer

options B)\(a^2b^4c^4d^2\) sign will always positive
so option B cannot be the answer

option c) \(a^3b^4c^5d\) sign depends on \(a^3c^5d\)
case 1) a<0,b<0,c<0,d>0 then \(a^3c^5d\)>0

option d) \(a^3b^4c^4d\) sign depends on \(a^3d\)
in all four sign of a & d is not same so \(a^3d\) will always negative
so option d is the answer

option e) \(a^3b^5c^4d^2\) sign depends on \(a^3b^5\)
cases in which a & b signs are same in those cases signs will be positive
so case 1,2 & 3 sign of this option will be positive

step 1:
let a= -ve, b=+ve, c=+ve, and d= +ve

c,d,e are positive

step 2:
a= -ve, b= -ve, c=-ve, d=+ve
c, and e becomes positive,

thus option D is correct

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

\(abc < 0\) can be + + - ; + - + ; - + + ; - - -
\(bcd > 0\) can be - - + ; - + - ; + - - ; + + +

A. \(a^2b^3c^3d\) - dependent on b,c,d - Can be positive
B. \(a^2b^4c^4d^2\) - All squares - Can be positive
C. \(a^3b^4c^5d\) - dependent on a,c,d - Can be positive
D. \(a^3b^4c^4d\) - - dependent on a,d - will always be of alternate sign
E. \(a^3b^5c^4d^2\) - dependent on a,b - Can be positive

Ans: D

We can see 2 possibilities: i) either a,b,c are all negative or ii) only one of a,b,c is negative and for b or c negative, d is also negative.
We have to see if the options can be made positive to prove 'definitely negative'.

Option A and B : If we consider only a negative by using possibility (ii), then the whole expression becomes positive (since a^2)
Option C and E : Considering all a,b,c negative, this option can be made positive.

Therefore Option D is the answer.

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

we need odd powers of a, b & c and even power of d to be sure that it is a negative number, only E seems to be the correct option with error in the option as c must also have odd power.

The answer is E. a3b5c4d2

bb Bunuel

I got this question correct, however, 10 points for answering correctly isn't added. Pls check for the bug.

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