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

abc<0 and bcd>0



Anything to the even power is always positive. Since, we are concerned only whether the option is positive or negative we can try and keep the powers of the elements in the products to be the lowest. Ignore the even powered elements and take the power of 1 of the odd powers of each elements( for larger odd powers divide by 2 and take only remainder(1) as power) in the products.

a < 0, b < 0, c < 0. Then d > 0

a < 0, b > 0, c > 0. Then d > 0

a > 0, b < 0, c > 0. Then d < 0

a > 0, b > 0, c < 0. Then d < 0


From the above, we can note the pattern that if a > 0, then d < 0 and if a < 0, then d > 0. We can narrow down choices by considering the resulting term in the product, if it contains only a and d.


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

Simplifying we can decide just by bcd.We know bcd > 0, so definitely a^2b^3c^3d is positive. But we are looking for the products that result in negative value. Eliminate


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

All terms in the product are even powers. So the result will be positive. Eliminate

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

We can simplify the above product of terms as acd.

2 solutions are possible that, when a and c are both positive, d can be either positive or negative.
So we can't determine a solution. Eliminate

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

We can simplify the above product of terms as ad.
When a is positive, d is negative, then the resulting product is negative.
When a is negative, d is positive, then the resulting product is negative.

Hence, this option will always result in negative number. KEEP


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

We can simplify the above product of terms as ab.

So there is no way we can identify if the result be positive or negative, as multiple solutions are possible.
Eliminate

Hence the best answer choice is D.

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

If we solve the prompt (D)

it leads to

(a^3) (b^4) (c^4) (d)

= (abc) (bcd) (b^2) (c^2)

Here, (b^2) (c^2) > 0;.........(1)
abc < 0;;.........(2)
bcd > 0;;.........(3)

product of (1), (2) & (3) < 0

Thus, (D) will be correct

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.

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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