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

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

abc < 0
bcd > 0
ab^2c^2d < 0
ad < 0

A. \(a^2b^3c^3d = a^2b^2c^2* bcd > 0\)
B. \(a^2b^4c^4d^2 = a^2b^4c^4d^2 > 0\)
C. \(a^3b^4c^5d = a^2b^4c^4* acd\)
D. \(a^3b^4c^4d = a^2b^4c^4 * ad < 0\): DEFINITELY NEGATIVE
E. \(a^3b^5c^4d^2 = a^2b^4c^4d^2 * ab\)

IMO D

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

*If you want to solve it easily: take bc together as bc are common in both*

Combining condition 1 and 2: Either a will be - , bc + , d - or a + , bc -, and d+

Trying this in options too by combining b and c will give that DEFINATELY d will be negative.

A) (abc)^2(bcd) => + x + is +
B) (abc)^2 (bcd)^2 => + x + is +
C) (abc)^3 (bcd) d => - x + x sign of d = will depend on sign of d
D) (abc)^3 ((bcd))= - x + = always -ve
E) (abc)^3 (bcd) (bd)= - x + x sign of bd = will depend on sign of bd

given If \(abc < 0\) and \(bcd > 0\)

abc<0 and bcd>0
possible cases
case 1 :a -ve; b-ve ; c-ve ; d+ve
case 2 :a -ve ; b+ve; c +ve ; d +ve
case 3 : a +ve ; b + ; c- ; d -
case 4 : a + ; b- ; c + ; d -

definitely negative will be
A. \(a^2b^3c^3d\) ; does not hold true for case 1
B. \(a^2b^4c^4d^2\) ; does not hold true for case 1
C. \(a^3b^4c^5d\) ; does not hold true for case 1
D. \(a^3b^4c^4d\) ; true for all cases
E. \(a^3b^5c^4d^2\) does not hold true for case 1

OPTION D is correct

Let's try to isolate B*C*D from each of the formulas, as this multiple will always be positive, regardless of the power.
Additionally, let's group other multiples based on their powers:

A. \((bcd)*(abc)^2\)
B. \((bcd)^2*(abc)^2\)
C. \((bcd)*(abc)^3*c\)
D. \((bcd)*(abc)^3\)
E. \((bcd)^2*(abc)^2*ab\)

As you can see, both A and B will always be positive, because A*B*C have positive powers.

Then, C and E are uncertain, because there are components, whose sign we cannot determine for sure: in C it's 'c', and in E it's 'ab'

Therefore, the correct option is D - where \((abc)^3\) is always negative and \((bcd) \) is always positive.

12 Days of Christmas GMAT Competition with Lots of Fun

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

A. \(a^2b^3c^3d\)-----(abc)^2 (bcd)= + x += +
B. \(a^2b^4c^4d^2\)--- Even powers always positive
C. \(a^3b^4c^5d\)-----(bcd) (abc)^3 c - we dont know sign of c
D. \(a^3b^4c^4d\)---- (abc)^3 (bcd)= - x + = - Yes
E. \(a^3b^5c^4d^2\)----(abc)^2 (bcd)^2 ab We dont know sign of ab


Ans D

Let the following be true,

a,b,c is -ve.
d is +ve.

In this case, only option D will be negative. The rest will be positive.

Hence, IMO D.

Solution:

Given, \(abc < 0\) and \(bcd > 0\)

Taking cases:

abc<0 & bcd>0
Case 1. (-++) (+++)
Case 2. (+-+) (-+-)
Case 3. (++-) (+--)
Case 4. (---) (--+)

Now checking options, to find which one is always (definitely) negative:
A) bcd will determine the sign of the expression. In order of the cases listed, the expression will be ++++. Always positive. OUT.
B) Always positive. OUT.
C) acd will determine the sign of the expression. In order of the cases listed, the expression will be --++. OUT.
D) ad will determine the sign of the expression. In order of the cases listed, the expression will be ----. Always negative. CORRECT.
E) ab will determine the sign of the expression. In order of the cases listed, the expression will be --++. OUT.

ANSWER 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

A. No a,b,c could be negative and d positive
B. No. This will be positive irrespective of a,b,c,d
C. No. a,b positive and c,d - could be negative
D. Yes. Correct
E. No. a,b - positive and c and d could be negative

D is the answer

Probabilities:
a, b, c, d
+ + - -
+ - + -
- + + +
- - + +

a^3*b^4*c^4*d = It's product is definitely negative when substituted in all probabilities.

IMO D

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

It means none of a,b,c,d = 0

Evaluating option (D),

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

can be written as

abc * bcd * a^2 * b^2 * c^2

Given , abc<0.............(1)
bcd>0 .............(2)
also, a^2 * b^2 * c^2 > 0 (squares of non-zero values).............(3)

From (1), (2) and (3)

abc * bcd * a^2 * b^2 * c^2 < 0 (ie, definitely negative)

IMO, (D) will be correct

abc<0 and bcd>0
=> bcd > abc
=> bcd-abc > 0
=> bc (d-a) > 0
=> bc > 0 and d > a

To hold all above inequalities, d must be +ve and a must be -ve. Also, b and c must have same sign.
Accordingly, choice D is definitively negative.
Answer is D.

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

For the given conditions i.e. abc<0 and bcd>0 here are the 4 possible sign arrangements of a,b,c, & d: -



After analyzing the table, it is ONLY the expression in Choice D i.e., a^3b^4c^4d which is negative for all the 4 possible sign arrangements of a,b,c & d with respect to the given conditions abc<0 and bcd>0.

Answer: CHOICE D

If abc<0 and bcd>0,

a,b,c,d are all no-zero real numbers

after removing all square values (positive); we get

(A) bcd > 0
(B) all squares ; hence, >0
(C) acd (+ve or -ve ---> can't be determined)
(D) abc * bcd * sq (abc)
<0 * >0 * > 0
thus, < 0........correct option
(E) ab (+ve or -ve ---> can't be determined)

Hence, (D) is the answer

Answer is D
we have if bc > 0, then a < 0 and d > 0 or if bc < 0 then a > 0 and b < 0. Thus a and b have opposite sign. Only D fits.

the number raised to the power of even will result in positive hence we can eliminate the choice as following
A will depend on bcd --> given from question that bcd>0
B all the number are raised to the even power hence it will result in positive --> eliminate
C will depend on acd
suppose a+,b-,c+,d- it will result in negative
suppose a-,b+,c+,d+ it will result in -
suppose a+,b+,c-,d- it will result in +
D will depend on ad
suppose suppose a+,b-,c+,d- it will result in negative
suppose a-,b+,c+,d+ it will result in -
suppose a+,b+,c-,d- it will result in -
Hence it d
E will depend on ab

IMO D.

a.bc < 0 and d.bc > 0
if bc is +ve, then a is -ve, d is +ve
if bc is -ve, then a is +ve, d is -ve

A. positive in both cases
B. positive in both cases
C. unable to determine value of c
D. Negative in both cases
E. unable to determine value of b