12 Days of Christmas GMAT Competition - Day 11: If abc 0

Question

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 
B.  a^2b^4c^4d^2 
C.  a^3b^4c^5d 
D.  a^3b^4c^4d 
E.  a^3b^5c^4d^2

thisisit25 · Accepted Answer

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

A.  a^2b^3c^3d 
 = (abc)^2(bcd) 
= +ve and +ve
= +ve

B.  a^2b^4c^4d^2 
 = (abc)^2(bcd)^2 
= +ve and +ve
= +ve

C.  a^3b^4c^5d 
 = (abc)^3(bcd)(c) 
= can't say as we don't know the sign of c

D.  a^3b^4c^4d 
=  (abc)^3(bcd) 
= -ve
 
E.  a^3b^5c^4d^2 
=  (abc)^3(bcd)(bd) 
= can't say as we don't know the sign of bd

Lizaza · Answer

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.

Kushchokhani · Answer

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.

sivatx2 · Answer

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.

shubhim20 · Answer

hi please explain is not b,c and d always positive since they are greater than 0 considering it not able to solve the question  Bunuel   bb   KarishmaB

Bunuel · Answer

No. bcd > 0 implies either all three are positive or exactly two are negative. So the possibilities are (+)(+)(+) > 0 or (+)(-)(-) > 0.

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

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