12 Days of Christmas GMAT Competition - Day 4: If abc < 0 and acd > 0

Make combinations of abc and acd for each of the five options.

A. \(a^4b^4c^3d^5\)
\((abc)^3abd^5\)
We don't know the individual signs of a,b,d or of the combination abd

B. \( a^4b^4c^4d^5\)
\((abc)^4d^5\)
We don't know the individual sign of d. Fifth power can be positive or negative depending on whether d is positive or negative.

C. \(a^5b^5c^4d^3\)
\((abc)^4abd^3\) -- similar to (A)

D. \(a^5b^3c^5d^5\)
\((acd)^5b^3\)
We don't know the individual sign of b. Third power can be positive or negative depending on whether d is positive or negative.

E. a^5b^2c^5d^5
(acd)^5b^2
acd>0 and square of anything non-zero is positive. (note that all are non-zero because the inequality in the question does not equal to zero ever)

Answer: E

given condition
If \(abc < 0\) and \(acd > 0\)
abc <0 ; possible
case 1 : all are -ve
case 2 : two are + and third is -ve

acd >0
case 1 : all are +ve
case 2 : two values are -ve and third is +ve

we can write it as abc <0 and acd>0
cases
1 . a ( -) b( +) c ( +) d ( -)
2. a ( -) b ( -) c ( -) d ( +)
3. a ( +) b ( +) c ( -) d (-)
4. a ( +) b ( -) c ( +) d (+)

answer options given which shall be always +ve

A. \(a^4b^4c^3d^5\)
will not be true in case 1, 2

B. \(a^4b^4c^4d^5\)
will not be true in case 1, 3,

C. \(a^5b^5c^4d^3\)
will not be true in case 3, 4

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

will not br true in case 2, 4

E. \(a^5b^2c^5d^5\)

True for all cases
OPTION E is correct

abcd can take the following signs respectively:
---+
++--
+-++
-++-

After comparing the signs in each answer choice we find that E is always + because the odd exponenets for -ve variables always multiply to give a positive result. The positive values are always raised to even exponent which does not affect the sign.

Ans E

To tackle this task, I found it easy to start by the most obvious thing:
We know for sure that \(a*c*d\) is positive. Well, let's try to find all \(a*c*d\) products if the options we have - because these products, no matter the power, will remain positive.
And then we will only have to assure that the remaining multiples have no risk whatsoever of being ever negative:

A. \(a^4b^4c^3d^5 =\)\((acd)^3\)\(*ab*\)\(d^2 \)

B. \(a^4b^4c^4d^5= \)\((acd)^4\)\(*\)\(b^4\)\(*d \)

C. \(a^5b^5c^4d^3= \)\((acd)^3\)\(*\)\(a^2\)\(*b^5*c \)

D. \(a^5b^3c^5d^5= \)\((acd)^5\)\(*b^3 \)

E. \(a^5b^2c^5d^5\)\(= \)\((acd)^5\)\(*\)\(b^2\)

In the formulas above, I've highlighted in green the parts that are always above zero - based on the even powers.
As we can see, in each of the options there are multiples that may well be negative - except for E, where each component is guaranteed to stay above zero.
Therefore, the right answer is E.

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.