Bunuel
Is cdef < 0?
(1) \(cd^2e^3f^4 < 0\)
(2) \(c^2d^3e^4f^5 > 0\)
Kudos for a correct solution.cdef<o if one letter has negative sign or 3 letters have negative sign.
Statement 1: We can deduce that either c or e is negative. d & f may be positive or negative . We can know because at end both will be positive due to even power (2 & 4)
if C = -, - - ++ , then, cdef <0 ......No
If C= -, - - + -, then, cdef <0.......Yes
Insuf
Statement 2: We can deduce that both d & f are negative. c & e may be positive or negative . We can know because at end both will be positive due to even power (2 & 4)
if C = -, - - +- , then, cdef <0 ......Yes
If C= +, +- +-, then, cdef <0.......No
Insuf
Combined:
We know from (2) that d &f are negative and either e or e is negative. So 3 letter with negative sign.
3 letter are negative and one is positive so cdef<0 is always yes
Answer: C