Mountain14 wrote:
Bunuel wrote:
rxs0005 wrote:
is x^2-8x+15 = 0
1 x not equal to 3
2 x-5 not equal to 0
1 x not equal to 3
2 x-5 not equal to 0
Solve x^2-8x+15=0 for x --> x=3 or x=5 --> so x^2-8x+15=0 is true if x=3 or x=5. Basically the question asks: is x=3 or x=5?
(1) x\neq{3}. x can still be 5. Not sufficient.
(2) x-5\neq{0} --> x\neq{5}. x can still be 3. Not sufficient.
(1)+(2) x\neq{3} and x\neq{5}, so the answer to the question is NO. Sufficient.
Answer: C.
Bunuel,
I did not understand the logic.
Stm1 - X can be 5, so x^2-8x+15 = 0 , Sufficient
Stm 2- X can be 3, So x^2-8x+15 = 0 , Sufficient..
Ans- D
Whats wrong in this.
QUESTION IS whether x^2-8x+15 = 0
statement 1:==> x\neq{3}.===>means x can be 4,5,6,7...anthing except 3
when x=5..the given equation equals to zero but with other numbers let say 4...16-32+15=-1
so insufficient.
statement 2:similarly herex cant be 5.....but at x=3==>equation equals to 0 BUT AT other numbers it is not equal to zero..
hence insufficient.
combining both you are neglecting both the roots of X hence that equation will never be zero .
hence sufficient C
HOPE IT HELPS