Hi,
Lets start with the easier stat 2. It is clearly insufficient but it tells us that :

c>b
c-b > 0
b-c < 0 {multiply by -1 and flip}

stat-1 : -> this is also insufficient alone as we don't know the values of B and C.

combining :
(a+b-c) /a < 0

from stat 2-> (b-c) will always be a -ve quantity.

Thus when a is negative the expression on the LHS of the inequality will always be positive.
suppose a = -6 , taking any -ve value of b-c you will get the negative sign on the numerator and denominator cancelled.

On the other hand if we take a as positive then the inequality will hold true until |b-c| > a
for example when b-c = -8 and a = 6 the expression is -1/3 {stat 1 does hold} whereas
b-c = -6 and a = 8 the expression is 1/3

we have 2 values when a is positive and a single value when a is negative hence answer is A is always negative. Lemme know if the trash language above is unclear.

knight247 wrote:If a≠0, is a<0?
(1)(a+b-c)/a<0
(2)c>b

OA is C.
Detailed explanations would be appreciated. Thanks

Hi,
Great job bblast. You have already solved this. Just my take:
It is clear that each statement alone is insufficient.
Both(1) and (2):
b<c => b-c<0
(a+b-c)/a<0 => 1+(b-c/a) < 0
So, (b-c)/a < -1
We know that (b-c) is negative. So, for (b-c)/a to be negative, a should be positive.
Hence, C

Frankenstein wrote:

knight247 wrote:If a≠0, is a<0?
(1)(a+b-c)/a<0
(2)c>b

OA is C.
Detailed explanations would be appreciated. Thanks

Hi,
Great job bblast. You have already solved this. Just my take:
It is clear that each statement alone is insufficient.
Both(1) and (2):
b<c => b-c<0
(a+b-c)/a<0 => 1+(b-c/a) < 0
So, (b-c)/a < -1
We know that (b-c) is negative. So, for (b-c)/a to be negative, a should be positive.
Hence, C

Now that was algebraic sweetness at its very best in the solution by Frank, and not full of BS like my post above.

IMO:C

From 1) a+b <c
From 2) c>b

From 1+2 a must be -ve -> a<0 sufficient!

knight247 wrote:If a≠0, is a<0?
(1)(a+b-c)/a<0
(2)c>b

OA is C.
Detailed explanations would be appreciated. Thanks

Hi -- Lets start with the easy one-- clearly this number is not sufficient..
now take the first statement

(a+b-c)/a<0 , LHS = (a/a)+ (b-c)/a = [1+ (b-c)/a] <0

This statement also does not solve our problem

From 2nd statement we know that c>b => that b-c will be negative
and a/a =1
Therefore, a has to be greater than zero to make (b-c)/a Negative..

HENCE C

C