st(1) d=-(12+e) Not Sufficient, as we don't know whether d is -ve or +ve;
st(2) d>e+12 Not Sufficient, as don't know whether e is -ve or ve;
Combined st(1&2): -(12+e)>e+12, -2e>24, e<-12 hence e is -ve. If e is -ve and placed LHS to -12 , from st(1) d=-(12+e), then d is +ve OR d>0. SO, our answer is No, d>0 Sufficient
IOM C
I think it is not possible to add equation and inequality right away, BUT we could substitute expression for d in st(1) into the expression for d in st(2). Yet for d to be 0, st(2) should have equality sign e-d=-12 BUT it has inequality sign, hence d > 0 and answer C.capthan wrote:Is d negative?
(1) e + d = -12
(2) e - d < -12
Can we add these two equations and get
d is either 0 or d is <0 to arrive on our answer E.
