Hi,
I think for the first question,only the first answer is sufficient because even if a number in the list is even integer,that doesn't make that integer positive
Q2. $6 is to be divided among x(say) Cupcake(denoted by C) and y(let) Dognuts(denoted by D)
So we have
xC+ yD = 6
now option-1
3C-2D = 0.1
(2)Option
C + D =2x 0.35= 0.70
So combining both the equations,we get
C= $ 0.3
D = $ 0.4
however we need the value of y
which can not be calculated unless x(no of cupcakes) is given