If y is a negative number greater than -8, is x greater than the average (arithmetic mean) of y and -8 ?
(1) On the number line, x is closer to -8 than it is to y.
(2) x = 4y
Hi there!
We know -8 < y < 0, and we must focus on the question: x > (y+(-8))/2 ?
(1) SUFFICIENT:
Draw a real line with "points" -8 (left), y (right) and their middle-point (say) M.
The question is, therefore, x > M ?
From sttm (1) we know that x is less than M, because it is (-8 or less) or (between -8 and M, M excluded).
(2) INSUFFICIENT:
Please note that the question is equivalent to y < 2x+8 or y < 8(y+1) or (y+1)/y < 1/8 ? (Remember that y<0)
> Take y negative but (below and) nearer 0, say y = -1. We get (y+1)/y = 0/-1 = 0 answering positively.
> Take y negative but (above and) nearer -8, say y = -7. We get (y+1)/y = -6/-7 = 6/7 answering negatively.
Regards,
Fabio.