himu wrote:Is xy > ab ?
b/x > y/a
(ab)^2 >(xy)^2
I received a PM asking me to explain why b/x > y/a does not imply that ab > xy.
The reason is that the signs of the denominators (a and x) are UNKNOWN.
To illustrate:
If a<0 and x>0, then ax<0.
Multiplying each side of an inequality by a negative value requires that the direction of the inequality change.
Thus, if each side of b/x > y/a is multiplied by ax<0 -- a NEGATIVE value -- then the direction of the inequality must CHANGE from > to <:
b/x
> y/a
ax * (b/x) * x
< ax * (y/a)
ab
< xy.
Thus, b/x > y/a does not necessarily imply that ab > xy.
When the signs of the denominators are unknown, it is usually safer to plug in values.
Statement 1: b/x > y/a.
Case 1:
Try to plug in values so that xy≥0 and ab<0, with the result that the answer to the question stem is YES: xy > ab.
If possible, plug in values that also satisfy statement 2.
If x=1 and y=1, and a=-2 and b=1, both statements are satisfied:
b/x > y/a --> 1/1 > 1/(-2)
(ab)² > (xy)² --> (-2*1)² > (1*1)².
In this case, xy>0 and ab<0, so xy > ab.
Case 2:
Try to plug in values so that xy<0 and ab≥0, with the result that the answer to the question stem is NO: xy < ab.
If possible, plug in values that also satisfy statement 2.
If x=1 and y=-1, and a=2 and b=1, both statements are satisfied:
b/x > y/a --> 1/1 > (-1)/2
(ab)² > (xy)² --> (2*1)² > (1 * (-1))².
In this case, xy<0 and ab>0, so xy < ab.
Since xy > ab in the first case and xy < ab in the second case, the two statements combined are INSUFFICIENT.
The correct answer is
E.
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