k.......t .......X
X=kt?

st(1) t<0, then k<0 too and kt>0 -hence X will be positive and to the right of t, Sufficient;
st(2) k<1, t can be +ve OR -ve OR even 0, Not Sufficient

IOM A

Tagne wrote: 1. On a number line, if the number k is to the left of the number t, the product kt to the right of t?
a. T < 0
b. K < 1

|x|>|y| ?
squaring both parts, we get x^2>y^2 ? explanation when mode is positive number - squared numbers will be positive and comparable

st(1) x^2>y^2 Sufficient
st(2) x>y Not Sufficient, compare x=2, y=-3 |x|<|y| BUT x>y. It can be also x=2 and y=-1, |x|>|y| And x>y

IOM A

Tagne wrote: 2. is IxI > IyI ?
a. X^2 > Y^2
b. X > Y

Using your logic / example, assuming x =2 and y =-3, x^2=4 and y^2=9. Therefore is st1 not insufficient?

nope in question we are asked about mods, and the mods are always |a| absolute values while we have squared parts adequate to mods in st(1) in st(2) we have the sign problem (-ve, +ve). I just tried to demonstrate the easiest way to tackle this problem without sign/inequality translations. When both sides are in mods always use the squaring techniques, as the mods in any case are absolute values only - positive values compared...

Tagne wrote:Using your logic / example, assuming x =2 and y =-3, x^2=4 and y^2=9. Therefore is st1 not insufficient?

hey check it here also good move for such nuisances https://gmatclub.com/forum/absolute-fund ... 10674.html