500 ds

[dunkin77]

Hi,

I thought the answer was A) but the correct answer is D)....

. If x&#8800;0, is |x| <1?
(1) x2<1
(2) |x| < 1/x

[gabriel]

Hi,

I thought the answer was A) but the correct answer is D)....

. If x&#8800;0, is |x| <1?
(1) x2<1
(2) |x| < 1/x

ok u already know that A is sufficient....

so let me focus on B ... it is given that mod(x)<1/x... so one thing is sure that x>0 .. else mod(x) wuld have been >1/x ...

now for any value greater than 1 ... 1/x will vary between 0 and 1 ... for eg. if x is 2 1/x=.5 .... so for x>1 ... 0 < 1/x < 1 ... and mod(x)>1 ... but this violates the condition given that mod (x)<1/x ..

so the only possible value for x is values between 0 and 1 ... when x is
between 0 and 1 ... 1/x>1 and mod(x) will vary between 0 and 1 ... and that fulfills the condition that mod(x)<1/x ...