Is [x] <1?
a) x / [x] < x
b) x < [x]
OA is C
I know the problem asks whether x is between -1 and 1. But I don't know how to solve or think the choices.Tksvm!
Complicated absolute value
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At first I was confused too, but it simply requires you to pick numbers.
try for instance:
stmt 1.
4/4 =1 holds
-2/2 = -1 not right
1/2 / 1/2 =1 not right
-1/2 / 1/2 = -1 holds
So we find that to satisfy stmt 1 x could either be >1 or -1<x<0.
Stmt 2 tells us what we need to add to stmt 1 to get C as answer.
Hope this helps!
try for instance:
stmt 1.
4/4 =1 holds
-2/2 = -1 not right
1/2 / 1/2 =1 not right
-1/2 / 1/2 = -1 holds
So we find that to satisfy stmt 1 x could either be >1 or -1<x<0.
Stmt 2 tells us what we need to add to stmt 1 to get C as answer.
Hope this helps!
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@El Cucu,
R u sure that u posted the entire question?
I think u missed "X is not equal to 0" Otherwise stmt 1 will fail and GMAC avoids infinity/indeterminate values.
Next time pls copy the problem entirely.
Let me assume that x is != 0 and solve the problem. (here ! implies "not")
stmt 1.
x/|x| < x
since x != 0, we can cross multiply.
x < x* |x|-----------------------1
case 1. x is +ve.
we can divide by x and < sign will remain the same.
so we get 1 < |x|
ie 1 < x (coz x is +ve)
case 2: x is -ve.
< will become ">"
Dividing by x,
1 > |x|
Now x is -ve so we get 1 > -x or -1 < x < 0 ( x < 0 coz we assumed that x is negative.. in case 2)
combining case 1 and 2. we get.
x > 1 OR -1 < x < 0 ====> ambiguous
Hence stmt 1 is insufficient.
stmt 2.
x < |x|
again .. intuitively one can say that x has to be -ve. for this to be true.
eg. let x = -2
so -2 < |-2| ==> -2 < 2 OK.
But it just says that x is -ve and nothing else. hence insufficent
Combining 1 and 2.
Apply x is -ve from stmt 2 to stmt 1.
we get -1 < x < 0.
hence x < 1
Answer is C.
HT Helps
R u sure that u posted the entire question?
I think u missed "X is not equal to 0" Otherwise stmt 1 will fail and GMAC avoids infinity/indeterminate values.
Next time pls copy the problem entirely.
Let me assume that x is != 0 and solve the problem. (here ! implies "not")
stmt 1.
x/|x| < x
since x != 0, we can cross multiply.
x < x* |x|-----------------------1
case 1. x is +ve.
we can divide by x and < sign will remain the same.
so we get 1 < |x|
ie 1 < x (coz x is +ve)
case 2: x is -ve.
< will become ">"
Dividing by x,
1 > |x|
Now x is -ve so we get 1 > -x or -1 < x < 0 ( x < 0 coz we assumed that x is negative.. in case 2)
combining case 1 and 2. we get.
x > 1 OR -1 < x < 0 ====> ambiguous
Hence stmt 1 is insufficient.
stmt 2.
x < |x|
again .. intuitively one can say that x has to be -ve. for this to be true.
eg. let x = -2
so -2 < |-2| ==> -2 < 2 OK.
But it just says that x is -ve and nothing else. hence insufficent
Combining 1 and 2.
Apply x is -ve from stmt 2 to stmt 1.
we get -1 < x < 0.
hence x < 1
Answer is C.
HT Helps