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rakeshd347
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rakeshd347 wrote:If x is not equal to 0, is |x| less than 1?
(1) x/|x|< x
(2) |x| > x
Simplify:
x ≠0
Is |x| < 1
or
Is -1 < x < 1 ? ... (i)
Solve:
St1:
Verify using very simple numbers:
x = -0.5
-0.5 / 0.5 < -0.5
-1 < -0.5 (i) is true
x = 2
2 / 2 < 2
1 < 2 (i) is false
INSUFFICIENT.
St2:
|x| > x
in other words x = -ive (This means x is always less than |+ive|, one can verify that only -ive values of x will satisfy this equation)
INSUFFICIENT.
St1 + St2:
We know that x is -ive we just need to check other smaller negative values by plugging numbers in the St1. This would narrow down the range if possible.
x = -1
-1 / 1 < -1
-1 < -1 (Invalid)
x = -0.5 (i) is always true as proved in St1.
x = -2
-2 / 2 < -2
-1 < -2 (Invalid)
Thus -1 < x < 0
We can definitely answer (i)
[spoiler]Answer : C[/spoiler]
Regards,
Vivek


















