mighty absolute

1. x/|x| has two possible values corresponding to two possible signs for x. If x is positive, then x/|x| = 1. Since x/|x| < x, that gives you that x > 1, so |x| > 1.
If x is negative, then x/|x| = -1. Since x is a negative and is greater than x/|x| = -1, you will have -1 < x < 0, with |x| < 1. So statement 1 by itself is not sufficient, since you get two different answers depending on the sign of x.

2. |x| > x means that x is a negative. This does not clarify if x < -1 (which would give you that |x| > 1), so statement 2 by itself is not sufficient.

But, taken together, the two statements are enough to do the job. Since x is a negative from stmt 2, you will have the second case for stmt 1, meaning |x| < 1.

x/|x|< x can be written as |x| > 1, and this the information we are looking for. Statement 1 is sufficient.

Techwiz is right. Answer is A.

techwiz wrote:x/|x|< x can be written as |x| > 1, and this the information we are looking for. Statement 1 is sufficient.

we cant write this unless x is positive.
if x is negative then you have to switch the direction of the < to >
because you are dividing by a negative number...

oa is c only

Can anyone tell me why A is not sufficient.

x/|x| <x - this is possible only when x lies between 0 and -1 (excluding 0 and -1 )which means |x|<1 .

Thanks

x/|x| can have different values depending on the sign of x. This is qhy you have two cases:
a. x is negative, making |x| = -x. This means that x/|x| = -1. Now, since -1 is smaller than x, this means that x > -1. However, since we assumed that x is negative, then you get that x is between -1 and 0, with |x| < 1.

b. x is positive, making |x| = x. This means that x/|x| = 1 and, since x/|x| < x, we get that 1 < x. This makes |x| > 1.

As you can see, there are two possible answers to this problem, depending on the sign of x. This is why 1 is not sufficient to answer the problem.
Ramp: you said that "x/|x| <x - this is possible only when x lies between 0 and -1 (excluding 0 and -1 )which means |x|<1 . "
Let me give you a counterexample:
take x = 5. This makes x/|x| = 5/|5| = 5/5 = 1, which will obviously be smaller than x = 5.