jeeteshsingh wrote:

ugimba wrote:

If x is not equal to 0, is |x| less than 1?

(1) \(x/|x| < x\)

(2) |x| > x

IMO A...

ST 1: can be written as \(x < x * |x|\) since multiplying by |x| which is always positive would not change the sign.

This gives x as -1<x<0.... therefore sufficient...

ST2: |x| > x gives value of x as x<0... hence could be -2,-3.... Therefore not sufficient...

Cheers!

JT

1) this only holds true if x is a negative fraction or greater than 1. The question however is asking if |x| is less than 1. The abs val of a number less than -1 will be greater than 1, whereas the abs val of a negative fraction will be less than 1.

x/|x| < x

try -2, -2/|-2| = -1, -1 is greater than -2, so does not hold

try -1/2, (-1/2)/|(-1/2)| = -1, -1 is less than -1/2 so this holds.

try 1/2, (1/2)/|(1/2)| = 1, 1 is greater than 1/2, so this does not hold

try 2, 2/|2| = 1, 1 is less than 2, so this holds

so, -1 < x < 0 and x > 1

the question asks: is |x| < 1?

-1 < x < 0 --> |x| will be a positive fraction, i.e. less than 1

x > 1 --> |x| will be greater than 1

hence insufficient.

2) the abs val of a positive number equals that number so this only holds true for negative numbers (including fractions). The question is asking if |x| is less than 1.

Same as in (1),

|x| > x

x < 0

try a negative integer, -2,

| -2 | = 2, --> |x| will be greater 1

try a negative fraction, -(1/2),

| -(1/2) | = 1/2 --> |x| will be less than 1

hence insufficient.

putting both together,

x < 0 and -1< x < 0 and x > 1

this limits x to -1< x < 0, thus |x| will be a positive fraction and will always be less than 1.

therefore the answer is C

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