Bunuel wrote:

Is x > 0?

(1) |3x - 30| < 20:

Get rid of the modulus: -20 < 3x - 30 < 20;

Add 30 to all three parts: 10 < 3x < 50;

Divide by 3: 10/2 < x < 50/3. Sufficient.

(2) 3x - 30 is greater than zero;

3x - 30 > 0;

3x > 30;

x > 10. Sufficient.

Answer: D.

Hi Bunuel,

Can you tell me how do you get rid of modulus.

The approach i use is: Like there are two scenarios for the |x| which is -x when x<0 and +x, when x>0.

Per the modulus rule:

when x<0,

the equation will be -3x+30<20; -3x<10; x>10/3

when x>0,

the equation becomes 3x-30<20;x<50/3

so x range is 10/3<x<50/3

Is this approach correct?