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# Is x > 0 ? (1) |3x - 30| < 20 (2) 3x - 30 is greater than zero

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Is x > 0 ? (1) |3x - 30| < 20 (2) 3x - 30 is greater than zero [#permalink]

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30 Nov 2017, 17:30
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Is x > 0 ?

(1) |3x - 30| < 20

(2) 3x - 30 is greater than zero
[Reveal] Spoiler: OA

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Re: Is x > 0 ? (1) |3x - 30| < 20 (2) 3x - 30 is greater than zero [#permalink]

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30 Nov 2017, 18:01
D

From st1 we can conclude 10/3<x<50/3. So x>0. Hence sufficient

From st2 we can conclude x>10. Hence sufficient

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Re: Is x > 0 (1) |3x - 30| < 20 (2) 3x - 30 is greater than zero [#permalink]

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30 Nov 2017, 19:53
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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.

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Re: Is x > 0 ? (1) |3x - 30| < 20 (2) 3x - 30 is greater than zero [#permalink]

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01 Dec 2017, 19:25
Is x > 0 ?

(1) |3x - 30| < 20

(2) 3x - 30 is greater than zero

From Stmn 1: |x-10| < $$\frac{20}{3}$$, which means x cannot be negative because of its negative modulus value will keep on increasing, it can only be positive.. Sufficient.
From Stmnt 2: 3 (x-10) > 0, which means x > 0.

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Re: Is x > 0 ? (1) |3x - 30| < 20 (2) 3x - 30 is greater than zero [#permalink]

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02 Dec 2017, 21:03
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.

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?

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Re: Is x > 0 ? (1) |3x - 30| < 20 (2) 3x - 30 is greater than zero [#permalink]

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02 Dec 2017, 22:20
shail2509 wrote:
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.

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?

1. |x| < a, where a is a non-negative number, means that -a < x < a. For example, |x| < 3, means that -3 < x < 3.

On the other hand: |x| > b, where b is a non-negative number, means that x > b or x < -b. For example, |x| > 5 means that x < -5 or x > 5.

2. Your approach is not correct.

When $$x \le 0$$ then $$|x|=-x$$, or more generally when $$\text{some expression} \le 0$$ then $$|\text{some expression}| = -(\text{some expression})$$. For example: $$|-5|=5=-(-5)$$;

When $$x \ge 0$$ then $$|x|=x$$, or more generally when $$\text{some expression} \ge 0$$ then $$|\text{some expression}| = \text{some expression}$$. For example: $$|5|=5$$;

Notice, that in |3x - 30| < 20, you should consider the ranges when 3x - 30 <= 0 and 3x - 30 > 0, not when x <=0 and x > 0, because the expression in modulus is 3x - 30 NOT x.

Finally, check below links for more:

10. Absolute Value

For more check Ultimate GMAT Quantitative Megathread

Hope it helps.
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Re: Is x > 0 ? (1) |3x - 30| < 20 (2) 3x - 30 is greater than zero   [#permalink] 02 Dec 2017, 22:20
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