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Is x > 0 ? (1) |x+3| < 4 (2) |x-3| < 4 Please assist with above prob

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Is x > 0 ? (1) |x+3| < 4 (2) |x-3| < 4 Please assist with above prob  [#permalink]

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New post 26 Sep 2016, 20:03
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Question Stats:

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Is x > 0 ?

(1) |x+3| < 4
(2) |x-3| < 4

Please assist with above problem
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Re: Is x > 0 ? (1) |x+3| < 4 (2) |x-3| < 4 Please assist with above prob  [#permalink]

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New post 26 Sep 2016, 21:32
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Re: Is x > 0? (1) |x + 3| < 4 (2) |x – 3| < 4  [#permalink]

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New post 23 Jan 2018, 23:04
DHINGRACHIRAG24 wrote:
Is x > 0?

(1) |x + 3| < 4
(2) |x – 3| < 4


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(1) |x-(-3)| < 4
This also means that distance of 'x' from '-3' on the number line is within 4 steps. So x is within 4 steps to right and left of -3. 4 steps to right of -3 is 1 and 4 steps to the left of -3 is -7. Thus x lies between -7 and 1. So x could be either positive or negative. Not sufficient.

(2) |x - 3| < 4
This also means that distance of 'x' from '3' on the number line is within 4 steps. So x is within 4 steps to right and left of 3. 4 steps to right of 3 is 7 and 4 steps to the left of 3 is -1. Thus x lies between -1 and 7. So x could be either positive or negative. Not sufficient.

After combining the two statements, x could lie between -1 and 1 (this is the only area on the number line which x can take if it has to satisfy both statement 1 and statement 2 conditions).
But still between -1 and 1, x could be negative or 0 or positive. So not sufficient.

Hence E answer
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Re: Is x > 0 ? (1) |x+3| < 4 (2) |x-3| < 4 Please assist with above prob  [#permalink]

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New post 31 May 2018, 02:52
Hi Bunuel,

I have a very basic doubt here.

On combining the statements 1 & 2, how did you get the range of x as -1 < x < 1 ? Why can't the range of x be -7 < x < 7 ?

Bunuel wrote:
Is x > 0 ?

(1) |x+3| < 4
-4 < x + 3 < 4
-7 < x < 1.

Not sufficient.

(2) |x-3| < 4
-4 < x - 3 < 4
-1 < x < 7.

Not sufficient.

(1)+(2) -1 < x < 1. Not sufficient.

Answer: E.
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Re: Is x > 0 ? (1) |x+3| < 4 (2) |x-3| < 4 Please assist with above prob  [#permalink]

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New post 31 May 2018, 03:10
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101mba101 wrote:
Hi Bunuel,

I have a very basic doubt here.

On combining the statements 1 & 2, how did you get the range of x as -1 < x < 1 ? Why can't the range of x be -7 < x < 7 ?

Bunuel wrote:
Is x > 0 ?

(1) |x+3| < 4
-4 < x + 3 < 4
-7 < x < 1.

Not sufficient.

(2) |x-3| < 4
-4 < x - 3 < 4
-1 < x < 7.

Not sufficient.

(1)+(2) -1 < x < 1. Not sufficient.

Answer: E.



HEllo

First statement concludes that -7 < x < 1. It means 'x' is a number which is greater than -7 but less than 1.
Second statement concludes that -1 < x < 7. This means that 'x' is a number which is greater than -1 but less than 7.

Now, combining the two statements. what is common about x? From first, x should be greater than -7 and from second x should be greater than -1. So if a number is both greater than -7 as well as greater than -1, then it has to be greater than -1 (which is the common part).
Similarly, from first x is less than 1 and from second x is less than 7. So if a number is lesser than 1 as well as lesser than 7, then it must be lesser than 1 (which is the common part).
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Is x > 0 ? (1) |x+3| < 4 (2) |x-3| < 4 Please assist with above prob  [#permalink]

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New post 31 May 2018, 03:37
Hey amanvermagmat,

Your explanation solved my doubt immediately. You kept it super simple. Thanks a lot!

Cheers!

amanvermagmat wrote:
HEllo

First statement concludes that -7 < x < 1. It means 'x' is a number which is greater than -7 but less than 1.
Second statement concludes that -1 < x < 7. This means that 'x' is a number which is greater than -1 but less than 7.

Now, combining the two statements. what is common about x? From first, x should be greater than -7 and from second x should be greater than -1. So if a number is both greater than -7 as well as greater than -1, then it has to be greater than -1 (which is the common part).
Similarly, from first x is less than 1 and from second x is less than 7. So if a number is lesser than 1 as well as lesser than 7, then it must be lesser than 1 (which is the common part).
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Re: Is x > 0 ? (1) |x+3| < 4 (2) |x-3| < 4 Please assist with above prob  [#permalink]

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New post 21 Jun 2018, 15:59
Hi Bunuel,

When statements (1)+(2) -1 < x < 1. that means x has to be "0", however it is not mentioned in the question that "X" is a Integer, so this is insufficient correct ?
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Re: Is x > 0 ? (1) |x+3| < 4 (2) |x-3| < 4 Please assist with above prob  [#permalink]

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New post 21 Jun 2018, 21:37
hero_with_1000_faces wrote:
Hi Bunuel,

When statements (1)+(2) -1 < x < 1. that means x has to be "0", however it is not mentioned in the question that "X" is a Integer, so this is insufficient correct ?


-1 < x < 1 means that x is any number from -1 and 1, not inclusive, not necessarily 0. For example, -3/100, -4/11, -0.000012, 0, 1/2, ... So, x can take infinitely many values, included 0. Hence we cannot say whether x is more than 0.
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Re: Is x > 0 ? (1) |x+3| < 4 (2) |x-3| < 4 Please assist with above prob  [#permalink]

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New post 05 Oct 2018, 03:29
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(1) INSUFFICIENT: We can solve this absolute value inequality by considering both the positive and negative scenarios for the absolute value expression |x + 3|.
If x > -3, making (x + 3) positive, we can rewrite |x + 3| as x + 3:
x + 3 < 4
x < 1
If x < -3, making (x + 3) negative, we can rewrite |x + 3| as -(x + 3):
-(x + 3) < 4
x + 3 > -4
x > -7
If we combine these two solutions we get -7 < x < 1, which means we can’t tell whether x is positive.

(2) INSUFFICIENT: We can solve this absolute value inequality by considering both the positive and negative scenarios for the absolute value expression |x – 3|.
If x > 3, making (x – 3) positive, we can rewrite |x – 3| as x – 3:
x – 3 < 4
x < 7
If x < 3, making (x – 3) negative, we can rewrite |x – 3| as -(x – 3) OR 3 – x
3 – x < 4
x > -1
If we combine these two solutions we get -1 < x < 7, which means we can’t tell whether x is positive.

(1) AND (2) INSUFFICIENT: If we combine the solutions from statements (1) and (2) we get an overlapping range of -1 < x < 1. We still can’t tell whether x is positive.
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Re: Is x > 0 ? (1) |x+3| < 4 (2) |x-3| < 4 Please assist with above prob   [#permalink] 05 Oct 2018, 03:29
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