Is x 0 ? (1) |x + 3|

Question

Is x > 0 ?

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

Bunuel · Accepted Answer

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) Overlap of the ranges from (1) and (2) is -1 < x < 1. So, x can be negative as well as positive. Not sufficient.

Answer: E.

amanvermagmat · Answer

(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

101mba101 · Answer

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  ?

amanvermagmat · Answer

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).

hero_with_1000_faces · Answer

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 ?

Bunuel · Answer

-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.

SUNILAA · Answer

(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.

Rajashree123 · Answer

But in this solution, if we can square both of the options we will get x<0. (Thus confirm no -C)
|x+3|<4 and |x-3|<4
thus squaring and adding:
x^2+6x+9-x^2+6x-9<16-16
=> 12x<0
=> x<0
what is the problem in my logic?

aritranandy8 · Answer

VeritasKarishma  please expain

Bunuel · Answer

You are not adding there, you are subtracting. We can only subtract inequalities, when their signs are in opposite directions, which is not the case here.

KarishmaB · Answer

(1) |x + 3| < 4

means "Distance of x from -3 is less than 4". So x will lie in the range -7 to 1 (exclusive).
x may be negative or positive. Not sufficient.

(2) |x - 3| < 4

means "Distance of x from 3 is less than 4". So x will lie in the range -1 to 7 (exclusive).
x may be negative or positive. Not sufficient.

Using both, x must satisfy both conditions. So x can only lie in the overlap region -1 to 1 (exclusive). It can again be negative or positive so not sufficient.

Answer (E)

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Is x > 0 ? (1) |x + 3| < 4 (2) |x - 3| < 4

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