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[quote="aniketb"]Is x > 0?

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

Statement I is insufficient:

x = -0.5 (NO)
x = 0.1 (YES)

Statement II is insufficient:

x = -0.5 (NO)
x = 0.1 (YES)

Combining is not necessary because x = -0.5 and x = 0.1 disprove statement I and II. Hence the answer is E.
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Is x > 0?

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

Statement I is insufficient:

x = -0.5 (NO)
x = 0.1 (YES)

Statement II is insufficient:

x = -0.5 (NO)
x = 0.1 (YES)

Combining is not necessary because x = -0.5 and x = 0.1 disprove statement I and II. Hence the answer is E.
Thanks for the reply PerfectScores, it would be helpful if you could please elaborate on what basis have you chosen the values? (For example -0.1 or -1 would also have done the job, why -0.5 and 0.1 only?) It will help, as I tend to pick wrong values some times and hence chose Bunuel's method.
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aniketb
Is x > 0?

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

Statement I is insufficient:

x = -0.5 (NO)
x = 0.1 (YES)

Statement II is insufficient:

x = -0.5 (NO)
x = 0.1 (YES)

Combining is not necessary because x = -0.5 and x = 0.1 disprove statement I and II. Hence the answer is E.
Thanks for the reply PerfectScores, it would be helpful if you could please elaborate on what basis have you chosen the values? (For example -0.1 or -1 would also have done the job, why -0.5 and 0.1 only?) It will help, as I tend to pick wrong values some times and hence chose Bunuel's method.

It is about Z O N E D

We will try to use Zero, One, Negative, Extreme (Even/Odd) (Small/Large), Decimals.

Here -0.5 and 0.1 seemed to eliminate both the answer options.
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we could also use visualization here:
|x + 3| < 4 is basically distance of point x from point (-3) is less than 0. So the extreme points are -3+4 = 1 and -3-4 = -7 (not inclusive)

Since 1 is not a valid point, lets pick something smaller, say 0.5 and -7 is not valid so something greater, say -0.5
st1.
|-0.5+3| = 2.5 <4 & |0.5+3|=3.5<4 -- not sufficient
St2.
|-0.5-3| = 3.5 <4 & |0.5-3| = 2.5 <4 -- not sufficient
Combining also you see both point taken suffice both conditions so Ans: E. (PS I calculated -0.5 as second point by eyeballing second eq & using above visualization approach).
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It is about Z O N E D

We will try to use Zero, One, Negative, Extreme (Even/Odd) (Small/Large), Decimals.

Here -0.5 and 0.1 seemed to eliminate both the answer options.


Hey PerfectScore Thanks alot, the ZONED technique is helpful... :yes
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rohanGmat
we could also use visualization here:
|x + 3| < 4 is basically distance of point x from point (-3) is less than 0. So the extreme points are -3+4 = 1 and -3-4 = -7 (not inclusive)

Since 1 is not a valid point, lets pick something smaller, say 0.5 and -7 is not valid so something greater, say -0.5
st1.
|-0.5+3| = 2.5 <4 & |0.5+3|=3.5<4 -- not sufficient
St2.
|-0.5-3| = 3.5 <4 & |0.5-3| = 2.5 <4 -- not sufficient
Combining also you see both point taken suffice both conditions so Ans: E. (PS I calculated -0.5 as second point by eyeballing second eq & using above visualization approach).
Hey Rohan, thanks for the reply. This is slightly off the topic, but I can see your Verbal score has improved a lot in second attempt. I am scoring around 650 with similar breakup of your first attempt. It will be very much helpful if you let me know how you did it? (The plan, last minute strategy, Mocks etc)

Thanks in advance... :)
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I don't see any question options... am I missing something? This is the second math question with which I've had this issue.
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I don't see any question options... am I missing something? This is the second math question with which I've had this issue.

This is a data sufficiency question. Options for DS questions are always the same.

The data sufficiency problem consists of a question and two statements, labeled (1) and (2), in which certain data are given. You have to decide whether the data given in the statements are sufficient for answering the question. Using the data given in the statements, plus your knowledge of mathematics and everyday facts (such as the number of days in July or the meaning of the word counterclockwise), you must indicate whether—

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
D. EACH statement ALONE is sufficient to answer the question asked.
E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

I suggest you to go through the following post ALL YOU NEED FOR QUANT.

Hope this helps.
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strc
Is x > 0?

(1) |x + 3| < 4
(2) |x – 3| < 4


lets see the statement straight as there is no info from the statements..
(1) |x + 3| < 4
we can form two eq..
x+3<4 so x<1..
x+3>-4, so x>-7..
so -7<x<1.. insuff

(2) |x – 3| < 4
again two eq..
x-3<4.. so x<7.
x-3>-4,, so x>-1..
so -1<x<7..
again insufff..

combined
therange of x is between -1 and 1.. insuff
E
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Is x > 0?

(1) |x + 3| < 4
(2) |x – 3| < 4

Required: Is x > 0?

Statement 1: |x + 3| < 4
On solving, we get
-4 < x + 3 < 4
-7 < x < 1
Since x is not restricted to integers, it can take any value.
INSUFFICIENT

Statement 2: |x – 3| < 4
On Solving,
-4 < x - 3 < 4
-1 < x < 7
Since x is not restricted to integers, it can take any value.
INSUFFICIENT

Statement 1 and Statement 2 Combined:
The common values of -7 < x < 1 and -1 < x < 7 are:
- 1 < x < 1
Still we cannot say if x > 0 or not.
INSUFFICIENT

Option E
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Is x > 0?

(1) |x + 3| < 4

(2) |x – 3| < 4
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Is x > 0?

(1) |x + 3| < 4

(2) |x – 3| < 4

Official Answer/Explanation
(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.

The correct answer is E.
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Is x > 0?

(1) |x + 3| < 4

(2) |x – 3| < 4

Official Answer/Explanation
(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.

The correct answer is E.

The highlighted area shows that x = 0. Thus, x is not > 0. Shouldn't the answer for this be C?
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tommannanchery
Is x > 0?

(1) |x + 3| < 4

(2) |x – 3| < 4

Merging topics. Please refer to the discussion above.
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tommannanchery
Is x > 0?

(1) |x + 3| < 4

(2) |x – 3| < 4

Official Answer/Explanation
(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.

The correct answer is E.

The highlighted area shows that x = 0. Thus, x is not > 0. Shouldn't the answer for this be C?

No. We are not told that x is an integer. It can be any number between -1 and 1, not necessarily 0.
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