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Re: Is |x - 3| < 7 ? (1) x > 0 (2) x < 10 [#permalink]
yashikaaggarwal wrote:
Is |x - 3| < 7 ?

(1) x > 0
Case 1: put x= 1
|1-3|<7
2<7 (sufficient)

Case 2: put x = 12
|12-3|>7
9>7 (insufficient)

Case 3: put x=9
|9-3|=7
7=7(insufficient)

Statement 1: (insufficient alone)

(2) x < 10
Case 1: x=9
|9-3|=7
7=7 (insufficient)

Case 2: put x= 1
|1-3|<7
2<7 (sufficient)

Case 3: put x = -12
|-12-3|>7
15>7 (insufficient)
Statement 2 is insufficient alone.

Statement 1 and 2:
0<x<10
Case 1: x=9
|9-3|=7
7=7 (insufficient)

Case 2: put x= 1
|1-3|<7
2<7 (sufficient)

(Not sufficient)

Answer is E

Posted from my mobile device


Statement 1 and 2:
0<x<10
Case 1: x=9
|9-3|=7
7=7 (insufficient)

|9-3| is 6 - it's sufficient :) :)
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Re: Is |x - 3| < 7 ? (1) x > 0 (2) x < 10 [#permalink]
Anuragjn wrote:
yashikaaggarwal wrote:
Is |x - 3| < 7 ?

(1) x > 0
Case 1: put x= 1
|1-3|<7
2<7 (sufficient)

Case 2: put x = 12
|12-3|>7
9>7 (insufficient)

Case 3: put x=9
|9-3|=7
7=7(insufficient)

Statement 1: (insufficient alone)

(2) x < 10
Case 1: x=9
|9-3|=7
7=7 (insufficient)

Case 2: put x= 1
|1-3|<7
2<7 (sufficient)

Case 3: put x = -12
|-12-3|>7
15>7 (insufficient)
Statement 2 is insufficient alone.

Statement 1 and 2:
0<x<10
Case 1: x=9
|9-3|=7
7=7 (insufficient)

Case 2: put x= 1
|1-3|<7
2<7 (sufficient)

(Not sufficient)

Answer is E

Posted from my mobile device


Statement 1 and 2:
0<x<10
Case 1: x=9
|9-3|=7
7=7 (insufficient)

|9-3| is 6 - it's sufficient :) :)

Yes got my mistake
Thank you.
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Re: Is |x - 3| < 7 ? (1) x > 0 (2) x < 10 [#permalink]
Bunuel wrote:
Is |x - 3| < 7 ?

(1) x > 0
(2) x < 10

DS21224


Is |x - 3| < 7

We get 2 cases.

Case 1: if x - 3 > 0
=> x- 3 < 7
=> x < 10.

Case 2: if x - 3 < 0
=> - x + 3 < 7
=> - x < 4
=> x > -4.

Thus we get : -4 < x < 10.
Now,

Statement 1: x > 0.
It satisfies x< 10.
but what if x > = 10. The the inequality wont hold true.

So Insufficient.


Statement 2: x < 10
It satisfies both : x < 10 x > -4.
But what if x < -4 also,

So this statement is also Insufficient.


Statement 1 + Statement 2:
0 < x < 10.
It satisfies both the sides of Inequality.
Thus Sufficient.


Answer C.
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Is |x - 3| < 7 ? (1) x > 0 (2) x < 10 [#permalink]
Given: |x - 3| < 7
Lets analyze the qtn statement: x is 7 points away from 3 on the number line
=> -4<x<10

Let analyze the statemnets:
I: x > 0 i.e x can be 3,5,10 or even 20 but we know x<10 NOT SUFFICIENT
II: x < 10 i.e. x can be 5,3,0 ,-4 or even -10 but we know -4<x NOT SUFFICIENT

Lets combine both:

from I: x>0
from II: x<10
==> 0<x<10

This is well within the range from out initial analysis of -4<x<10. Hence SUFFICIENT

C
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Re: Is |x - 3| < 7 ? (1) x > 0 (2) x < 10 [#permalink]
1
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|x - 3| < 7 can be re-written as below:

~ -7<x-3 < 7
~ -4 < x < 10

(1): x>0: x>10 does not make the inequality correct >> INSUFFICIENT
(2): x<10: if x<-4 >> does not make the inequality correct >> INSUFFICIENT
(1) + (2): 0<x<10 (within the range from -4 to 10) is totally Sufficient.

Hence C
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Re: Is |x - 3| < 7 ? (1) x > 0 (2) x < 10 [#permalink]
Is |x - 3| < 7 ?

Rephrased: -4 < x < 10?

(1) x > 0
Clearly insufficient.
e.g. x = 11 or 100 then the answer is NO.
e.g. x = 5 then YES

(2) x < 10
Clearly insufficient.
x = 5 then YES
x = -100 then NO

C: Sufficient b/c 0 < x < 10 overlaps with the defined region.
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Re: Is |x - 3| < 7 ? (1) x > 0 (2) x < 10 [#permalink]
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Re: Is |x - 3| < 7 ? (1) x > 0 (2) x < 10 [#permalink]
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