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Is x > 0 ? (1) /x - 3/ < 5 (2) /x + 2/ > 5

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Manager
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Is x > 0 ? (1) /x - 3/ < 5 (2) /x + 2/ > 5 [#permalink] New post 10 Jun 2007, 18:23
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A
B
C
D
E

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Is x > 0 ?
(1) /x - 3/ < 5
(2) /x + 2/ > 5
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 [#permalink] New post 10 Jun 2007, 19:10
St1:
Positive case: x-3 <5> x<8
Negative case: -x+3 <5> x>-2

Insufficient.

St2:
Positive case: x+2 > 5 --> x > 3
Negative case: -x-2 > 5 --> x < -7

Insufficient.

Using both: Insufficient as permissible range is -2 to 3.

Ans E
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 [#permalink] New post 11 Jun 2007, 21:59
Do you have an OE on this ? I am trying to work out the exact maths to solve these kinds of problems. After doing some modulus maths I am arriving at the following :

Stmt 1 tells us :
when x > 3 ---> x < 8
when x <3> x > -2

Stmt 2 tells us :
when x > -2 ---> x > 3
when x <2> x <7> 0 ? Clearly when x < -2 it is NOT < 0 hence the answer should be E ?
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 [#permalink] New post 11 Jun 2007, 22:32
I get:

statement 1: -2 < x < 8

statement 2: x<-7 and x>3.

Original stem says x > 0, so I suppose we can discard x<-7. This leaves us with 3 < x < 8.

C.

Last edited by sludge on 11 Jun 2007, 22:34, edited 1 time in total.
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 [#permalink] New post 11 Jun 2007, 22:33
(C) for me :)

X > 0 ?

From 1
|x - 3| < 5

o If x >= 3, then
|x - 3| < 5
<=> x-3 < 5
<=> x < 8 >>>> So 3 =< x < 8

o If x < 3, then
|x - 3| < 5
<=> -(x-3) < 5
<=> x > -2 >>>> So -2 < x < 3

Finally,
-2 < x < 8

INSUFF.

From 2
|x + 2| > 5

o If x >= -2, then
|x + 2| > 5
<=> x+2 > 5
<=> x > 3 >>>> Ok (>= -2)

o If x < -2, then
|x + 2| > 5
<=> -(x+2) > 5
<=> x < -7 >>>> Ok (< -2)

Finally,
x < -7 or x > 3

INSUFF.

Combining (1) with (2)
We have:
-2 < x < 8
AND
x < -7 or x > 3

This implies that 3 < x < 8, and so x > 0.

SUFF.
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 [#permalink] New post 12 Jun 2007, 00:18
dam* it. such simple question and still it took me two tries as i was changing signs of 5.
anyway for me answer is C
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 [#permalink] New post 12 Jun 2007, 01:24
1) abs(x-3)<5> -2<x<8>5 => x<7>3 INSUF

Both together => 3<x<8, x is positive, SUF

C
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 [#permalink] New post 12 Jun 2007, 05:08
Combining (1) with (2)
We have:
-2 < x < 8
AND
x <7> 3

This implies that 3 < x <8> 0.
-------------------


Can you please explain WHY did you discard x < -7 ?
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 [#permalink] New post 12 Jun 2007, 05:29
Combining (1) with (2)
We have:
-2 < x < 8
AND
x < -7 or x > 3

(1) ----------------- -2 xxxxxxxxxxxxxxxxxxx 8 -----------

(2) xxxxxx -7 ---------------- 3 xxxxxxxxxxxxxxxxxxxxxx

(1) and (2):

--------- -7 ----- -2 ------- 3 xxxxxxxxxxxx 8 -----------

Where x is where x "can be", you need to see where X is in both lines, if it where "or" you just need to find in each line seperately.

Hope that helps
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 [#permalink] New post 12 Jun 2007, 05:32
kwam wrote:
Combining (1) with (2)
We have:
-2 < x < 8
AND
x < -7 or x > 3

(1) ----------------- -2 xxxxxxxxxxxxxxxxxxx 8 -----------

(2) xxxxxx -7 ---------------- 3 xxxxxxxxxxxxxxxxxxxxxx

(1) and (2):

--------- -7 ----- -2 ------- 3 xxxxxxxxxxxx 8 -----------

Where x is where x "can be", you need to see where X is in both lines, if it where "or" you just need to find in each line seperately.

Hope that helps


Yes.... It gives the clear view :)... Well done :)
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 [#permalink] New post 12 Jun 2007, 06:39
From 1:
8 > x > -2 INSUFF

From 2:
x > 3; x <7> x > 3
C.
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 [#permalink] New post 12 Jun 2007, 14:47
I have them same question bsd had. why are we discarding the expressions w/negatives? I didn't get kwam's explanation.
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 [#permalink] New post 12 Jun 2007, 16:35
Hi ggarr,

Now that I understand this, let me try and explain. Once we have narrowed it down to 2 different number lines (see the number lines in kwams explanation), we can only accept values of x that are common for BOTH number lines and discard the rest.

An analogy would be if Stmt 1 says x = 2 or 3 or 4
and Stmt 2 says x = 2 or 5 or 7. In this case x can only be 2 because it is the common value and all the rest of the values are discarded.

Similarly on that number line the only values for x that are possible are the common ones (i.e. between 3 and 8)

Hope this helps.
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 [#permalink] New post 13 Jun 2007, 09:40
good diagram kwam, that explains the and/or really well
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 [#permalink] New post 16 Jun 2007, 10:35
My answer is C.
Draw a number line and select the common range for X.
Common range comes to 3<X<8> 0
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 [#permalink] New post 16 Jun 2007, 19:09
A simplifies to
x>-2 or x<8

B simplifies to
x<7>3

A and B are both insufficient
Together - insufficient as x >-2 or x <8
E
  [#permalink] 16 Jun 2007, 19:09
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