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

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Manager
Joined: 07 May 2007
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Is x > 0 ? (1) /x - 3/ < 5 (2) /x + 2/ > 5 [#permalink]  10 Jun 2007, 18:23
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Is x > 0 ?
(1) /x - 3/ < 5
(2) /x + 2/ > 5
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5078
Location: Singapore
Followers: 22

Kudos [?]: 184 [0], given: 0

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
CEO
Joined: 17 May 2007
Posts: 2994
Followers: 59

Kudos [?]: 467 [0], given: 210

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 ?
Manager
Joined: 23 Dec 2006
Posts: 136
Followers: 1

Kudos [?]: 11 [0], given: 0

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.
SVP
Joined: 01 May 2006
Posts: 1805
Followers: 8

Kudos [?]: 99 [0], given: 0

(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.
Manager
Joined: 17 Oct 2006
Posts: 52
Followers: 0

Kudos [?]: 1 [0], given: 0

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
VP
Joined: 09 Jan 2007
Posts: 1045
Location: New York, NY
Schools: Chicago Booth Class of 2010
Followers: 10

Kudos [?]: 157 [0], given: 3

1) abs(x-3)<5> -2<x<8>5 => x<7>3 INSUF

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

C
CEO
Joined: 17 May 2007
Posts: 2994
Followers: 59

Kudos [?]: 467 [0], given: 210

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 ?
VP
Joined: 09 Jan 2007
Posts: 1045
Location: New York, NY
Schools: Chicago Booth Class of 2010
Followers: 10

Kudos [?]: 157 [0], given: 3

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
SVP
Joined: 01 May 2006
Posts: 1805
Followers: 8

Kudos [?]: 99 [0], given: 0

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
Director
Joined: 06 Sep 2006
Posts: 745
Followers: 1

Kudos [?]: 17 [0], given: 0

From 1:
8 > x > -2 INSUFF

From 2:
x > 3; x <7> x > 3
C.
Director
Joined: 12 Jun 2006
Posts: 536
Followers: 1

Kudos [?]: 29 [0], given: 1

I have them same question bsd had. why are we discarding the expressions w/negatives? I didn't get kwam's explanation.
CEO
Joined: 17 May 2007
Posts: 2994
Followers: 59

Kudos [?]: 467 [0], given: 210

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.
VP
Joined: 22 Oct 2006
Posts: 1443
Schools: Chicago Booth '11
Followers: 8

Kudos [?]: 154 [0], given: 12

good diagram kwam, that explains the and/or really well
Intern
Joined: 11 Jun 2007
Posts: 22
Followers: 0

Kudos [?]: 14 [0], given: 0

Draw a number line and select the common range for X.
Common range comes to 3<X<8> 0
Senior Manager
Joined: 21 Jun 2006
Posts: 287
Followers: 1

Kudos [?]: 30 [0], given: 0

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
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