Is x 0 ? (1) /x - 3/ 5

Question

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

ywilfred · Answer

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

bsd_lover · Answer

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 ?

sludge · Answer

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.

Fig · Answer

(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.

shahrukh · Answer

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

kwam · Answer

1) abs(x-3)<5> -25 => x<7>3 INSUF Both together => 3

bsd_lover · Answer

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 ?

kwam · Answer

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

Fig · Answer

Yes.... It gives the clear view

... Well done

asaf · Answer

From 1: 
8 > x > -2 INSUFF

From 2:
x > 3; x <7> x > 3
C.

ggarr · Answer

I have them same question bsd had. why are we discarding the expressions w/negatives? I didn't get kwam's explanation.

bsd_lover · Answer

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.

terp26 · Answer

good diagram kwam, that explains the and/or really well

gk2k2 · Answer

My answer is C.
Draw a number line and select the common range for X. 
Common range comes to 3<X<8> 0

ArvGMAT · Answer

A simplifies to 
x>-2 or x3

A and B are both insufficient
Together - insufficient as x >-2 or x <8 
E

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

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