Problem from GMAT prep

Is |x| < 1
(i) |x+1| = 2|x-1|
(ii) |x-3| <> ZERO ( '<>' is NOT equal to)

Answer is C

----------------------------------
Here is what I know about this problem

|x| < 1 means -1 < x < 1
----------------------------------

I have NO idea of what to do with the rest of the info....

ok- first RECOGNIZE the type of problem. It is of the type:

|Quantity1| = |Quantity2|. So, what is/are the solution(s)?

1. Quantity1=Quantity2 AND/OR
2. Quantity1= -Quantity2

Reason for AND/OR: One has to plug the answer choices back in the equation to check whether each one works or not.
So, let's look at 1:
(i) |x+1| = 2|x-1|
CASE I: x +1 = 2 (x-1) => x = 3
CASE II x + 1 = -2 (x-1) => x = 1/3 ..you can do the algebra and verify that both choices work in the original eqn.

NOTE: Since we have ABS on both the sides, the solutions will work but doesn't take much to check.

Now, is X between -1 and 1? No-Case1; Yes-Case II => insufficient.

(ii) |x-3| <> ZERO ( '<>' is NOT equal to)
This says that x does not equal 3 (right, coz this means that the distance between X and 3 is not zero)=> insufficient.

BUT taken together, we see that x is 1/3. So, both are needed.

Let me know if anything is unclear.

BAD NEWS: one has to to be able to quickly recongnize the type of problem and the approach.

GOOD NEWS:Once you have seen and understood this, nothing of this sort will [or should] bother you.

How come there is no
3. -Quantity1= Quantity2

Rohit, 5 mins after posting my last reply it dawned on me why (3) is non existent

Because (3) yields same answer as (2) which is x = 1/3

I ll assume that (1) & (2) holds good for any such question on GMAT involving.... |something| = |something|


|Quantity1| = |Quantity2|. So, what is/are the solution(s)?

1. Quantity1=Quantity2 AND/OR
2. Quantity1= -Quantity2

guys when modulus is on both sides I would prefer this approach:

St. 1

|x+1| = 2|x-1|

squaring both sides (all positive nos.)

(x+1)^2 = 4 . (x-1)^2

this works out to

(x-3)(3x-1) = 0.

So Either x =3 (Thus |x| >1)

or x = 1/3. (Thus |x| <1)

Not sufficient.

St. 2

|x-3| <> ZERO

clearly Not sufficient.


combining.

(x-3)(3x-1) = 0. From St. 1

|x-3| <> ZERO From St. 2

Thus x = 1/3.

so sufficient.

Ans. C

So you've got it vinversa? Others?

Instead of squaring, one alternate method is (fewer steps, saves time):

when you have MODs on both side, there are just 2 possibilities:
(this is explained very well in equations strategy guide there are in all four possibilities:
both sides negative
LHS negative
RHS negative
both sides positve

One will observe, that of this four, 2 give out exactly same results. So we stick to only 2 unique solutions)

|x+1| = 2|x-1|

#1. both same signs

x+1 = 2(x-1)
x=3

#2. any one side negative
-(x+1) = 2(x-1)
x = 1/3

Rest remains same as explained by adiagr

(i) |x+1| = 2|x-1|
If x> 1, then
x+1 = 2x-2
x=3 ... (a)
If -1<x< 1, then
x+1 = -2(x-1)
x+1 = -2x+2
3x=1=>x=1/3 ... (b)

If -1>x, then
-(x+1) = -2(x-1)
-x-1=-2x+2
X=3 not possible
Based on (a) and (b), Based on this , we cannot determine whether |x| < 1
Hence statement 1 is insufficient
(ii) |x-3| <> ZERO ( '<>' is NOT equal to)
 X<3 or x>3
Based on this , we cannot determine whether |x| < 1

So , using both , we can discard the possibility of X=3 from (a) based on statement 2.
The answer is x=1/3 . This implies |X|<1

there is no such equation, because it's the same equation as #2. (the equation A = -B is the same as the equation -A = B; if you don't see why, just multiply one of them by -1 on both sides.)
for the same reason, you also don't have to write
-(Quantity1) = -(Quantity2)
because that's the same as equation #1.

for this test, the derivation of these equations is of absolutely no importance at all -- you won't be tested on derivations, and you certainly shouldn't be deriving these results on the exam (you won't have the time for that!). therefore, i don't think we should take up forum space with the derivations; they can probably be found in many other math-related places on the internet, if you are sufficiently curious.

--

for more information about the absolute value, click here
http://www.manhattangmat.com/thursdays-with-ron.cfm
and watch the FEBRUARY 18 workshop