Let x>-3selango wrote:Is x > 0?
(1) |x + 3| = 4x - 3
(2) |x + 1| = 2x - 1
OA later
Bhai..dont u think the question is itself dubious?kvcpk wrote:Let x>-3selango wrote:Is x > 0?
(1) |x + 3| = 4x - 3
(2) |x + 1| = 2x - 1
OA later
then
x+3 = 4x-3
3x=6
x=2...YES
let x<-3
-x-3=4x-3
ILLEGAL Expression
Hence x=2 SUFF
|x + 1| = 2x - 1
Let x>-1
x+1=2x-1
x=2...YES
Let x<-1
-x-1=2x-1
ILLEGAL EXPRESSION
SUFF
Pick D
Yeah.. I am seeing this type of question for the first time.. But if the source is MGMAT, then it might be correct. :roll:gmatmachoman wrote: Bhai..dont u think the question is itself dubious?
Ok try putting x= 0 in st 1..u would see a illlogical condition..selango wrote:This question is from MGMAT cat test.
I solved it as below,
From stmt1,
|x + 3| = 4x - 3
There are 2 ways to solve this,
x+3=4x-3
x=2
-(x+3)=4x-3 or -x-3=4x-3
x=0
Either x is 2 or 0.But only x=2 satisfies the Inequality.
So x=2.Sufficient.
From stmt2,
|x + 1| = 2x - 1
x+1=2x-1,x=2
-(x+1)=2x-2,x=0
But only x=2 satisfies the Inequality.
So x=2.Sufficient.
Hence D
First I solved the equation I thought to select option E as there are 2 values for x.
But I checked the Inequality with x values,only x=2 satisfy the equation.First time I am experiencing this strange thing.
Great tip! Thanks!Testluv wrote:Hi,
I've discussed this question before but it was taking too long to find my old post. We can solve more quickly by reasoning through the concept of absolute value rather than relying on pure algebra:
Is x > 0?
(1) |x + 3| = 4x - 3
We see the absolute value bars on the left. We know that absolute value is always positive or zero. We also know that, by definition, the left hand side of an equation must equal the right hand side. Therefore:
4x - 3 >= 0
x >= 3/4
If x is greater than or equal to 3/4, then it is definitely greater than 0.
SUFFICIENT.
(2) |x + 1| = 2x - 1
Similarly,
2x - 1>=0
x>= 1/2
SUFFICIENT.
Choose D.
