DS(Inequality)

[rintoo22]

Please find the attachment.

The Inequality is |x|+|x-1|=1
First outcome x+x-1=1 => 2x=2 => x=1.......(1)
Second outcome -x-x+1=1 => -2x=0 => x=0 .....(2)

Statement 1 x>=0 does not satisfy. Hence insufficient
Statement 2 x<=1 does nto satisfy. Hence Insufficient

St 1 and 2 together. Should be insufficient because x could be any value between 0 and 1 included.
The answer in my opinion shoudl be E. However QA is C.
Please assist and let me know where I am going wrong.

[Anju@Gurome]

Is |x| + |x - 1| = 1?

(1) x ≥ 0
(2) x ≤ 1

|x| = Distance of x from zero on the number line
|x - 1| = Distance of x from 1 on the number line

Now, distance between 0 and 1 on the number line is (1 - 0) = 1
Hence, if the sum of the distances of x from 0 and 1 on the number line is equal to 1, then x must lie between 0 and 1 on the number line, inclusive 0 and 1.

So the question reduces down to "Is 0 ≤ x ≤ 1?"

Clearly none of the statements are individually sufficient.

1 & 2 Together: Now, we know that 0 ≤ x ≤ 1

Sufficient

The correct answer is C.

[Anju@Gurome]

The Inequality is |x|+|x-1|=1
First outcome x+x-1=1 => 2x=2 => x=1.......(1)
Second outcome -x-x+1=1 => -2x=0 => x=0 .....(2)

I'll repeat what I always post in this forum while solving absolute value problems : Do not blindly follow algebra, i.e. do not simply write case 1: |x| = x and case 2: |x| = -x etc. and whenever possible think in terms of distances on the number line.

Here, we have have two terms with absolute values : |x| and |x - 1|
Hence, we have two critical points (value of x where sign change occurs) : x = 0 and x = 1
Hence, we need to analyze the equation in three regions : x < 0, 0 ≤ x ≤ 1, and x > 1

Case 1: x < 0 ---> |x| = -x and |x - 1| = -(x - 1)
So, -x - (x - 1) = 1
--> -2x = 0
--> x = 0 --> Contradiction with our initial assumption of x < 0
--> No solution in this region

Case 2: 0 ≤ x ≤ 1 ---> |x| = x and |x - 1| = -(x - 1)
So, x - (x - 1) = 1
--> 1 = 1
--> This is always true meaning all value in this region will satisfy our equation

Case 3: x > 1 ---> |x| = -x and |x - 1| = (x - 1)
So, x + (x - 1) = 1
--> 2x = 2
--> x = 1 --> Contradiction with our initial assumption of x > 1
--> No solution in this region

Hence, the question is asking whether 0 ≤ x ≤ 1 or not not just whether x = 0 and x = 1.

Hope that helps.

[rintoo22]

Super Thanks Anju.

[GMATGuruNY]

Does |x| + |x-1| = 1?

Statement 1: x≥0
If x=0, then |x| + |x-1| = 0+1 = 1.
If x=10, then |x| + |x-1| = 10 + 9 = 19.
INSUFFICIENT.

Statement 2: x≤1
If x=1, then |x| + |x-1| = 1 + 0 = 1.
If x=-10, then |x| + |x-1| = 10 + 9 = 19.
INSUFFICIENT.

Statements combined: 0≤x≤1.
When we evaluated each statement on its own, we learned the following:
If x=0 or x=1, then |x| + |x-1| = 1.
The one case left to try is a value BETWEEN 0 and 1:
If x=1/2, then |x| + |x-1| = 1/2 + 1/2 = 1.
Since |x| + |x-1| = 1 in each case, SUFFICIENT.

The correct answer is C.