Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

If y=|x-1|+|x+1|, then y=?

1) x > -1
2) x < 1

|x-1| = the distance between x and 1.
|x+1| = the distance between x and -1.
-1<--|x+1|-->x<--|x-1|-->1
Here, since the total distance between the two endpoints is 2, the sum of the 2 distances in red is 2.
If x is any value between the two endpoints, then the sum of the two distances will be 2.
Examples:
If x=-1, then y = |x-1| + |x+1| = 2.
If x=0, then y = |x-1| + |x+1| = 2.
If x=1/2, then y = |x-1| + |x+1| = 2.
In each case, since x is between -1 and 1, inclusive, the sum of the two distances is 2:
y = |x+1| + |x-3| = 2.

If x is positioned OUTSIDE -1 and 1, the sum of the two distances will INCREASE.
If x=-2, then y = |x-1| + |x+1| = 4.
If x=3, then y = |x-1| + |x+1| = 6 .

Statement 1:
If x is such that -1 < x ≤ 1, then y=2.
If x NOT such that -1 < x ≤ 1, then y>2.
INSUFFICIENT.

Statement 2:
If x is such that -1 ≤ x < 1, then y=2.
If x NOT such that -1 ≤ x < 1, then y > 2.
INSUFFICIENT.

Statements combined:
Since -1 < x < 1, y=2.
SUFFICIENT.

The correct answer is C.

Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

If y=|x-1|+|x+1|, then y=?

1) x > -1
2) x < 1

$$y = \left| {x - 1} \right| + \left| {x + 1} \right|$$
$$? = y$$

$$\left( 1 \right)\,\,x > - 1\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,x = 0\,\,\,\, \Rightarrow \,\,\,? = 2\, \hfill \cr
\,{\rm{Take}}\,\,x = 2\,\,\,\, \Rightarrow \,\,\,? = 4\, \hfill \cr} \right.$$
$$\left( 2 \right)\,\,x < 1\,\,\,\left\{ \matrix{
\,\left( {{\mathop{\rm Re}\nolimits} } \right){\rm{Take}}\,\,x = 0\,\,\,\, \Rightarrow \,\,\,? = 2\, \hfill \cr
\,{\rm{Take}}\,\,x = - 2\,\,\,\, \Rightarrow \,\,\,? = 4\, \hfill \cr} \right.$$
$$\left( {1 + 2} \right) - 1 < x < 1\,\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{
\,\left| {x - 1} \right| = 1 - x \hfill \cr
\,\left| {x + 1} \right| = x + 1 \hfill \cr} \right.\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 2$$


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

There are three ranges of values of x to consider.
If x > 1, then y = | x - 1 | + | x + 1 | = x - 1 + x + 1 = 2x and we don't have a unique value of y.
If -1 ≤ x ≤ 1, then y = | x - 1 | + | x + 1 | = - ( x - 1 ) + x + 1 = 2 and we have a unique value of y.
If x < 1, then y = | x - 1 | + | x + 1 | = -( x - 1 ) - ( x + 1 ) = -2x and we don't have a unique value of y.

Asking for the value of y is equivalent asking if -1 ≤ x ≤ 1.
Both conditions yield the inequality -1 < x < 1, when applied together. Therefore, both conditions are sufficient, when taken together.

In inequality questions, the law "Question is King" tells us that if the solution set of the question includes the solution set of the condition, then the condition is sufficient.
Therefore, C is the answer.
Answer: C