If |x – 6| = 5, what is the value of x?

Hey kuldeepdubai1986 ,

You missed very important point of modulus.

|x - 6 | = x - 6 if x >= 6
|x - 6 | = -(x-6) if x < 6

Now, in option A. We are told x > 0. That could mean x could be 1,2 or value more than 6.

Hence, when you open the modulus, you will get

|x-6| = 5

x-6 = 5 => x = 11

or -(x-6) = 5 => -x+6=5 => x = 1

Hence, two values of x for x > 0. Therefore, A is insufficient.

Does that make sense?

If |x – 6| = 5
By removing the modulus we will have,
x-6 = 5 or x-6=-5. It will have two values of X i.e. 1 and 11.
Coming to options,
Option 1: x>0. As both the values are greater than 0. Option is not sufficient.
Option 2: x^2 = 121. X will have two values either 11 or -11. Sufficient.

Answer B.

Can someone explain to me how the answer is B?

Statement 2 gives us two answers: -11 and 11, therefore it should be NS.

If we combine both statements we get 11. So shouldnt the answer be C?

It does. Looks like I should have read the question stem more closely.

i am still unable to understand why the answer is not D but B.
We should check for both the negative and positive values of x.For x<0,-(x-6)=5 or -x+6=5 or x=1.hence not satisfied.
For x>0,(x-6)=5 or x=11.
from statement 1: only x=11 will satisfy the value.
from statement 2 also: x=11 will satisfy the value.
hence D.
can someone please help where am i going wrong?

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.

\(|x-6| = 5\)
\(⇔ x-6 = ±5\)
\(⇔ x = 6 ±5\)
\(⇔ x = 11\) or \(x = 1\)

Condition 1)
Since we have \(x = 11\) or \(x = 1\) from \(x > 0\), we don't have a unique solution.
Thus condition 1) is not sufficient.

Condition 2)
\(x^2 = 121\)
\(⇔ x = ± 11\)
Since we have a solution 11 only, condition 2) is sufficient.

Therefore, B is the answer.

C-trap problem

If |x – 6| = 5, what is the value of x?
x = 11,1

(1) x > 0
Insufficient b/c x = 11 or 1

(2) x^2 = 121

|x| = 11
x = 11, -11
Sufficient b/c x MUST be 11 to satisfy the mod above.

Answer is B.