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Re: If |x – 6| = 3, what is the value of x? [#permalink]

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06 Dec 2016, 04:13

I don't know whether this is even 600 level. Given that |x – 6| = 3. Removing modulus, We will get two equation, x-6=3 or x-6=-3. Solving the two equation we will two values of X. i.e. 9 or 3. Option 1: x is divisible by 9. Sufficient. Option 2: x is divisible by 3. Both the values 3 and 9 are divisible by 3. Not sufficient.

Re: If |x – 6| = 3, what is the value of x? [#permalink]

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09 Jul 2017, 00:13

so answer is A was my first thought.

|x-6| = 3

so obv x = 9 or - 3

1) x dvble 9 : so I substituted

|9-6| = 3; also |27/9 - 6| as both are dvble by 9. so X i thought could be either 27 or 9 . Insuff

2) x dvble by 3.

|x-6| =3 I put x as 9 and 18 --> |18/3 - 6| and |9/3 - 6| . so x again had two values

Answer is A. can you explain what is wrong above? Also, based on reading the explanations above, it appears we consider x as dvble by something when it leaves behind a remainder one (that is why 9/3 = 3 and 3/3 = 1 hence insufficient; while 9/9 = 1 was seen as sufficient. But why can't we consider 27/9 here?)Would appreciate any reply to this!

Re: If |x – 6| = 3, what is the value of x? [#permalink]

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09 Jul 2017, 00:18

1

This post received KUDOS

Madhavi1990 wrote:

so answer is A was my first thought.

|x-6| = 3

so obv x = 9 or - 3

1) x dvble 9 : so I substituted

|9-6| = 3; also |27/9 - 6| as both are dvble by 9. so X i thought could be either 27 or 9 . Insuff

2) x dvble by 3.

|x-6| =3 I put x as 9 and 18 --> |18/3 - 6| and |9/3 - 6| . so x again had two values

Answer is A. can you explain what is wrong above? Also, based on reading the explanations above, it appears we consider x as dvble by something when it leaves behind a remainder one (that is why 9/3 = 3 and 3/3 = 1 hence insufficient; while 9/9 = 1 was seen as sufficient. But why can't we consider 27/9 here?)Would appreciate any reply to this!

Solving Question stem gives:

|x-6| = 3 x-6=3 OR x-6=-3 x=9 OR x=3 So, possible value of X can be 9 or 3, any condition which makes one value as certain will be true.

1. Div by 9 means X=9 it can't be 3 - So, it is sufficient 2. Div by 3 means X=9 or X=3 - So, it is insufficient

|9-6| = 3; also |27/9 - 6| as both are dvble by 9. so X i thought could be either 27 or 9 . Insuff

2) x dvble by 3.

|x-6| =3 I put x as 9 and 18 --> |18/3 - 6| and |9/3 - 6| . so x again had two values

Answer is A. can you explain what is wrong above? Also, based on reading the explanations above, it appears we consider x as dvble by something when it leaves behind a remainder one (that is why 9/3 = 3 and 3/3 = 1 hence insufficient; while 9/9 = 1 was seen as sufficient. But why can't we consider 27/9 here?)Would appreciate any reply to this!

The logic there is not right.

If |x – 6| = 3, what is the value of x?

|x – 6| = 3 means that x is either 9 or 3. So, the question asks to find out which one of them is x.

(1) x is divisible by 9. x must be 9 because out of 9 and 3 only 9 is divisible by 9. Sufficient.

(2) x is divisible by 3. Both 9 and 3 are divisible by 3, so we cannot say whether x is 9 or 3. Not sufficient.