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What is x?

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Re: What is x?  [#permalink]

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New post 31 Aug 2016, 13:26
Donnie84 wrote:
St1: |x| < 2

This implies that -2 < x < 2.

St2: |x| = 3x – 2

Using one of Bunuel's legendary tricks:

Since absolute value is always >= 0, the expression on right hand side will be >= 0.

3x - 2 >= 0
x >= 2/3, meaning x is positive.

So, we can only check for one condition.

If x > 0

x = 3x - 2
2x = 2
x = 1 -> satisfies the condition that x is positive.

Answer (B).

The trick is legendary because it saves us from checking the condition x is negative. Let's see what will happen if we do check:

If x < 0

x = 2 - 3x
4x = 2
x = 1/2 -> does not satisfy the condition that x is negative.

As you can see, the trick saves time and can make a big difference in the actual exam.

Sorry to be critical but the trick IS NOT saving any time per say. If you follow the usual method, you have to check for 2 conditions - When |x| = +ve & when |x| = -ve. While in this method, we are first checking the range of values that x can take and then based upon it, we are calculating only when |x| = +ve. The number of check points is same in both the cases for This Particular question, but it may not be the case for any other question. We may get the range of x lying in both the quadrants in which case, Bunuel's method would ultimately lead to a longer solution.

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Re: What is x?  [#permalink]

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New post 01 Sep 2017, 19:49
rohitgoel15 wrote:
What is x?

(1) |x| < 2
(2) |x| = 3x – 2

This question may be a C trap

Stmnt 1

X could be -1,0,1

Stmnt 2

x= 3x- 2
x +2 = 3x
2= 2x
x =1

-x= 3x-2
-x +2 =3x
2= 4x
x =1/2

1/2 does not satisfy the equation so x must be 1

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Re: What is x?  [#permalink]

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New post 03 Sep 2017, 16:35
rohitgoel15 wrote:
What is x?

(1) |x| < 2
(2) |x| = 3x – 2

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.


In the original condition, there is 1 variable \(x\) and 0 equation. So you need 1 equation.

Condition 1)
Since it is an inequality, there is no equation. Thus we can't identify the variable x and this is not sufficient.

Condition 2)
i) \(x \ge 0\)
\(|x| = 3x - 2\) is equivalent to \(x = 3x - 2\) or \(2x - 2 = 0\).
Thus we have \(x = 1\).
This is sufficient.

ii) \(x < 0\)
\(|x| = 3x - 2\) is equivalent to \(-x = 3x - 2\) or \(4x - 2 = 0\).
Thus we have \(x = 1/2\). However \(x = 1/2 > 0\).
There is no negative solution.

Therefore, we have a unique solution \(x = 1\).
This condition is sufficient.

The answer is B.

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.

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Re: What is x?  [#permalink]

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New post 10 Nov 2017, 20:29
rohitgoel15 wrote:
What is x?

(1) |x| < 2
(2) |x| = 3x – 2

No where it is given that X is an integer. Hence, x can be any number between 0 and 2 by statement1.
From statement 2, we know that x is either 1 or 1/2. If 1/2, then the overall equation becomes negative and modulus can't be negative.

Answer is B
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Re: What is x?  [#permalink]

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New post 10 Nov 2017, 22:14

A. |x|<2 meant

-2<x<2 leading to x= 0,1,-1,-0.5,-1.5

B. |x|=3x-2

|x| is positive.

So, unique value.

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Re: What is x?  [#permalink]

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New post 13 Nov 2018, 09:47
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Re: What is x? &nbs [#permalink] 13 Nov 2018, 09:47

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