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by Deepthi Subbu » Mon Jan 10, 2011 10:29 pm
If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4

(2) (x - 2)^2 > 9

OAD

I understand that this is a simple problem , but I am missing something somewhere .
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Source: — Data Sufficiency |

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by Anurag@Gurome » Mon Jan 10, 2011 10:46 pm
Deepthi Subbu wrote:If x is positive, is x > 3 ?

(1) (x - 1)² > 4
(2) (x - 2)² > 9
Statement 1: (x - 1)² > 4
This implies |x - 1| > 2
=> (x - 1) > 2 ........ OR ........ (x - 1) < -2
=> x > 3 ................ OR ........ x < -1

As x is positive, it is not possible that x < -1.
Hence, x > 3

Sufficient

Statement 1: (x - 2)² > 9
This implies |x - 2| > 3
=> (x - 2) > 3 ........ OR ........ (x - 2) < -3
=> x > 5 ................ OR ........ x < -1

As x is positive, it is not possible that x < -1.
Hence, x > 5, which implies x is obviously greater than 3

Sufficient

The correct answer is D.
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by gmatmachoman » Tue Jan 11, 2011 2:18 am
Deepthi Subbu wrote:If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4

(2) (x - 2)^2 > 9

OAD

I understand that this is a simple problem , but I am missing something somewhere .
Trick would have been if the statement doesn't say X is positive!!

Then pick E
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by MAAJ » Tue Jan 11, 2011 1:53 pm
To solve this, you must know 2 things:
  • The square root of a number which is squared equals the absolute value of that number. This is: sqrt(x²) = |x|
    Absolute values (most of them) have two solutions, one positive and one negative [sometimes the negative solution is not valid]


If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4
  • Note that (x-1)² is our squared number/expression
    You can take the square root on both side, because both are perfect squares
    After taking the square root the expression will turn into: |x-1| > 2
    We have to solve for (x-1)>2 and for -(x-1)>2
    So x > 3 OR x < -1
    Because x must be positive, then x > 3
    Statement 1 is Sufficient
(2) (x - 2)^2 > 9
  • Note that (x-2)² is our squared number/expression
    You can take the square root on both side, because both are perfect squares
    After taking the square root the expression will turn into: |x-2| > 3
    We have to solve for (x-2)>3 and for -(x-2)>3
    So x > 5 OR x < -1
    Because x must be positive, then x > 5 (Is x > 3? YES, because its always greater then 5)
    Statement 2 is Sufficient
Correct Answer is [spoiler]D[/spoiler]
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