This solution is more on the logical side than algebric. I am sure other s may have an algebric solution
x is positive x can be a positive fraction or a postive integer
Stmt I
(x-1) ^2>4 only possible if x>3 (cant be a positive fraction less than 3 or a pos intger less than 3)
SUFF
Stmt II
(x-2 ^2>9 only possible if x>3 (cant be a positive fraction less than 3 or a pos intger less than 3)
suff
D)
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GMAT PREP - Inequalities
Source: Beat The GMAT — Data Sufficiency |
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Toph@GMAT_REBOOT
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If X is positive, is X > 3.
Statement 1: (X-1)^2 > 4
Statement 2: (X-2)^2 > 9
The given information that X is positive is necessary to solve this problem correctly.
Statement 1:
(X-1)^2 > 4 = X-1 > 2 or X - 1 < -2, so x > 3 or X < -1. However we know that X has to be positive, so ignore the possibility that X < -1. So we know that X > 3. (Sufficient)
Statement 2:
(X-2)^2 > 9 = X-2 > 3 or X - 2 < -3, so X > 5 or X < -1. However, once again, we know that X has to be positive, so ignore the possibility that X < -1. So we know that X > 5 (and therefore X > 3). (Sufficient).
Both statements are sufficient, so the answer is D.
Statement 1: (X-1)^2 > 4
Statement 2: (X-2)^2 > 9
The given information that X is positive is necessary to solve this problem correctly.
Statement 1:
(X-1)^2 > 4 = X-1 > 2 or X - 1 < -2, so x > 3 or X < -1. However we know that X has to be positive, so ignore the possibility that X < -1. So we know that X > 3. (Sufficient)
Statement 2:
(X-2)^2 > 9 = X-2 > 3 or X - 2 < -3, so X > 5 or X < -1. However, once again, we know that X has to be positive, so ignore the possibility that X < -1. So we know that X > 5 (and therefore X > 3). (Sufficient).
Both statements are sufficient, so the answer is D.
Last edited by Toph@GMAT_REBOOT on Fri Nov 28, 2008 12:14 pm, edited 1 time in total.
RipRop,(X-1)^2 > 4 = X-1 > +/- 2,
I may be missing something here. Should this not be
-(x-1) >2
Then u get x<-1
In your case if x-1>-2
x>-1
Did I miss something here?
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Toph@GMAT_REBOOT
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Thanks, you're right. I was lazy with that part. The solutions were right but the way I displayed the formula was incocrect. I fixed it above for the OP.
Thanks RipRop!
This is my way for getting to the 2 solutions (explaining it just for FYI purposes)
(x-1) ^ 2 > 4
can be
(x-1) ^ 2 > 2 ^ 2 (case 1)
or (x-1) ^ 2 > (-2) ^ 2 case 2
Case 1:
(x-1) ^ 2 > 2 ^ 2
sqrt((x-1)^2) > sqrt (2^2)
|x-1| > |2| (definition of sqrt(x^2) = |x| |2| is 2)
x-1>2
x>3
-(x-1)>2
x<-1
case 2:
(x-1) ^ 2 > (-2) ^ 2
sqrt((x-1)^2) > sqrt ((-2)^2)
|x-1| > |-2| (definition of sqrt(x^2) = |x| |-2| is 2)
x-1>2
x>3
-(x-1)>2
x<-1
Comments/Critiques/Suggestions welcome!
This is my way for getting to the 2 solutions (explaining it just for FYI purposes)
(x-1) ^ 2 > 4
can be
(x-1) ^ 2 > 2 ^ 2 (case 1)
or (x-1) ^ 2 > (-2) ^ 2 case 2
Case 1:
(x-1) ^ 2 > 2 ^ 2
sqrt((x-1)^2) > sqrt (2^2)
|x-1| > |2| (definition of sqrt(x^2) = |x| |2| is 2)
x-1>2
x>3
-(x-1)>2
x<-1
case 2:
(x-1) ^ 2 > (-2) ^ 2
sqrt((x-1)^2) > sqrt ((-2)^2)
|x-1| > |-2| (definition of sqrt(x^2) = |x| |-2| is 2)
x-1>2
x>3
-(x-1)>2
x<-1
Comments/Critiques/Suggestions welcome!
Cramya,cramya wrote:Thanks RipRop!
This is my way for getting to the 2 solutions (explaining it just for FYI purposes)
(x-1) ^ 2 > 4
can be
(x-1) ^ 2 > 2 ^ 2 (case 1)
or (x-1) ^ 2 > (-2) ^ 2 case 2
Case 1:
(x-1) ^ 2 > 2 ^ 2
sqrt((x-1)^2) > sqrt (2^2)
|x-1| > |2| (definition of sqrt(x^2) = |x| |2| is 2)
x-1>2
x>3
-(x-1)>2
x<-1
case 2:
(x-1) ^ 2 > (-2) ^ 2
sqrt((x-1)^2) > sqrt ((-2)^2)
|x-1| > |-2| (definition of sqrt(x^2) = |x| |-2| is 2)
x-1>2
x>3
-(x-1)>2
x<-1
Comments/Critiques/Suggestions welcome!
would like to suggest a similar yet simple approach. May be you already know this.
given from 1
(x-1)^2>4
or |x-1| > 2
1 is the critical point whr x-1=0,
the distance 2 units greater from 1 on both sides.
it gives x<-1 or x>3 but we have x>0 so, x<-1 is ignored.
similarly
(x-2)^2 > 9
|x-2| > 3
2 is the critical point.
the distance is 3 units greater from 2 on both sides.
it gives x<-1 or x>5
again we can ignore x<-1 as x is positive.
Hence ans is D.
Suggestions welcome.
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