IMO E
It has to be E....BTW what is the OA???
GMAT prep: If x is positive, is x>3?
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Source: Beat The GMAT — Problem Solving |
IMO : D (Each statement alone is sufficient)
(1) (x-1)^2>4
x^2-2x+1>4 --> Eqn.1
x^2-2x-3>0
x=-1,3
substitute values from the number line in the intervals (-∞,-1), (-1,3)and (3,+∞)
take -2 in (-∞,-1) and substitute in Eqn.1
9>4
so true
take 0 in (-1,3)
1>4 not true
take 4 in the interval (3,+∞)
9>4
true
so x < -1 and x> 3
it is enough to check in this (3,+∞) interval for this problem
so stmt 1 is sufficient
(2) (x-2)^2>9
x^2-4x+4>9--> Eqn. 2
x^2-4x-5>0
x=-1 or 5
take 6 in the interval (3,+∞) and substiute in Eqn.2
16>9
so x> 5 true
stmt 2 is also sufficient
so answer is D ( EACH statement ALONE is sufficient)
Let me know if i committed any mistakes.
(1) (x-1)^2>4
x^2-2x+1>4 --> Eqn.1
x^2-2x-3>0
x=-1,3
substitute values from the number line in the intervals (-∞,-1), (-1,3)and (3,+∞)
take -2 in (-∞,-1) and substitute in Eqn.1
9>4
so true
take 0 in (-1,3)
1>4 not true
take 4 in the interval (3,+∞)
9>4
true
so x < -1 and x> 3
it is enough to check in this (3,+∞) interval for this problem
so stmt 1 is sufficient
(2) (x-2)^2>9
x^2-4x+4>9--> Eqn. 2
x^2-4x-5>0
x=-1 or 5
take 6 in the interval (3,+∞) and substiute in Eqn.2
16>9
so x> 5 true
stmt 2 is also sufficient
so answer is D ( EACH statement ALONE is sufficient)
Let me know if i committed any mistakes.
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bumblenbumble01
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a quicker way to solve would be:
1) (x-1)² > 4 --> (x-1) > 2 or (x-1) < -2
=> x > 3 or x < -1. From the question stem we see that x > 0 ("if x is positive"), so the only solution is x > 3
A sufficient
2) (x-2)² > 9 --> (x-2) > 3 or (x-2) < - 3
=> x> 5 or x < -1. From the question stem we see that x > 0 ("if x is positive") x > 0, so the only solution is x > 5
B sufficient
1) (x-1)² > 4 --> (x-1) > 2 or (x-1) < -2
=> x > 3 or x < -1. From the question stem we see that x > 0 ("if x is positive"), so the only solution is x > 3
A sufficient
2) (x-2)² > 9 --> (x-2) > 3 or (x-2) < - 3
=> x> 5 or x < -1. From the question stem we see that x > 0 ("if x is positive") x > 0, so the only solution is x > 5
B sufficient
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missrochelle
- Master | Next Rank: 500 Posts
- Posts: 117
- Joined: Wed Jun 09, 2010 7:02 am
Can someone tell me the general rule for when you are supposed to "factor" the left side of an inequality versus when you can just take the square root of both sides, and then double check the positive and negative case?
I think the factoring , then checking regions method is wayyy to long and tedious. So I'm wondering when, if ever, it MUST come into play?
My guess is that if its a perfect square on the right side of the equation - you can take the root! Am I right?
I think the factoring , then checking regions method is wayyy to long and tedious. So I'm wondering when, if ever, it MUST come into play?
My guess is that if its a perfect square on the right side of the equation - you can take the root! Am I right?
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missrochelle
- Master | Next Rank: 500 Posts
- Posts: 117
- Joined: Wed Jun 09, 2010 7:02 am
Can someone tell me the general rule for when you are supposed to "factor" the left side of an inequality versus when you can just take the square root of both sides, and then double check the positive and negative case?
I think the factoring , then checking regions method is wayyy to long and tedious. So I'm wondering when, if ever, it MUST come into play?
My guess is that if its a perfect square on the right side of the equation - you can take the root! Am I right?
I think the factoring , then checking regions method is wayyy to long and tedious. So I'm wondering when, if ever, it MUST come into play?
My guess is that if its a perfect square on the right side of the equation - you can take the root! Am I right?
