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GMAT prep: If x is positive, is x>3?

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by abhinav85 » Wed Jun 03, 2009 4:45 am
IMO E

It has to be E....BTW what is the OA???
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by crejoc » Wed Jun 03, 2009 4:59 am
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,+&#8734;) 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,+&#8734;) 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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by bumblenbumble01 » Wed Jun 03, 2009 8:25 am
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
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by mike22629 » Wed Jun 03, 2009 10:45 am
IMO D.

Statement 1:)

(x-1)^2 >4 = lx-1l > 2

Since x must be positive, x>3

Same approach with second statment
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by nhai2003 » Wed Jun 03, 2009 4:52 pm
OA: D
Thanks guys!
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by missrochelle » Mon Aug 23, 2010 4:19 pm
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?
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by missrochelle » Mon Aug 23, 2010 4:20 pm
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?
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