If X is positive, is x > 3 ?

[vinversa]

This one is from GMAT Prep software
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If x is positive, is x > 3 ?
(1) (x-1)^2 > 4
(2) (x-2)^2 > 9

GMAT/GMAC correct answer is D - Each statement alone is sufficient.
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Manhattan GMAT - please help us understand why the solution is D...instead of E.

I though solution would be E because
(1)
(x-1)^2 > 4
(x-1) > +/- 2 (no definite answer here)

(2)
(x-2)^2 > 9
(x-2) > +/- 3 (no definite answer here)

Hence my solution = E
GMAT solution = D D D D (Each statement alone is sufficient)

[rash.patil]

stmt 1:
(x-1)^2 > 4
=> x^2 - 2x + 1 > 4
=> x(x - 2) > 3
Since it is given that x is a +ve integer, we can conclude that, to satisfy the above inequality, minimum value of x should be 4.

Hence 1 is sufficient.

stmt 2
again this is same as stmt 1.
(x-2)^2 > 9
=> x(x-4) > 5
To satisfy this inequality, minimum value of x is 6.
So stmt 2 is also sufficient.

Ans : D

[vinversa]

Thank you for your response

Please suggest if I m understanding this correctly...

stmt 1:
(x-1)^2 > 4
=> x^2 - 2x + 1 > 4
=> x(x - 2) > 3 --- equation 1
Since it is given that x is a +ve integer, we can conclude that, to satisfy the above inequality, minimum value of x should be 4.

Why is Min value of X = 4 for X a +ve int
IS it because - you are considering the equation 1 in two parts

X>3 &&
(X-2) > 3

For X>3, X has to be 4, 5, 6 to satisfy equation
For (x-2) > 3, X has to be 6, 7, 8, etc to satisfy equation...

[rash.patil]

x(x - 2) > 3 doesn't mean
x > 3
and (x-2) > 3.

Here i am considering left side of equation 1 as product of 2 numbers .
Minimum value greater than 3 is 4. also x(x-2) is positive.
so if we have x = 1 then x(x-2) is -1
if x = 2, x(x-2) = 0
if x=3, x(x-2) = 3.
In all these cases the ans is not greater than 3.
so in order to satisfy the given statement (x-1)^2 > 4 ( which i have rephrased as x(x-2) > 3 )
minimum value of x is 4.

Same logic applies for stmt 2 also.
Hope this helps :)

[alwyntp]

x is positive, is x > 3 ?
(1) (x-1)^2 > 4
(2) (x-2)^2 > 9

Stem 1 -
X > 0 (since x is positive)
X - 1 > -1
so x -1 is not negative.
Therefore we can safely take square root on both the sides.
x - 1 > 2
x > 3
sufficient.

Stem 2
X > 0 (since x is positive)
X-2 > -2
if we take values of x -2 = -1 or x -2 = 0, the condition in stem2 is not met. that means x -2 is positive.
square root on both the sides. x -2 > 3. x > 5
sufficient.

[RonPurewal]

to solve this problem, first make sure that you have a basic understanding of ABSOLUTE VALUE INEQUALITIES and SQUARE INEQUALITIES. (i've grouped the two of these together, since their solutions are so similar.)

make sure that you can produce the following solutions from memory, without having to do any algebra. (these are just examples -- they work for other numbers besides just "4" and "16". i just didn't want to write them with a lot of other variables.)


#1
if you have |quantity| < 4
--> -4 < quantity < 4.
#2
if you have quantity^2 < 16
--> -4 < quantity < 4.

#3
if you have |quantity| > 4
--> quantity > 4 OR quantity < -4.
#4
if you have quantity^2 > 16
--> quantity > 4 OR quantity < -4.

the statements here are both examples of type number 4.

therefore, you can solve them immediately:

statement one
(x - 1)^2 > 4
therefore
x - 1 > 2 or x - 1 < -2
x > 3 or x < -1
we can disregard the second of these, since we know that x is not negative.
therefore, x > 3.
sufficient.

statement two
(x - 2)^2 > 9
therefore
x - 2 > 3 or x - 2 < -3
x > 5 or x < -1
we can disregard the second of these, since we know that x is not negative.
therefore, x > 5.
sufficient.

[I_need_a_700plus]

How do we know that x in an integer in this problem?

[mondegreen]

How do we know that x in an integer in this problem?

Why do you need to know whether x is an integer or not?

[RonPurewal]

x doesn't have to be an integer.

[rustom.hakimiyan]

to solve this problem, first make sure that you have a basic understanding of ABSOLUTE VALUE INEQUALITIES and SQUARE INEQUALITIES. (i've grouped the two of these together, since their solutions are so similar.)

make sure that you can produce the following solutions from memory, without having to do any algebra. (these are just examples -- they work for other numbers besides just "4" and "16". i just didn't want to write them with a lot of other variables.)


#1
if you have |quantity| < 4
--> -4 < quantity < 4.
#2
if you have quantity^2 < 16
--> -4 < quantity < 4.

#3
if you have |quantity| > 4
--> quantity > 4 OR quantity < -4.
#4
if you have quantity^2 > 16
--> quantity > 4 OR quantity < -4.

the statements here are both examples of type number 4.

therefore, you can solve them immediately:

statement one
(x - 1)^2 > 4
therefore
x - 1 > 2 or x - 1 < -2
x > 3 or x < -1
we can disregard the second of these, since we know that x is not negative.
therefore, x > 3.
sufficient.

statement two
(x - 2)^2 > 9
therefore
x - 2 > 3 or x - 2 < -3
x > 5 or x < -1
we can disregard the second of these, since we know that x is not negative.
therefore, x > 5.
sufficient.

Hi Ron,

I have noticed that in some instances, we need to FOIL out (x-2)^2 instead of taking the sqroot of both sides from the very beginning. In this case, it happens to come out to the same values of x>5 and x<-1, but why is it that sometimes, we are required to go through the process of (x-2)(x-2)>9 etc...?

Thanks.

[RonPurewal]

That should work, too.

(x - 2)(x - 2) > 9
x^2 - 4x + 4 > 9
x^2 - 4x - 5 > 0
(x - 5)(x + 1) > 0

Now you'll have to split the number line into the part to the left of -1, the part between -1 and 5, and the part to the right of 5 (as usual for this type of inequality). You'll get the same answer.

If you're doing valid algebra, then it should work.

[ShriramC110]

Hi Ron,

can you please tell what should be the answer for this ""left of -1, the part between -1 and 5, and the part to the right of 5""

What will be the final inequalities for this??

Will it be x > 5 or x < -1.

Thanks

[tim]

You've offered a false dichotomy. Both x>5 and x<-1 work in Ron's example. The part that doesn't work is the portion between -1 and 5. Test numbers in each interval as Ron implied and you'll arrive at that conclusion.