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# If x is positive, is x > 3

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If x is positive, is x > 3  [#permalink]

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Updated on: 14 Oct 2015, 16:14
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Question Stats:

61% (00:47) correct 39% (00:53) wrong based on 742 sessions

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If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4
(2) (x - 2)^2 > 9

Can someone point a mistake in my method?
(1)
Taking one of the equations:
(x - 1)^2 > 4
x^2 + 1 - 2x > 4
x^2 + 1 - 2x - 4 > 0
x^2 - 3x + 1x - 3 > 0
(x-3) (x+1) > 0
x > 3 and x > -1

while in the explanation given the ans is coming out to be x > 3 and x < -1

Originally posted by rohitgoel15 on 16 Apr 2012, 09:42.
Last edited by ENGRTOMBA2018 on 14 Oct 2015, 16:14, edited 1 time in total.
Formatted the question
Math Expert
Joined: 02 Sep 2009
Posts: 47923
If x is positive, is x > 3  [#permalink]

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16 Apr 2012, 11:58
9
15
rohitgoel15 wrote:
If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4
(2) (x - 2)^2 > 9

Can someone point a mistake in my method?
(1)
Taking one of the equations:
(x - 1)^2 > 4
x^2 + 1 - 2x > 4
x^2 + 1 - 2x - 4 > 0
x^2 - 3x + 1x - 3 > 0
(x-3) (x+1) > 0
x > 3 and x > -1

while in the explanation given the ans is coming out to be x > 3 and x < -1

If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4 --> $$(x+1)(x-3)>0$$ --> roots are -1 and 3. Now, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: $$x<-1$$ or $$x>3$$. Since given that $$x$$ is positive then only one range is valid: $$x>3$$. Sufficient.

(2) (x - 2)^2 > 9 --> $$(x+1)(x-5)>0$$ --> roots are -1 and 5. Again, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: $$x<-1$$ or $$x>5$$. Since given that $$x$$ is positive then only one range is valid: $$x>5$$. Sufficient.

Solving inequalities:
http://gmatclub.com/forum/x2-4x-94661.html#p731476 (check this one first)
http://gmatclub.com/forum/inequalities-trick-91482.html
http://gmatclub.com/forum/data-suff-ine ... 09078.html
http://gmatclub.com/forum/range-for-var ... me#p873535
http://gmatclub.com/forum/everything-is ... me#p868863

Another approach:

If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4 --> since both sides of the inequality are non-negative then we can take square root from both parts: $$|x-1|>2$$. $$|x-1|$$ is just the distance between 1 and $$x$$ on the number line. We are told that this distance is more than 2: --(-1)----1----3-- so, $$x<-1$$ or $$x>3$$. Since given that $$x$$ is positive then only one range is valid: $$x>3$$. Sufficient.

(2) (x - 2)^2 > 9 --> $$|x-2|>3$$. The same here: $$|x-2|$$ is just the distance between 2 and $$x$$ on the number line. We are told that this distance is more than 3: --(-1)----2----5-- so, $$x<-1$$ or $$x>5$$. Since given that $$x$$ is positive then only one range is valid: $$x>5$$. Sufficient.

Hope it helps.
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Re: If x is positive, is x > 3  [#permalink]

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20 Jul 2013, 06:57
Hi bunuel,

Just wanted to clarify in the alternative approach you mentioned "non-negative" so if the other side of the inequality has a negative number, the only way to proceed with the problem is by expansion?
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Posts: 47923
Re: If x is positive, is x > 3  [#permalink]

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20 Jul 2013, 07:32
1
6
fozzzy wrote:
Hi bunuel,

Just wanted to clarify in the alternative approach you mentioned "non-negative" so if the other side of the inequality has a negative number, the only way to proceed with the problem is by expansion?

If it were (x - 1)^2 > -4, it would simply mean that x can take any value.

As for general rules for inequalities: taking the square root, squaring, ...

You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

RAISING INEQUALITIES TO EVEN/ODD POWER:

A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
$$2<4$$ --> we can square both sides and write: $$2^2<4^2$$;
$$0\leq{x}<{y}$$ --> we can square both sides and write: $$x^2<y^2$$;

But if either of side is negative then raising to even power doesn't always work.
For example: $$1>-2$$ if we square we'll get $$1>4$$ which is not right. So if given that $$x>y$$ then we can not square both sides and write $$x^2>y^2$$ if we are not certain that both $$x$$ and $$y$$ are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
$$-2<-1$$ --> we can raise both sides to third power and write: $$-2^3=-8<-1=-1^3$$ or $$-5<1$$ --> $$-5^2=-125<1=1^3$$;
$$x<y$$ --> we can raise both sides to third power and write: $$x^3<y^3$$.

For multiplication check here: help-with-add-subtract-mult-divid-multiple-inequalities-155290.html#p1242652

THEORY ON INEQUALITIES:

x2-4x-94661.html#p731476
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html
everything-is-less-than-zero-108884.html
graphic-approach-to-problems-with-inequalities-68037.html
inequations-inequalities-part-154664.html
inequations-inequalities-part-154738.html

QUESTIONS:

All DS Inequalities Problems to practice: search.php?search_id=tag&tag_id=184
All PS Inequalities Problems to practice: search.php?search_id=tag&tag_id=189

700+ Inequalities problems: inequality-and-absolute-value-questions-from-my-collection-86939.html

Hope it helps.
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Joined: 08 Dec 2012
Posts: 40
Re: If x is positive, is x > 3  [#permalink]

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05 Sep 2013, 03:47
Bunuel wrote:
rohitgoel15 wrote:
If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4
(2) (x - 2)^2 > 9

Can someone point a mistake in my method?
(1)
Taking one of the equations:
(x - 1)^2 > 4
x^2 + 1 - 2x > 4
x^2 + 1 - 2x - 4 > 0
x^2 - 3x + 1x - 3 > 0
(x-3) (x+1) > 0
x > 3 and x > -1

while in the explanation given the ans is coming out to be x > 3 and x < -1

If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4 --> $$(x+1)(x-3)>0$$ --> roots are -1 and 3. Now, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: $$x<-1$$ or $$x>3$$. Since given that $$x$$ is positive then only one range is valid: $$x>3$$. Sufficient.

(2) (x - 2)^2 > 9 --> $$(x+1)(x-5)>0$$ --> roots are -1 and 5. Again, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: $$x<-1$$ or $$x>5$$. Since given that $$x$$ is positive then only one range is valid: $$x>5$$. Sufficient.

Solving inequalities:
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.

Another approach:

If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4 --> since both sides of the inequality are non-negative then we can take square root from both parts: $$|x-1|>2$$. $$|x-1|$$ is just the distance between 1 and $$x$$ on the number line. We are told that this distance is more than 2: --(-1)----1----3-- so, $$x<-1$$ or $$x>3$$. Since given that $$x$$ is positive then only one range is valid: $$x>3$$. Sufficient.

(2) (x - 2)^2 > 9 --> $$|x-2|>3$$. The same here: $$|x-2|$$ is just the distance between 2 and $$x$$ on the number line. We are told that this distance is more than 3: --(-1)----2----5-- so, $$x<-1$$ or $$x>5$$. Since given that $$x$$ is positive then only one range is valid: $$x>5$$. Sufficient.

I used x= +6 and -6 ..Which is true in both the cases..it shud be E..
Math Expert
Joined: 02 Sep 2009
Posts: 47923
Re: If x is positive, is x > 3  [#permalink]

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05 Sep 2013, 03:52
SUNGMAT710 wrote:
Bunuel wrote:
rohitgoel15 wrote:
If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4
(2) (x - 2)^2 > 9

Can someone point a mistake in my method?
(1)
Taking one of the equations:
(x - 1)^2 > 4
x^2 + 1 - 2x > 4
x^2 + 1 - 2x - 4 > 0
x^2 - 3x + 1x - 3 > 0
(x-3) (x+1) > 0
x > 3 and x > -1

while in the explanation given the ans is coming out to be x > 3 and x < -1

If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4 --> $$(x+1)(x-3)>0$$ --> roots are -1 and 3. Now, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: $$x<-1$$ or $$x>3$$. Since given that $$x$$ is positive then only one range is valid: $$x>3$$. Sufficient.

(2) (x - 2)^2 > 9 --> $$(x+1)(x-5)>0$$ --> roots are -1 and 5. Again, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: $$x<-1$$ or $$x>5$$. Since given that $$x$$ is positive then only one range is valid: $$x>5$$. Sufficient.

Solving inequalities:
x2-4x-94661.html#p731476 (check this one first)
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.

Another approach:

If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4 --> since both sides of the inequality are non-negative then we can take square root from both parts: $$|x-1|>2$$. $$|x-1|$$ is just the distance between 1 and $$x$$ on the number line. We are told that this distance is more than 2: --(-1)----1----3-- so, $$x<-1$$ or $$x>3$$. Since given that $$x$$ is positive then only one range is valid: $$x>3$$. Sufficient.

(2) (x - 2)^2 > 9 --> $$|x-2|>3$$. The same here: $$|x-2|$$ is just the distance between 2 and $$x$$ on the number line. We are told that this distance is more than 3: --(-1)----2----5-- so, $$x<-1$$ or $$x>5$$. Since given that $$x$$ is positive then only one range is valid: $$x>5$$. Sufficient.

I used x= +6 and -6 ..Which is true in both the cases..it shud be E..

Stem says that x is a positive number, thus x cannot be -6.

Hope it's clear.
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Posts: 813
Re: If x is positive, is x > 3  [#permalink]

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27 Oct 2013, 22:43
Bunuel wrote:

If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4 --> $$(x+1)(x-3)>0$$ --> roots are -1 and 3. Now, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: $$x<-1$$ or $$x>3$$. Since given that $$x$$ is positive then only one range is valid: $$x>3$$. Sufficient.

(2) (x - 2)^2 > 9 --> $$(x+1)(x-5)>0$$ --> roots are -1 and 5. Again, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: $$x<-1$$ or $$x>5$$. Since given that $$x$$ is positive then only one range is valid: $$x>5$$. Sufficient.

I have a question if the sign was "<" instead of ">" what would the solution be? (x+1)(x-3) < 0
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If x is positive, is x > 3  [#permalink]

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27 Oct 2013, 22:57
fozzzy wrote:
Bunuel wrote:

If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4 --> $$(x+1)(x-3)>0$$ --> roots are -1 and 3. Now, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: $$x<-1$$ or $$x>3$$. Since given that $$x$$ is positive then only one range is valid: $$x>3$$. Sufficient.

(2) (x - 2)^2 > 9 --> $$(x+1)(x-5)>0$$ --> roots are -1 and 5. Again, ">" sign indicates that the solution lies to the left of a smaller root and to the right of the larger root: $$x<-1$$ or $$x>5$$. Since given that $$x$$ is positive then only one range is valid: $$x>5$$. Sufficient.

I have a question if the sign was "<" instead of ">" what would the solution be? (x+1)(x-3) < 0

$$(x+1)(x-3)<0$$ --> $$-1<x<3$$.
$$(x+1)(x-5)<0$$ --> $$-1<x<5$$.
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Re: If x is positive, is x > 3  [#permalink]

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28 Dec 2015, 02:46
Key thing to note is that X is positive.

Statement 1 - If (x-1)^2 > 4, x has to be greater than 3. If x = 1, 2 or 3, then the expression will only be < or equal to 4. Hence 1 is sufficient.

Statement 2 - If (x - 2)^2 > 9, x has to be greater than 3. If x = 1, 2 or 3, then the expression will not be > 9. Hence 2 is also sufficient.

Bunuel - Is this really a 700 level question ?
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Re: If x is positive, is x > 3  [#permalink]

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06 Jul 2018, 00:41
rohitgoel15 wrote:
If x is positive, is x > 3 ?

(1) (x - 1)^2 > 4
(2) (x - 2)^2 > 9

Given x > 0, asked is x > 3?

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

Hence lx - 1l > 2, so we have x > 3 or x < -1

Since x > 0, there fore x > 3. Answer is YES.

Statement 1 alone is Sufficient.

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

Hence lx - 2l > 3, so we have x > 5 or x < -1

Since x > 0, therefore x > 5 & hence x > 3. Answer is YES.

Statement 2 alone is Sufficient.

Thanks,
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Re: If x is positive, is x > 3  [#permalink]

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22 Jul 2018, 06:46
Bunuel niks18 pushpitkc chetan2u KarishmaB amanvermagmat pikolo2510

I understood alternate approach by Bunuel since taking square root
of a squared variable and turning it in to absolute value came more naturally to me.

Where I faltered is doing this is less than 2 mins as per below steps:
|x-1| > 2
x-1 >2 or -x+1 > 2
x>3 or -x>1 ie x<-1

Similar painstaking calculations I had to do for St 2
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If x is positive, is x > 3  [#permalink]

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22 Jul 2018, 07:41

remember a small rule and this question can be done in 30 seconds.

See my approach.
Attachment:

IMG-20180722-WA0005.jpg [ 84.46 KiB | Viewed 847 times ]

Bunuel niks18 pushpitkc chetan2u KarishmaB amanvermagmat pikolo2510

I understood alternate approach by Bunuel since taking square root
of a squared variable and turning it in to absolute value came more naturally to me.

Where I faltered is doing this is less than 2 mins as per below steps:
|x-1| > 2
x-1 >2 or -x+1 > 2
x>3 or -x>1 ie x<-1

Similar painstaking calculations I had to do for St 2

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Joined: 27 Oct 2017
Posts: 673
Location: India
Concentration: International Business, General Management
GPA: 3.64
WE: Business Development (Energy and Utilities)
Re: If x is positive, is x > 3  [#permalink]

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22 Jul 2018, 08:24
Your approach is perfect .
Practice more of such questions so that the steps come like a reflex and save time.

Bunuel niks18 pushpitkc chetan2u KarishmaB amanvermagmat pikolo2510

I understood alternate approach by Bunuel since taking square root
of a squared variable and turning it in to absolute value came more naturally to me.

Where I faltered is doing this is less than 2 mins as per below steps:
|x-1| > 2
x-1 >2 or -x+1 > 2
x>3 or -x>1 ie x<-1

Similar painstaking calculations I had to do for St 2

_________________
Re: If x is positive, is x > 3 &nbs [#permalink] 22 Jul 2018, 08:24
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