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Pls help : (x - 2)^2 > x^2

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Source: — Data Sufficiency |

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by Brent@GMATPrepNow » Thu Oct 02, 2014 8:12 am
himu wrote:Is (x - 2)² > x²?

(1) x² > x
(2) (1/x) > 0
Target question: Is (x - 2)² > x²?

This is a great candidate for rephrasing the target question.
Aside: We have a free video with tips on rephrasing the target question: https://www.gmatprepnow.com/module/gmat- ... cy?id=1100

Take (x - 2)² > x² and expand the left side to get: x² - 4x + 4 > x²
Subtract x² from both sides to get: -4x + 4 > 0
Add 4x to both sides to get: 4 > 4x
Divide both sides by 4 to get: 1 > x
So, we can REPHRASE our target question....
REPHRASED target question: Is x < 1? [Is x less than 1?]

Statement 1: x² > x
First, since x² > x, we can conclude that x ≠ 0
So, we know that x² is POSITIVE
So, let's divide both sides of the inequality by x² to get: 1 > 1/x
This means that EITHER x < 0 OR x > 1
So there are two possible cases to consider.
case a: If x < 0, then it IS the case that x < 1
case b: If x > 1, then it is NOT the case that x < 1
Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: (1/x) > 0
If 1 divided by x equals some POSITIVE value, we can conclude that x is POSITIVE
If x is POSITIVE, then x could be greater than 1, or x could be less than 1
Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that that EITHER x < 0 OR x > 1
Statement 2 tells us that x is POSITIVE
So, we can eliminate the possibility that x < 0
This means it MUST be the case that x > 1
So, we can conclude that x is NOT less than 1
Since we can answer the REPHRASED target question with certainty, the combined statements are SUFFICIENT

Answer = C

Cheers,
Brent

For even more information on rephrasing the target question, you can read this article I wrote for BTG: https://www.beatthegmat.com/mba/2014/06/ ... t-question
Last edited by Brent@GMATPrepNow on Sun Feb 08, 2015 9:59 am, edited 1 time in total.
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by [email protected] » Thu Oct 02, 2014 11:24 am
Hi himu,

This question is perfect for TESTing VALUES:

We're asked if (x - 2)^2 is greater than x^2? This is a YES/NO question.

Fact 1: X^2 > X

If....
X = 2, then (0)^2 is NOT greater than 2^2 and the answer to the question is NO.
X = -1, then (-3)^2 IS greater than (-1)^2 and the answer to the question is YES.
Fact 1 is INSUFFICIENT

Fact 2: (1/X) > 0

This means that X MUST be POSITIVE. We can use our first TEST from Fact 1...
If....
X = 2, then (0)^2 is NOT greater than 2^2 and the answer to the question is NO.
X = 1/2 then (-1.5)^2 IS greater than (1/2)^2 and the answer to the question is YES.
Fact 2 is INSUFFICIENT

Combined, we know....
X is POSITIVE (from Fact 2)
X > 1 (knowing that X is POSITIVE and combining that with the inequality in Fact 1).

Since X > 1, the value of (X-2)^2 will ALWAYS be LESS than X^2. You can prove it rather easily:

If...
X = 1.1... (.9)^2 vs. (1.1)^2
X = 2..... 0^2 vs. 2^2
X = 5.... 3^2 vs. 5^2
Etc.
The second term is ALWAYS bigger, so the answer to the question is ALWAYS NO.
Combined, SUFFICIENT

Final Answer: C

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by j_shreyans » Fri Feb 06, 2015 11:03 am
Hi,

i am confused in statement 1 .

Statement 1 says x^2>x

Can we cancel out one x from the above?

x>1 ( can we do that)

Please suggest me.

Thanks

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by DavidG@VeritasPrep » Fri Feb 06, 2015 11:09 am
You need to be very careful when dividing by a variable. First, we can only divide by a variable if we know that variable doesn't equal 0. In this case, we're okay, because x^2 couldn't be greater than x if x were 0.

However, we don't know if x is positive or negative. If it's negative, the sign will flip when we divide both sides by x. If it's positive, it won't.

So if x >0 then x > 1;

but if x < 0, the sign would flip and x < 1. (But, of course, if x < 0 , we already know it's less than 1.) So x > 1 or x < 0.
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by [email protected] » Fri Feb 06, 2015 11:12 am
Hi j_shreyans,

You have to be very careful about "dividing out" a variable, since it usually means that you're removing some of the possible solutions.

Here, we're dealing with X^2 > X

LOTS of different values will fit this inequality, including:
1) ANYTHING > 1
2) ANY Negative

By dividing both sides by X, you're left with X > 1. Notice how you've now lost ALL of the negative solutions? That's why you can't divide both sides by X here.

The GMAT is going to test you on this concept (possibly several times and almost certainly in DS), so you have to be careful with your arithmetic. You can add or subtract variables away, but you have to be careful about multiplying or dividing them away.

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by DavidG@VeritasPrep » Fri Feb 06, 2015 11:21 am
We can also think about it algebraically. If x^2 > x, we can subtract x from both sides:

x^2 - x > 0

Now factor out an x to get

x(x - 1) > 0

Anytime you have a positive product, both elements of the product must be +, or they both must be -

If both x and (x - 1) are positive, then x > 1.
If both x and (x - 1) are negative, then x < 0.
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by utkalnayak » Fri Feb 06, 2015 11:37 am
Shreyans,
In an inequality comparison we can not simply divide any variable without knowing whether it is positive or negative.
example (-5)^2 > -5 but dividing -5 both sides gives -5 > 1, which is not true. however X^2 is always positive, hence we can divide that on either side.

Regards,
Utkal

Never mind, just saw that the experts have already answered your query.
Thanks,
Utkal
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by j_shreyans » Fri Feb 06, 2015 11:09 pm
Hi All ,

Thanks for your reply.

One more thing need to be cleared.

Question is (x-2)^2>x^2

now i should rephrase the question.

x^2+4-4x>x^2

now subtract x^2 both side

so we left 4-4x^2>0

now add 4x both side

now we left with 4>4x

divide 4 both side

so the finally should be 1>x or x<1 right?


Please correct me if am wrong.

Thanks

Shreyans
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by [email protected] » Sat Feb 07, 2015 10:27 am
Hi j_shreyans,

Yes, all of your math "steps" in rewriting the question are correct.

At it's simplest, the question is asking "is 1 > X?"

Keep in mind that this is still the QUESTION, it's not a fact.

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by j_shreyans » Sat Feb 07, 2015 9:46 pm
Hi Rich ,

Thanks for your reply.

Target question is still x<1 right?

In Brent's solution after rephrasing the question the target question is x>1

Please advise experts.

Thanks,

Shreyans
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by GMATGuruNY » Sun Feb 08, 2015 4:29 am
j_shreyans wrote: so the finally should be 1>x or x<1 right?

In Brent's solution after rephrasing the question the target question is x>1
You are correct.
In his solution, Brent inadvertently flipped the inequality symbols in red:
Subtract x² from both sides to get: -4x + 4 > 0
Add 4x to both sides to get: 4 < 4x
Divide both sides by 4 to get: 1 < x
So, we can REPHRASE our target question....
REPHRASED target question: Is x > 1?
The last three steps should read:
Add 4x to both sides to get: 4 > 4x
Divide both sides by 4 to get: 1 > x .
REPHRASED target question: Is x < 1?
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by Brent@GMATPrepNow » Sun Feb 08, 2015 10:01 am
j_shreyans wrote:Hi Rich ,

Thanks for your reply.

Target question is still x<1 right?

In Brent's solution after rephrasing the question the target question is x>1

Please advise experts.

Thanks,

Shreyans
Hi Shreyans,

I didn't realize that I had incorrectly rephrased the target question as "Is x > 1?"
I've edited my response so that it's "Is x < 1?"
Sorry for the confusion.

Cheers,
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by Matt@VeritasPrep » Mon Feb 09, 2015 12:19 am
j_shreyans wrote:
so the finally should be 1>x or x<1 right?


It would look like this:

(x - 2)² > x²
x² - 4x + 4 > x²
-4x + 4 > 0
4 > 4x
1 > x

Since we limited ourselves to adding, subtracting, and dividing by a positive number, we don't have to flip the sign.

One other thing worth noting: if you know that x ≠ 0, you can safely divide by x², or x to any even power, since a nonzero x² is always positive.

For instance, if we have

x² > x³

We can divide the whole inequality by x², since it's positive, and get 1 > x (and x ≠ 0). Students often don't think of this, but it's useful on certain inequality problems.
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by j_shreyans » Mon Feb 09, 2015 8:48 am
Hi All ,

Thanks for your reply. Now all clear...

Thanks

shreyans
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