Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: Is x^2+y^2>100?? (1) 2xy<100 (2) (x+y)^2 [#permalink]
28 Oct 2012, 08:44

Yes there is something dicey. In the solution he has added (1) and (2) and proved that X^2+y^2>100. Request some expert to reply!
_________________

I've failed over and over and over again in my life and that is why I succeed--Michael Jordan Kudos drives a person to better himself every single time. So Pls give it generously Wont give up till i hit a 700+

Re: Is x^2 + y^2 > 100? [#permalink]
31 Oct 2012, 14:49

It's B for me.

(x + y)^2 = x^2 + 2xy + y^2

The largest possible value 2xy can reach is (x + y)^2/2, that only occurs when x = y.

When x = y, it turns out that x^2 + 2xy + y^2 > 200 is x^2 + 2xx + x^2 > 200, which can be rearranged in 2x^2 + 2x^2 > 200, meaning that x^2 + y^2 is at least half of the stated value, while 2xy can be at most the other half.

Any value above 200 will require that x^2 + y^2 > 100, making 2) sufficient.

Re: Is x^2 + y^2 > 100? [#permalink]
18 Feb 2013, 18:58

Bunuel wrote:

(2) (x + y)^2 > 200 --> x^2+2xy+y^2>200. Now, as (x-y)^2\geq{0} (square of any number is more than or equal to zero) then x^2+y^2\geq{2xy}

Hi Bunuel,

could you please explain why you followed with (x-y)^2 instead of (x+y)^2? Shouldn't (x-y)^2 be distributed as x^2-2xy+y^2 in which case it will be different from the original expression? Also, when you transfer 2xy to the other side from x^2+2xy+y^2, why do you keep it positive?

Re: Is x^2 + y^2 > 100? [#permalink]
19 Feb 2013, 04:43

Expert's post

LinaNY wrote:

Bunuel wrote:

(2) (x + y)^2 > 200 --> x^2+2xy+y^2>200. Now, as (x-y)^2\geq{0} (square of any number is more than or equal to zero) then x^2+y^2\geq{2xy}

Hi Bunuel,

could you please explain why you followed with (x-y)^2 instead of (x+y)^2? Shouldn't (x-y)^2 be distributed as x^2-2xy+y^2 in which case it will be different from the original expression? Also, when you transfer 2xy to the other side from x^2+2xy+y^2, why do you keep it positive?

We have (x + y)^2 > 200 which is the same as x^2+2xy+y^2>200.

Now, we need to find the relationship between x^2+y^2 and 2xy.

Next, we know that (x-y)^2\geq{0} --> x^2-2xy+y^2\geq{0} --> x^2+y^2\geq{2xy}.

So, we can safely substitute 2xy with x^2+y^2 in x^2+2xy+y^2>200 (as x^2+y^2 is at least as big as 2xy then the inequality will still hold true) --> x^2+(x^2+y^2)+y^2>200 --> 2(x^2+y^2)>200 --> x^2+y^2>100. Sufficient.

(1) 2xy < 100 --> clearly insufficient: if x=y=0 then the answer will be NO but if x=10 and y=-10 then the answer will be YES.

(2) (x + y)^2 > 200 --> x^2+2xy+y^2>200. Now, as (x-y)^2\geq{0} (square of any number is more than or equal to zero) then x^2+y^2\geq{2xy} so we can safely substitute 2xy with x^2+y^2 (as x^2+y^2 is at least as big as 2xy then the inequality will still hold true) --> x^2+(x^2+y^2)+y^2>200 --> 2(x^2+y^2)>200 --> x^2+y^2>100. Sufficient.

Answer: B.

Are you sure the OA is C?

Brunel I have one confusion,

As you said that x^2+y^2 is at least as big as 2xy and so 2(x^2 + y^2)>200 but if x^2 +y^2 is = 2xy then wont 2(x^2+y^2)<200 as 4xy<200. (2xy<100, First statement, two statements cannot contradice)

So till the time we are not sure how big is the difference between x^2+y^2 and 2xy can we really say if 2(x^2+y^2)>200

(1) 2xy < 100 --> clearly insufficient: if x=y=0 then the answer will be NO but if x=10 and y=-10 then the answer will be YES.

(2) (x + y)^2 > 200 --> x^2+2xy+y^2>200. Now, as (x-y)^2\geq{0} (square of any number is more than or equal to zero) then x^2+y^2\geq{2xy} so we can safely substitute 2xy with x^2+y^2 (as x^2+y^2 is at least as big as 2xy then the inequality will still hold true) --> x^2+(x^2+y^2)+y^2>200 --> 2(x^2+y^2)>200 --> x^2+y^2>100. Sufficient.

Answer: B.

Are you sure the OA is C?

Brunel I have one confusion,

As you said that x^2+y^2 is at least as big as 2xy and so 2(x^2 + y^2)>200 but if x^2 +y^2 is = 2xy then wont 2(x^2+y^2)<200 as 4xy<200. (2xy<100, First statement, two statements cannot contradice)

So till the time we are not sure how big is the difference between x^2+y^2 and 2xy can we really say if 2(x^2+y^2)>200

No. If x^2 +y^2 = 2xy, then we can simply substitute 2xy to get the same: x^2+(x^2+y^2)+y^2>200 --> 2(x^2+y^2)>200.

Re: Is x^2 + y^2 > 100? [#permalink]
31 Mar 2013, 11:34

If 2(x^2+y^2)>200 and (x^2+y^2)=2xy then 2(2xy)>200 and 2xy>100 but the statement one says that 2xy<100. So even though we are not considering 1st statement, the answer cannot contradict the first statement.

Re: Is x^2 + y^2 > 100? [#permalink]
31 Mar 2013, 11:44

Expert's post

anujtsingh wrote:

If 2(x^2+y^2)>200 and (x^2+y^2)=2xy then 2(2xy)>200 and 2xy>100 but the statement one says that 2xy<100. So even though we are not considering 1st statement, the answer cannot contradict the first statement.

Pls correct me if I am wrong

Frankly I don't understand your point. I just wanted to show that we CAN substitute 2xy with x^2+y^2 if x^2+y^2\geq{2xy}.
_________________

(1) 2xy < 100 --> clearly insufficient: if x=y=0 then the answer will be NO but if x=10 and y=-10 then the answer will be YES.

(2) (x + y)^2 > 200 --> x^2+2xy+y^2>200. Now, as (x-y)^2\geq{0} (square of any number is more than or equal to zero) then x^2+y^2\geq{2xy} so we can safely substitute 2xy with x^2+y^2 (as x^2+y^2 is at least as big as 2xy then the inequality will still hold true) --> x^2+(x^2+y^2)+y^2>200 --> 2(x^2+y^2)>200 --> x^2+y^2>100. Sufficient.

Answer: B.

Are you sure the OA is C?

Hello Bunuel, I guess the logic you ve used in the second statement justification is slightly flawed. As you ve stated 2xy <= x^2 + y^2 ; now the max value 2xy can take is x^2 + y^2 ; and we cannot substitue this in x^2 + y^2 +2xy > 200 , as it would mean we are putting the max value of LHS element to check greater than condition which will not suffice, it would be enough if it were a less than condition. So I guess the OA is correct i.e. C ...

(1) 2xy < 100 --> clearly insufficient: if x=y=0 then the answer will be NO but if x=10 and y=-10 then the answer will be YES.

(2) (x + y)^2 > 200 --> x^2+2xy+y^2>200. Now, as (x-y)^2\geq{0} (square of any number is more than or equal to zero) then x^2+y^2\geq{2xy} so we can safely substitute 2xy with x^2+y^2 (as x^2+y^2 is at least as big as 2xy then the inequality will still hold true) --> x^2+(x^2+y^2)+y^2>200 --> 2(x^2+y^2)>200 --> x^2+y^2>100. Sufficient.

Answer: B.

Are you sure the OA is C?

Hello Bunuel, I guess the logic you ve used in the second statement justification is slightly flawed. As you ve stated 2xy <= x^2 + y^2 ; now the max value 2xy can take is x^2 + y^2 ; and we cannot substitue this in x^2 + y^2 +2xy > 200 , as it would mean we are putting the max value of LHS element to check greater than condition which will not suffice, it would be enough if it were a less than condition. So I guess the OA is correct i.e. C ...

There is no flaw in my reasoning. Statement (2) says: x^2+2xy+y^2>200. Next, we know that x^2+y^2\geq{2xy} is true for any values of x and y. So we can manipulate and substitute 2xy with x^2+y^2 in (2) (because x^2+y^2 is at least as large as 2xy): x^2+(x^2+y^2)+y^2>200 --> x^2+y^2>100.

(1) 2xy < 100 --> clearly insufficient: if x=y=0 then the answer will be NO but if x=10 and y=-10 then the answer will be YES.

(2) (x + y)^2 > 200 --> x^2+2xy+y^2>200. Now, as (x-y)^2\geq{0} (square of any number is more than or equal to zero) then x^2+y^2\geq{2xy} so we can safely substitute 2xy with x^2+y^2 (as x^2+y^2 is at least as big as 2xy then the inequality will still hold true) --> x^2+(x^2+y^2)+y^2>200 --> 2(x^2+y^2)>200 --> x^2+y^2>100. Sufficient.

Answer: B.

Are you sure the OA is C?

Hello Bunuel, I guess the logic you ve used in the second statement justification is slightly flawed. As you ve stated 2xy <= x^2 + y^2 ; now the max value 2xy can take is x^2 + y^2 ; and we cannot substitue this in x^2 + y^2 +2xy > 200 , as it would mean we are putting the max value of LHS element to check greater than condition which will not suffice, it would be enough if it were a less than condition. So I guess the OA is correct i.e. C ...

Now you know that 2xy is less than or equal to x^2 + y^2. If 2xy is equal to (x^2 + y^2), (x^2 + y^2) will be greater than 100 since the total sum is greater than 200. If 2xy is less than (x^2 + y^2), then anyway (x^2 + y^2) will be greater than 100 (which is half of 200).

So statement 2 is sufficient alone.
_________________

Re: Is x^2 + y^2 > 100? [#permalink]
16 Jan 2014, 03:04

Bunuel wrote:

joe26219 wrote:

Bunuel wrote:

Is x^2 + y^2 > 100?

(1) 2xy < 100 --> clearly insufficient: if x=y=0 then the answer will be NO but if x=10 and y=-10 then the answer will be YES.

(2) (x + y)^2 > 200 --> x^2+2xy+y^2>200. Now, as (x-y)^2\geq{0} (square of any number is more than or equal to zero) then x^2+y^2\geq{2xy} so we can safely substitute 2xy with x^2+y^2 (as x^2+y^2 is at least as big as 2xy then the inequality will still hold true) --> x^2+(x^2+y^2)+y^2>200 --> 2(x^2+y^2)>200 --> x^2+y^2>100. Sufficient.

Answer: B.

Are you sure the OA is C?

Hello Bunuel, I guess the logic you ve used in the second statement justification is slightly flawed. As you ve stated 2xy <= x^2 + y^2 ; now the max value 2xy can take is x^2 + y^2 ; and we cannot substitue this in x^2 + y^2 +2xy > 200 , as it would mean we are putting the max value of LHS element to check greater than condition which will not suffice, it would be enough if it were a less than condition. So I guess the OA is correct i.e. C ...

There is no flaw in my reasoning. Statement (2) says: x^2+2xy+y^2>200. Next, we know that x^2+y^2\geq{2xy} is true for any values of x and y. So we can manipulate and substitute 2xy with x^2+y^2 in (2) (becausex^2+y^2 is at least as large as 2xy): x^2+(x^2+y^2)+y^2>200 --> x^2+y^2>100.

Thank you Bunuel.Got it.Your explanation was really helpful (the highlighted line esp.) I guess any value of 2xy should be assumed to satisfy the given condition, as it is given.

gmatclubot

Re: Is x^2 + y^2 > 100?
[#permalink]
16 Jan 2014, 03:04