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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

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:

If x^2 + y^2 = 29, what is the value of (x-y)^2 (1) xy = 10 [#permalink]
28 Nov 2010, 12:10

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

70% (01:35) correct
30% (00:47) wrong based on 170 sessions

If x^2 + y^2 = 29, what is the value of (x-y)^2

(1) xy = 10 (2) x = 5

Hi guys, I'd like some help here. The answer is A. I see why (1) is sufficient, but I don't get why (2) is not. We can just simply plug x = 5 into the equation, find y, and then solve for (x-y)^2

Re: #73 OG Quant Review 2nd [#permalink]
28 Nov 2010, 12:25

2

This post received KUDOS

Expert's post

krit wrote:

If x^2 + y^2 = 29, what is the value of (x-y)^2

(1) xy = 10 (2) x = 5

Hi guys, I'd like some help here. The answer is A. I see why (1) is sufficient, but I don't get why (2) is not. We can just simply plug x = 5 into the equation, find y, and then solve for (x-y)^2

Thanks

Given: \(x^2+y^2=29\). Question: \((x-y)^2=x^2-2xy+y^2=(x^2+y^2)-2xy=29-2xy=?\) So, basically we should find the value of \(xy\)

(1) xy = 10. Directly gives us the value we needed. Sufficient.

(2) x = 5. Now even if we substitute the value of \(x\) in \(x^2+y^2=29\) we'll get two values for \(y\): 2 and -2, hence two values for \((x-y)^2\): 9 and 49. Not sufficient.

Re: Quant Review 2nd Edition DS 73 - Help [#permalink]
04 Dec 2010, 18:04

1

This post received KUDOS

Expert's post

jscott319 wrote:

Looking at this problem I was able to see that according to this equation, \(x^2 + y^2 = 29\), X or Y could either be +/-5 or +/-2.

Another thing, you can't assume that the equation will have only integral values. x = 3 and y = root(20) also satisfies this equation as do many other values. _________________

Re: #73 OG Quant Review 2nd [#permalink]
28 Nov 2010, 12:25

Stmt 1 works because you know the sign of both of x and y is the same either both negative or both positive - this affects the result of (x-y)^2 as -5-(2) =-7 or -5-(-2)=-3. With stmt 2 you don't know the sign of 2 hence can't arrive at one answer.

Re: #73 OG Quant Review 2nd [#permalink]
28 Nov 2010, 12:26

Yes you are right you CAN just plug x into the equation. BUT, you are forgetting that y can be POSITIVE or NEGATIVE.

For example: 2^2 = 4 ; -2^2 = 4 If you plug X = 5 into the equation that you are asked to find the value for, (x-y)^2 Then you will see, when you have 2 different values for Y you will get 2 different answers --> (5-2)^2 = 9 OR (5- (-2))^2 = 49

Therefore, with B, you have two possible answers and it is INSUFFICIENT

Quant Review 2nd Edition DS 73 - Help [#permalink]
04 Dec 2010, 17:21

If \(x^2 + y^2 = 29\), what is the value of \((x - y)^2\) ?

1) xy = 10 2) x = 5

I am unsure about my approach. It led me to the wrong answer but I do not know where it is flawed. Looking at this problem I was able to see that according to this equation, \(x^2 + y^2 = 29\), X or Y could either be +/-5 or +/-2.

Depending on which, we would have different answers to \((x - y)^2\)

Statement 1) Fits into the equation for 5, 2 or -5, -2 Statement 2) If x = 5 then y MUST be 2

I chose D

There is a much different explanation in the book relating to factoring out \((x - y)^2\) --> \(x^2 -2xy + y^2\) and then plugging in 29 to \(x^2 + y^2\) as given in the problem.

Re: Quant Review 2nd Edition DS 73 - Help [#permalink]
04 Dec 2010, 17:28

Expert's post

jscott319 wrote:

If \(x^2 + y^2 = 29\), what is the value of \((x - y)^2\) ?

1) xy = 10 2) x = 5

I am unsure about my approach. It led me to the wrong answer but I do not know where it is flawed. Looking at this problem I was able to see that according to this equation, \(x^2 + y^2 = 29\), X or Y could either be +/-5 or +/-2.

Depending on which, we would have different answers to \((x - y)^2\)

Statement 1) Fits into the equation for 5, 2 or -5, -2 Statement 2) If x = 5 then y MUST be 2

I chose D

There is a much different explanation in the book relating to factoring out \((x - y)^2\) --> \(x^2 -2xy + y^2\) and then plugging in 29 to \(x^2 + y^2\) as given in the problem.

Any thoughts?

I thought of the explanation that's given in the book too.

Quote:

If x = 5 then y MUST be 2

- Why so? Why can't it be -2? \((5)^2 + (-2)^2 = 29\) as well.

Re: Quant Review 2nd Edition DS 73 - Help [#permalink]
04 Dec 2010, 18:19

Expert's post

VeritasPrepKarishma wrote:

jscott319 wrote:

Looking at this problem I was able to see that according to this equation, \(x^2 + y^2 = 29\), X or Y could either be +/-5 or +/-2.

Another thing, you can't assume that the equation will have only integral values. x = 3 and y = root(20) also satisfies this equation as do many other values.

Good point Karishma!

Quote:

Hmm good point , ha. What could lead me out of this train of thought and into thinking more in line with what led you to the solution your way?

Haha, I'm an engineer and I went to a retardedly hard high school. It's kind of second nature now. But I think when you get a question like this, think about common mistakes. As I mentioned, assuming that positive and positive, and negative and negative go together is not right. Also, you cannot assume only integral values. So just think of all possible combinations. I'd actually advise you to get the GMAT Club tests and go through as many as you can, while noting down the stuff that you get wrong, especially if its stuff like this will help you keep track of stuff.

If x^2+y^2=29, what is the value of (x-y)^2? (1) xy=10 (2) x=5

The OA says the answer is A whereas I think statement 2 is also sufficient as given below. x=5 hence y^2=29-25=4-->y=+2 or -2. Hence (x-y)^2= (5-2)^2 or (5+2)^2

Is it because of two answer choices, Statement 2 is insufficient?

If x^2+y^2=29, what is the value of (x-y)^2? (1) xy=10 (2) x=5

The OA says the answer is A whereas I think statement 2 is also sufficient as given below. x=5 hence y^2=29-25=4-->y=+2 or -2. Hence (x-y)^2= (5-2)^2 or (5+2)^2

Is it because of two answer choices, Statement 2 is insufficient?

Precisely! When the question is asking for the value of something, then "sufficient" means you can determine a unique individual unambiguous value. There is absolutely only one place on the entire infinity of the number line where that value could be --- that's what "sufficient" means. If you can narrow things down to two values, it's not sufficient. Does that make sense?

If x^2+y^2=29, what is the value of (x-y)^2? (1) xy=10 (2) x=5

The OA says the answer is A whereas I think statement 2 is also sufficient as given below. x=5 hence y^2=29-25=4-->y=+2 or -2. Hence (x-y)^2= (5-2)^2 or (5+2)^2

Is it because of two answer choices, Statement 2 is insufficient?

Merging similar topics. Please refer to the solutions above.

Re: If x^2 + y^2 = 29, what is the value of (x-y)^2 (1) xy = 10 [#permalink]
11 Jul 2014, 13:38

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: If x^2 + y^2 = 29, what is the value of (x-y)^2 (1) xy = 10 [#permalink]
23 Jul 2014, 01:56

Bunuel wrote:

krit wrote:

If x^2 + y^2 = 29, what is the value of (x-y)^2

(1) xy = 10 (2) x = 5

Hi guys, I'd like some help here. The answer is A. I see why (1) is sufficient, but I don't get why (2) is not. We can just simply plug x = 5 into the equation, find y, and then solve for (x-y)^2

Thanks

Given: \(x^2+y^2=29\). Question: \((x-y)^2=x^2-2xy+y^2=(x^2+y^2)-2xy=29-2xy=?\) So, basically we should find the value of \(xy\)

(1) xy = 10. Directly gives us the value we needed. Sufficient.

(2) x = 5. Now even if we substitute the value of \(x\) in \(x^2+y^2=29\) we'll get two values for \(y\): 2 and -2, hence two values for \((x-y)^2\): 9 and 49. Not sufficient.

Answer: A.

Hope it's clear.

And I thought we consider only the positive value of a square root!!! Not so?

Re: If x^2 + y^2 = 29, what is the value of (x-y)^2 (1) xy = 10 [#permalink]
23 Jul 2014, 02:01

Expert's post

Kconfused wrote:

Bunuel wrote:

krit wrote:

If x^2 + y^2 = 29, what is the value of (x-y)^2

(1) xy = 10 (2) x = 5

Hi guys, I'd like some help here. The answer is A. I see why (1) is sufficient, but I don't get why (2) is not. We can just simply plug x = 5 into the equation, find y, and then solve for (x-y)^2

Thanks

Given: \(x^2+y^2=29\). Question: \((x-y)^2=x^2-2xy+y^2=(x^2+y^2)-2xy=29-2xy=?\) So, basically we should find the value of \(xy\)

(1) xy = 10. Directly gives us the value we needed. Sufficient.

(2) x = 5. Now even if we substitute the value of \(x\) in \(x^2+y^2=29\) we'll get two values for \(y\): 2 and -2, hence two values for \((x-y)^2\): 9 and 49. Not sufficient.

Answer: A.

Hope it's clear.

And I thought we consider only the positive value of a square root!!! Not so?

When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root. That is, \(\sqrt{4}=2\), NOT +2 or -2.

In contrast, the equation \(x^2=4\) has TWO solutions, +2 and -2.

This week went in reviewing all the topics that I have covered in my previous study session. I reviewed all the notes that I have made and started reviewing the Quant...

I was checking my phone all day. I wasn’t sure when I would receive the admission decision from Tepper. I received an acceptance from Goizueta in the early morning...

I started running as a cross country team member since highshcool and what’s really awesome about running is that...you never get bored of it! I participated in...