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

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

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

Hope it's clear.
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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
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Re: #73 OG Quant Review 2nd [#permalink]  28 Nov 2010, 12:30
wow...okay. It's clear now. I completely forgot the -2. Thank you all
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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.

Any thoughts?
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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.
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Re: Quant Review 2nd Edition DS 73 - Help [#permalink]  04 Dec 2010, 17:45
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?
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Re: Quant Review 2nd Edition DS 73 - Help [#permalink]  04 Dec 2010, 18:04
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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.
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Re: Quant Review 2nd Edition DS 73 - Help [#permalink]  04 Dec 2010, 18:19
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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.
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Algebra [#permalink]  23 Dec 2012, 16:46
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?
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Re: Algebra [#permalink]  23 Dec 2012, 19:00
Expert's post
mtripathy wrote:
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?

Mike
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Re: Algebra [#permalink]  23 Dec 2012, 23:29
Expert's post
mtripathy wrote:
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.

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

Hope it's clear.

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

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.

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