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# If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11,

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Joined: 16 Jun 2013
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If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11,  [#permalink]

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Updated on: 28 Nov 2013, 05:40
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49% (01:23) correct 51% (01:31) wrong based on 1045 sessions

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If $$x = u^2 - v^2$$, $$y = 2uv$$ and $$z = u^2 + v^2$$, and if $$x = 11$$, what is the value of z?

(1) y = 60
(2) u = 6

my approach towards statement 2:
x=11 and u=6,
thus 11=6^2-v^2
25=v^2
v=5 or v=-5
Since z=u^2+v^2, it does not matter if v is 5 or -5, so z=55
Please correct me if something is wrong!/spoiler]

Originally posted by vogelleblanc on 27 Nov 2013, 23:26.
Last edited by Bunuel on 28 Nov 2013, 05:40, edited 2 times in total.
Renamed the topic and edited the question.
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If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11,  [#permalink]

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17 Apr 2014, 20:33
15
5
jlgdr wrote:
Could someone please elaborate a little bit more on this problem?

How do you get to the values of (u,v) ? Is it just by ballparking? Or is there an elegant algebraic approach for this one?

Thanks
Cheers!
J

Given in the stem:

$$11 = u^2 - v^2 = (u + v)(u - v)$$
$$y = 2uv$$
$$z = u^2 + v^2$$

$$z + y = u^2 + v^2 + 2uv = (u + v)^2$$
$$z - y = u^2 + v^2 - 2uv = (u - v)^2$$

$$(z + y)(z - y) = (u + v)^2*(u - v)^2 = 11^2 = 121$$

121 can be written as product of two numbers in 2 ways: 1, 121 and 11, 11
So (z + y)(z - y) = (61 + 60)*(61 - 60) or = (11 + 0)(11 - 0)
So z can be 61 or 11 depending on whether y is 60 or 0.

Statement 1 gives y = 60. So z must be 61. Sufficient

Statement 2 gives u is 6 which means 2uv (= y) is not 0. So z must be 61. Sufficient

_________________

Karishma
Veritas Prep GMAT Instructor

Save up to $1,000 on GMAT prep through 8/20! Learn more here > GMAT self-study has never been more personalized or more fun. Try ORION Free! ##### Most Helpful Community Reply Verbal Forum Moderator Joined: 10 Oct 2012 Posts: 615 Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink] ### Show Tags 01 Dec 2013, 02:33 19 5 vogelleblanc wrote: If $$x = u^2 - v^2$$, $$y = 2uv$$ and $$z = u^2 + v^2$$, and if $$x = 11$$, what is the value of z? (1) y = 60 (2) u = 6 Another approach, for F.S 1, in which you don't have to calculate for anything We know that $$x = u^2 - v^2$$ = 11,Squaring on both sides, we have $$u^4 + v^4 - 2*v^2*u^2$$ = 121 Thus, adding$$4*v^2*u^2$$ on both sides, we have $$(u^2 + v^2)^2 = 121+4*v^2*u^2 = z^2$$ Hence, this statement is sufficient to calculate the value of z. _________________ ##### General Discussion Intern Joined: 19 Oct 2013 Posts: 1 Re: If x=u^2-v^2, y=2uv and z=u^2+v^2, and if x=11, what is z? [#permalink] ### Show Tags 28 Nov 2013, 01:33 1 Since y=60 in st (1) then uv=30, and the values of u*v is 6,5 or -6,-5 if x=11.I.E 6^2- 5^2=11 and (-6)^2-(-5)^2=11. Therfore the value of z=36+25=61 Posted from my mobile device Intern Joined: 21 Oct 2012 Posts: 1 Schools: Insead '14 Re: If x=u^2-v^2, y=2uv and z=u^2+v^2, and if x=11, what is z? [#permalink] ### Show Tags 28 Nov 2013, 02:06 vogelleblanc wrote: If x=u^2-v^2, y=2uv and z=u^2+v^2, and if x=11, what is the value of z? (1) y=60 (2) u=6 my approach towards statement 2: x=11 and u=6, thus 11=6^2-v^2 25=v^2 v=5 or v=-5 Since z=u^2+v^2, it does not matter if v is 5 or -5, so z=55 Please correct me if something is wrong! I do not understand statement 1, please help me Answer has to be "B" ... its not explicitly mentioned that U and V are integer so. A is not option.... B because X and U are given now... so V can be calculated and so the Z Intern Joined: 16 Jun 2013 Posts: 16 GMAT 1: 540 Q34 V30 GMAT 2: 700 Q43 V42 Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink] ### Show Tags 28 Nov 2013, 06:46 1 @sethiprashant i also think you are right with your reasoning that u and v do not have to be integers and thus 6 and 5 are not the only possible combination. However, the OA is D... :/ What are we overlooking? Just to prove that i quote the question correctly, i attached a screenshot of the question Attachments Screenshot_28.11.13_21_48.png [ 86.68 KiB | Viewed 34375 times ] Math Expert Joined: 02 Sep 2009 Posts: 48037 If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink] ### Show Tags 29 Nov 2013, 10:05 8 3 vogelleblanc wrote: If $$x = u^2 - v^2$$, $$y = 2uv$$ and $$z = u^2 + v^2$$, and if $$x = 11$$, what is the value of z? (1) y = 60 (2) u = 6 @sethiprashant i also think you are right with your reasoning that u and v do not have to be integers and thus 6 and 5 are not the only possible combination. However, the OA is D... :/ What are we overlooking? Just to prove that i quote the question correctly, i attached a screenshot of the question For (1) we have: $$11 = u^2 - v^2$$; $$60= 2uv$$; and we need to find the value of $$u^2 + v^2$$. If you solve $$11 = u^2 - v^2$$ and $$60= 2uv$$, you get (u, v) = (-6, -5) or (u, v) = (6, 5). In either case $$u^2 + v^2=36+25$$. Thus the answer is D. Hope it's clear. _________________ Intern Joined: 16 Jun 2013 Posts: 16 GMAT 1: 540 Q34 V30 GMAT 2: 700 Q43 V42 Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink] ### Show Tags 01 Dec 2013, 02:16 1 ... ah. So, the fact that x=11, limits the possible values of u and v to (6,5) or (-6,-5), right? bunuel, thanks a lot so far! Director Joined: 25 Apr 2012 Posts: 701 Location: India GPA: 3.21 WE: Business Development (Other) Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink] ### Show Tags 03 Dec 2013, 01:26 1 vogelleblanc wrote: If $$x = u^2 - v^2$$, $$y = 2uv$$ and $$z = u^2 + v^2$$, and if $$x = 11$$, what is the value of z? (1) y = 60 (2) u = 6 my approach towards statement 2: x=11 and u=6, thus 11=6^2-v^2 25=v^2 v=5 or v=-5 Since z=u^2+v^2, it does not matter if v is 5 or -5, so z=55 Please correct me if something is wrong!/spoiler] I do not understand statement 1, please help me Answer is correct. We have that x= u^2-v^2 and x =11 z= u^2+ v^2 and y = 2 uv from St 1 we have 2uv= 60 -----> uv =30-------> u=30/v Putting the above in equation of x we have 11= (30/v)^2 - v^2---------> Solving, we get an equation in degree 4 v^4- 11v^2-900=0, let v^2= a....so the equation becomes a^2+ 11a-900=0 Solving for quadratic equation, we get roots as = (-11 +/- \sqrt{11^2 +4*1*900}) / 2*1 or (-11 +/- 61)/ 2 so possible values of a = -72/2 or 50/2....since a = v^2 and therefore neglecting negative value as square of a number is greater than or equal to zero therefore v^ 2= 25 or v = +5/-5 So whether v= 5 or v= -5 it does not change the value of z as z = u^2 +v^2 or (30/+ or -5)^2 + (+/- 5)^2 --------> z= 61 So St1 is sufficient Clearly St 2 is also sufficient as u = 6 then 11= 36 - v^2 or v^2 =25 or +/5 Ans D _________________ “If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.” Director Joined: 03 Feb 2013 Posts: 903 Location: India Concentration: Operations, Strategy GMAT 1: 760 Q49 V44 GPA: 3.88 WE: Engineering (Computer Software) Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink] ### Show Tags 29 Mar 2014, 07:26 1 mau5 wrote: vogelleblanc wrote: If $$x = u^2 - v^2$$, $$y = 2uv$$ and $$z = u^2 + v^2$$, and if $$x = 11$$, what is the value of z? (1) y = 60 (2) u = 6 Another approach, for F.S 1, in which you don't have to calculate for anything We know that $$x = u^2 - v^2$$ = 11,Squaring on both sides, we have $$u^4 + v^4 - 2*v^2*u^2$$ = 121 Thus, adding$$4*v^2*u^2$$ on both sides, we have $$(u^2 + v^2)^2 = 121+4*v^2*u^2 = z^2$$ Hence, this statement is sufficient to calculate the value of z. As Z can be negative as well as +ve, Z value cannot be uniquely determined by using the above method. It is better to solve equations and determine. _________________ Thanks, Kinjal My Application Experience : http://gmatclub.com/forum/hardwork-never-gets-unrewarded-for-ever-189267-40.html#p1516961 Linkedin : https://www.linkedin.com/in/kinjal-das/ Please click on Kudos, if you think the post is helpful SVP Joined: 06 Sep 2013 Posts: 1850 Concentration: Finance Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink] ### Show Tags 17 Apr 2014, 13:13 Could someone please elaborate a little bit more on this problem? How do you get to the values of (u,v) ? Is it just by ballparking? Or is there an elegant algebraic approach for this one? Thanks Cheers! J SVP Joined: 06 Sep 2013 Posts: 1850 Concentration: Finance Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink] ### Show Tags 02 May 2014, 16:58 VeritasPrepKarishma wrote: jlgdr wrote: Could someone please elaborate a little bit more on this problem? How do you get to the values of (u,v) ? Is it just by ballparking? Or is there an elegant algebraic approach for this one? Thanks Cheers! J Given in the stem: $$11 = u^2 - v^2 = (u + v)(u - v)$$ $$y = 2uv$$ $$z = u^2 + v^2$$ $$z + y = u^2 + v^2 + 2uv = (u + v)^2$$ $$z - y = u^2 + v^2 - 2uv = (u - v)^2$$ $$(z + y)(z - y) = (u + v)^2*(u - v)^2 = 11^2 = 121$$ 121 can be written as product of two numbers in 2 ways: 1, 121 and 11, 11 So (z + y)(z - y) = (61 + 60)*(61 - 60) or = (11 + 0)(11 - 0) So z can be 61 or 11 depending on whether y is 60 or 0. $$Statement 1 gives y = 60. So z must be 61. Sufficient$$ Statement 2 gives u is 6 which means 2uv (= y) is not 0. So z must be 61. Sufficient Answer (D) Hi Karishma, Assume you meant $$(z + y)(z - y) = (u + v)^2*(u - v)^2 = 11^2 = 121$$ Please confirm Thanks for this approach, its the best i've seen so far on this master problem Cheers! J Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8195 Location: Pune, India Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink] ### Show Tags 02 May 2014, 19:05 jlgdr wrote: VeritasPrepKarishma wrote: jlgdr wrote: Could someone please elaborate a little bit more on this problem? How do you get to the values of (u,v) ? Is it just by ballparking? Or is there an elegant algebraic approach for this one? Thanks Cheers! J Given in the stem: $$11 = u^2 - v^2 = (u + v)(u - v)$$ $$y = 2uv$$ $$z = u^2 + v^2$$ $$z + y = u^2 + v^2 + 2uv = (u + v)^2$$ $$z - y = u^2 + v^2 - 2uv = (u - v)^2$$ $$(z + y)(z - y) = (u + v)^2*(u - v)^2 = 11^2 = 121$$ 121 can be written as product of two numbers in 2 ways: 1, 121 and 11, 11 So (z + y)(z - y) = (61 + 60)*(61 - 60) or = (11 + 0)(11 - 0) So z can be 61 or 11 depending on whether y is 60 or 0. $$Statement 1 gives y = 60. So z must be 61. Sufficient$$ Statement 2 gives u is 6 which means 2uv (= y) is not 0. So z must be 61. Sufficient Answer (D) Hi Karishma, Assume you meant $$(z + y)(z - y) = (u + v)^2*(u - v)^2 = 11^2 = 121$$ Please confirm Thanks for this approach, its the best i've seen so far on this master problem Cheers! J I am not sure what the problem is. I have written exactly the same thing: $$(z + y)(z - y) = (u + v)^2*(u - v)^2 = 11^2 = 121$$ _________________ Karishma Veritas Prep GMAT Instructor Save up to$1,000 on GMAT prep through 8/20! Learn more here >

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Joined: 06 Sep 2013
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Concentration: Finance
Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11,  [#permalink]

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28 May 2014, 11:40
I am not sure what the problem is.
I have written exactly the same thing: $$(z + y)(z - y) = (u + v)^2*(u - v)^2 = 11^2 = 121$$[/quote]

Not so, you wrote.
$$(z + y)(z - y) = (u + v)^2*(u + v)^2 = 11^2 = 121$$

Anyways, just wanted to be clear as intended meaning. Do not worry, this tiny error will not change my high esteem for your work in quant

Thanks again Karishma

Cheers
J
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Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11,  [#permalink]

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29 Jun 2014, 05:37
VeritasPrepKarishma wrote:
jlgdr wrote:
Could someone please elaborate a little bit more on this problem?
..........
121 can be written as product of two numbers in 2 ways: 1, 121 and 11, 11
.........

Why do you assume that (u+v)^2 and (u-v)^2 are integers ?
If we knew that u,v are integers, then the simplest way to find the figures in my opinion is:
uv = 30 = 1*30 = 2*15 = 3*10 = 5*6 = 6*5 = ...
(u-v)(u+v) = 11

It seems that the only exact method to find u and v is solving for quadratic (4th power) equation, but it's also too time consuming. If someone knows quick method of finding u and v without assuming that tose numbers are integers please share it.
Math Expert
Joined: 02 Sep 2009
Posts: 48037
Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11,  [#permalink]

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29 Jun 2014, 12:56
7
2
VeritasPrepKarishma wrote:
jlgdr wrote:
Could someone please elaborate a little bit more on this problem?
..........
121 can be written as product of two numbers in 2 ways: 1, 121 and 11, 11
.........

Why do you assume that (u+v)^2 and (u-v)^2 are integers ?
If we knew that u,v are integers, then the simplest way to find the figures in my opinion is:
uv = 30 = 1*30 = 2*15 = 3*10 = 5*6 = 6*5 = ...
(u-v)(u+v) = 11

It seems that the only exact method to find u and v is solving for quadratic (4th power) equation, but it's also too time consuming. If someone knows quick method of finding u and v without assuming that tose numbers are integers please share it.

If $$x = u^2 - v^2$$, $$y = 2uv$$ and $$z = u^2 + v^2$$, and if $$x = 11$$, what is the value of z?

Given: $$u^2 - v^2=11$$ and $$y = 2uv$$.
Question: $$u^2 + v^2=?$$

(1) y = 60 --> $$2uv=60$$ --> $$4u^2v^2=3,600$$.

Square $$u^2 - v^2=11$$: $$u^4-2u^2v^2+v^2=121$$;

Add $$4u^2v^2$$ to both sides: $$u^4+2u^2v^2+v^4=121+3,600$$;

Apply $$a^2+2ab+b^2=(a+b)^2$$: $$(u^2+v^2)^2=121+3,600$$.

$$u^2+v^2=\sqrt{121+3,600}=61$$.

Sufficient.

(2) u = 6 --> $$36 - v^2=11$$ --> $$v^2=25$$ --> $$u^2+v^2=36+25=61$$. Sufficient.

Hope it helps.
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Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11,  [#permalink]

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29 Jun 2014, 20:58
1
Why do you assume that (u+v)^2 and (u-v)^2 are integers ?
If we knew that u,v are integers, then the simplest way to find the figures in my opinion is:
uv = 30 = 1*30 = 2*15 = 3*10 = 5*6 = 6*5 = ...
(u-v)(u+v) = 11

It seems that the only exact method to find u and v is solving for quadratic (4th power) equation, but it's also too time consuming. If someone knows quick method of finding u and v without assuming that tose numbers are integers please share it.

Also note that you are given that x is an integer (11). Also statement 1 tells you that y is an integer (60) and statement 2 tells you that u is an integer. So using any one of the statements gives you enough hint that we are dealing with integers only.
_________________

Karishma
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If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11,  [#permalink]

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15 Sep 2014, 12:53
Bunuel wrote:
vogelleblanc wrote:
If $$x = u^2 - v^2$$, $$y = 2uv$$ and $$z = u^2 + v^2$$, and if $$x = 11$$, what is the value of z?

(1) y = 60
(2) u = 6

sethiprashant i also think you are right with your reasoning that u and v do not have to be integers and thus 6 and 5 are not the only possible combination. However, the OA is D... :/
What are we overlooking?

Just to prove that i quote the question correctly, i attached a screenshot of the question

For (1) we have:
$$11 = u^2 - v^2$$;

$$60= 2uv$$;

and we need to find the value of $$u^2 + v^2$$.

If you solve $$11 = u^2 - v^2$$ and $$60= 2uv$$, you get (u, v) = (-6, -5) or (u, v) = (6, 5). In either case $$u^2 + v^2=36+25$$.

Hope it's clear.

Given that $$x =11 = u^2 - v^2 = (u+v)(u-v)$$
Thus there we can be 4 possibilities of (u,v) = (6,5)OR(6,-5)OR(-6,5)OR(-6,-5)
From this only we have sufficient info to answer " what is the value of z, given $$z = u^2 + v^2$$ "
Because in each case, answer will be 61.
So we do not even need any statement!
Please explain if i am missing anything here..
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Concentration: Finance, Economics
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If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11,  [#permalink]

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15 Sep 2014, 14:07
1
I don't know why, but I started to solve this DS-question like a PS-question without considering the information given in (1) and (2).

$$x = u^2 - v^2$$ ----> $$x = (u+v)*(u-v)$$

We're given $$x = 11$$ , since 11 is a prime number the only factors of $$x$$ are 1 and 11, this tells us $$u$$ and $$v$$ can take any combination of +-5 and+-6

Hence, $$z = u^2 + v^2$$ will always take the value 61

Therefore we don't need any data to solve the problem!

edit: saw tushain came to the same conclustion
_________________

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Joined: 02 Sep 2009
Posts: 48037
Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11,  [#permalink]

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15 Sep 2014, 21:05
tushain wrote:
Bunuel wrote:
vogelleblanc wrote:
If $$x = u^2 - v^2$$, $$y = 2uv$$ and $$z = u^2 + v^2$$, and if $$x = 11$$, what is the value of z?

(1) y = 60
(2) u = 6

sethiprashant i also think you are right with your reasoning that u and v do not have to be integers and thus 6 and 5 are not the only possible combination. However, the OA is D... :/
What are we overlooking?

Just to prove that i quote the question correctly, i attached a screenshot of the question

For (1) we have:
$$11 = u^2 - v^2$$;

$$60= 2uv$$;

and we need to find the value of $$u^2 + v^2$$.

If you solve $$11 = u^2 - v^2$$ and $$60= 2uv$$, you get (u, v) = (-6, -5) or (u, v) = (6, 5). In either case $$u^2 + v^2=36+25$$.

Hope it's clear.

Given that $$x =11 = u^2 - v^2 = (u+v)(u-v)$$
Thus there we can be 4 possibilities of (u,v) = (6,5)OR(6,-5)OR(-6,5)OR(-6,-5)
From this only we have sufficient info to answer " what is the value of z, given $$z = u^2 + v^2$$ "
Because in each case, answer will be 61.
So we do not even need any statement!
Please explain if i am missing anything here..

TehMoUsE wrote:
I don't know why, but I started to solve this DS-question like a PS-question without considering the information given in (1) and (2).

$$x = u^2 - v^2$$ ----> $$x = (u+v)*(u-v)$$

We're given $$x = 11$$ , since 11 is a prime number the only factors of $$x$$ are 1 and 11, this tells us $$u$$ and $$v$$ can take any combination of +-5 and+-6

Hence, $$z = u^2 + v^2$$ will always take the value 61

Therefore we don't need any data to solve the problem!

edit: saw tushain came to the same conclustion

Both of you incorrectly assume that the variables are integers only: 11 = u^2 - v^2 has infinitely many non-integer solutions for u and v.
_________________
Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, &nbs [#permalink] 15 Sep 2014, 21:05

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