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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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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=6^2v^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
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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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Re: If x=u^2v^2, y=2uv and z=u^2+v^2, and if x=11, what is z? [#permalink]
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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
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Re: If x=u^2v^2, y=2uv and z=u^2+v^2, and if x=11, what is z? [#permalink]
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28 Nov 2013, 02:06
vogelleblanc wrote: If x=u^2v^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=6^2v^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



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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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@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
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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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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.
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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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... 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!



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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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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.
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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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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=6^2v^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^2v^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^2900=0, let v^2= a....so the equation becomes a^2+ 11a900=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
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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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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.
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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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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



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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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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)
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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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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



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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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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: \((z + y)(z  y) = (u + v)^2*(u  v)^2 = 11^2 = 121\)
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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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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 (uv)^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 = ... (uv)(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.



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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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VadimKlimenko wrote: 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 (uv)^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 = ... (uv)(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 \(u^2  v^2=11\): \(u^42u^2v^2+v^2=121\); 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. Answer: D. 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
VadimKlimenko wrote: Why do you assume that (u+v)^2 and (uv)^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 = ... (uv)(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.
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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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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 \(u^2 + v^2=36+25\). Thus the answer is D. Hope it's clear. Given that \(x =11 = u^2  v^2 = (u+v)(uv)\) 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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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
I don't know why, but I started to solve this DSquestion like a PSquestion without considering the information given in (1) and (2). \(x = u^2  v^2\) > \(x = (u+v)*(uv)\) 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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Re: If x = u^2  v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]
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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 \(u^2 + v^2=36+25\). Thus the answer is D. Hope it's clear. Given that \(x =11 = u^2  v^2 = (u+v)(uv)\) 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 DSquestion like a PSquestion without considering the information given in (1) and (2).
\(x = u^2  v^2\) > \(x = (u+v)*(uv)\)
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 noninteger solutions for u and v.
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Re: If x = u^2  v^2, y = 2uv and z = u^2 + v^2, and if x = 11,
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15 Sep 2014, 21:05



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