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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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30 Mar 2015, 07:54
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.[/quote]
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.[/quote]
Hi Bunuel
Stmtn 1 Can we use Differences of Square approach
Property  The difference between squares grows as the squares themselves get larger
Hence given u^2  v^2 = x = 11
Thus the only numbers that fit this equation are when u = 6 and v = 5. Hence we can find Z  Sufficient
Similalry Stmtn 2 = Suff
Hence D
Pls mention if this approach is correct
Thanks



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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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10 Feb 2016, 22:49
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 Responding to a pm: Quote: I have a quick question about how ( u + v)^2 * ( u  v)^2 = 11^2 in the problem below. If (u  v)^2 = (u + v) * (u  v) = 11 , then does ( u + v)^2 * ( u  v)^2 = 11^2 ? Is this correct? If not, can you please help me understand this?
Note that \(x = u^2  v^2\) and x is given as 11. So, u^2  v^2 = 11 \((u + v)*(u  v) = 11\) So \([(u + v)*(u  v)]^2 = 11^2\) \((u+v)^2 * (u  v)^2 = 11^2\) Does this help?
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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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14 Mar 2016, 11:44
kinjiGC wrote: 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. Z can't be negative as its a sum of squares....



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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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17 Feb 2017, 17:57
x=u^2−v^2..........(1).......x=(u+v)(uv).........(u+v)=x/(uv) y=2uv.........(2) z=u^2+v^2.......(3) Adding eqn (2)+(3) results....... z+y=u^2+v^2+2uv=>>>>>(u+v)^2=>>>>((x/(uv))^2=>>>>>x^2/u^22uz+v^2 z+y=x^2/zy (z+y)(zy)=x^2 z^2y^2=x^2
(1) y = 60 z^2y^2=x^2 z^260^2=11^2 z^2=60^2+11^2 we get the value of z
(2) u = 6 From eqn (1) we get the value of v and substituting v in eqn (2) we get y using z^2y^2=x^2 we get the value of z.
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,
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16 Aug 2017, 17:52
If u^2  v^2 = 11 > difference of squares > 36  25 = 11 > z = u^2 + v^2 = 36 + 25
Thus z is known before either statements 1 or 2. Correct?



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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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02 Mar 2018, 07:13
Analyze: z = 2u^2 + 11 (2) u=6 => v^2 = 11 6^2 ==> z solved (1) y=30 => u = 30/v put it in v^2 + 11 = u^2 => (v^2)^2 + 11v^2  900 = 0. Analyze: P = product of 2 roots =  900, S = sum of roots = 11 => Equation has 2 roots : one v^2 is positive and one v^2 is negative. But v^2 is always positive => v^2 is only one value ==> z solved ==> D



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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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02 Mar 2018, 19:22
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) I solved it as follows. Please let me know if my method is incorrect. Statement 1. y=60 uv = 30 v=30/u substituting v=30/u in z = u2+v2 ..we get the value of u and subsequently use that value in uv = 30 to get the value of v.Hence sufficient Statement 2 u given we know uv=30 hence sufficient. Hence D.



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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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16 May 2018, 07:23
Hi all,
I thought statement one was much easier to prove sufficient than stated above.
We know 60=2uv, 11=u^2v^2, and Z=u^2+v^2. Thus, 30/u=v. Then plug in 30/u for v in 11=u^2v^2> ie 11= u^2  (30/u)^2. There is one variable, thus we know we can solve for u. Then theoretically, we plug u in to find v. Then we would plug the solved values for u and v in Z=u^2+v^2. Thus, we could solve for z.
Can someone please correct me if this reasoning is incorrect?



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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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06 Nov 2018, 19:07
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. Can sb help me understand why (u, v) = (6, 5) or (u, v) = (6, 5) can't be also (30,1) or (30,1)?



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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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09 Dec 2018, 11:58
What information do we know? There are 5 variables \(x,y,z,u,v\). The question gives 4 equations so we should only need one more equation to solve for \(z\).
Cond 1) \(y = 60, y=2uv \implies 30 = uv \implies z = u^2 + v^2 = u^2 + (\frac{30}{u})^2 \implies u^4  11u^2  900 = 0.\) We know that this equation has the form \((u^2  Q)(u^2 +R) = 0\) for two numbers \(Q,R\). (To verify this is actually the case we can verify that the Determinant > 0 (i.e. \(D = b^2  4ac > 0, \text{ where quadratic is given by } ax^2 + bx + c\)). Next, since \(u^2\) is always positive, we know that the answer has to be \(u^2 = Q\).
Since we know \(x, u^2\), we know \(v^2\). Since we now know \(v^2\) and \(u^2\) we now know \(z\).
Sufficient.
Cond 2) Since we know \(u\), we know \(u^2\). Since we know \(x, u^2\), we know \(v^2\). Since we now know \(v^2\) and \(u^2\) we now know \(z\).
Sufficient.
(D)



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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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10 Dec 2018, 01:29
Hi Bunuel, VeritasKarishma and chetan2uI took a different approach to prove that S(1) is sufficient... I would appreciate it if you could confirm my logic:) We know \(11=(U+V)*(UV)\) And we are given that \(30=U*V\), so \(U=\frac{30}{V}\) Plugging that into the equation above we get \(11=\frac{30^2}{V^2}V^2\), which simplifies into \(V^4+11^2900=0\) then I replaced \(V^2\) in the equation with \(A\), so \(V^2=A\) \(A^2+11A900=0\) > \((A+36)*(A25)\) so \(V^2\) can bei either \(36\) or \(25\), but since a power of \(2\) can never result in a negative number \(V^2=25\) Is my reasoning above correct? Thank you in advance!
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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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10 Dec 2018, 01:52
T1101 wrote: Hi Bunuel, VeritasKarishma and chetan2uI took a different approach to prove that S(1) is sufficient... I would appreciate it if you could confirm my logic:) We know \(11=(U+V)*(UV)\) And we are given that \(30=U*V\), so \(U=\frac{30}{V}\) Plugging that into the equation above we get \(11=\frac{30^2}{V^2}V^2\), which simplifies into \(V^4+11^2900=0\) then I replaced \(V^2\) in the equation with \(A\), so \(V^2=A\) \(A^2+11A900=0\) > \((A+36)*(A25)\) so \(V^2\) can bei either \(36\) or \(25\), but since a power of \(2\) can never result in a negative number \(V^2=25\) Is my reasoning above correct? Thank you in advance! Yes, you are correct.. v would give you two values and accordingly u will take two values. However z will come out same in both cases. Of course you don't have to find each case... z=u^2+v^2...u^2=11+v^2, so z=11+v^2+v^2=11+2v^2 Your reasoning is correct.
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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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04 Jan 2019, 04:07
without solving for the values i chose to count the variables and the equations and surprisingly my ans is right.Please correct me if i am wrong x = u^2 – v^2, y = 2uv, z = u^2 + v^2 X=11
1. y=60 11= u^2 – v^2, since x=11 60= 2uv, since y = 60 z = u^2 + v^2 no. of variables (u, v and z) is equal to number of equation.Hence Sufficient.
2.U=6 Same as above.Hence answer is D




Re: If x = u^2  v^2, y = 2uv and z = u^2 + v^2, and if x = 11, &nbs
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