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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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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

[Reveal] Spoiler:
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 :)
[Reveal] Spoiler: OA

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^2-v^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^2-v^2, y=2uv and z=u^2+v^2, and if x=11, what is z? [#permalink]

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New post 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:
[Reveal] Spoiler:
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

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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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New post 03 Dec 2013, 01:26
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

[Reveal] Spoiler:
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
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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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New post 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!
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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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New post 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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New post 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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New post 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
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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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New post 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.

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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 (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.

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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VadimKlimenko wrote:
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.
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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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New post 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 \(u^2 + v^2=36+25\).

Thus the answer is D.

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

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New post 15 Sep 2014, 14:07
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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Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]

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New post 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 \(u^2 + v^2=36+25\).

Thus the answer is D.

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.
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Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11,   [#permalink] 15 Sep 2014, 21:05

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