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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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Updated on: 28 Nov 2013, 05:40
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47% (02:25) correct 53% (02:38) wrong based on 1073 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
17
6
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

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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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01 Dec 2013, 02:33
21
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.
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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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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

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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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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!

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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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
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Screenshot_28.11.13_21_48.png [ 86.68 KiB | Viewed 37142 times ]

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Posts: 49915
If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11,  [#permalink]

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29 Nov 2013, 10:05
12
4
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.
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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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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!
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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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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]

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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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.
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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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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

Hi Karishma,

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

Thanks for this approach, its the best i've seen so far on this master problem

Cheers!
J
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Posts: 8385
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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

Hi Karishma,

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

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 (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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Joined: 02 Sep 2009
Posts: 49915
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.
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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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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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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.
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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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