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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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Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]
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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Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]
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]
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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If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]
I just assumed that if there are 5 variables then we need 5 different equations to solve it. Question already has 3 equation + 1 value of variable = 4 equations. So getting any one variable's value would suffice to get all other variables's values.
Are there any flaws in my approach?
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If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]
Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1

Originally posted by avigutman on 06 Oct 2020, 18:34.
Last edited by avigutman on 30 Nov 2020, 14:00, edited 1 time in total.
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If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]
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]

This is a really interesting question.

My approach:

We are given that $$x = u^2 - v^2$$, $$x = 11$$

Statement 1 tells us y = 60.

=> $$60 = 2uv$$

Square on both sides

$$y^2 = (2uv)^2$$

$$(60)^2 = (2uv)^2$$

$$3600 = 4u^2v^2$$

Take $$x = u^2 - v^2$$ & square on both sides

we get, $$(11)^2 = (u^2 - v^2)^2$$
$$121= u^4 + v^4 - 2u^2v^2$$

Add $$y^2$$ on both sides to get a positive value for $$- 2u^2v^2$$.

$$121 + 3600 = u^4 + v^4 - 2u^2v^2 + 4u^2v^2$$

$$121 + 3600 = u^4 + v^4 + 2u^2v^2$$

Here, we can write $$u^4 + v^4 + 2u^2v^2$$ as $$(u^2 + v^2)^2$$, since $$(a + b)^2 = a^2 + b^2 + 2ab$$

$$(u^2 + v^2)^2 = 3721$$

$$u^2 + v^2 = \sqrt{3721}$$

$$z = \sqrt{3721}$$

$$z = 61$$

Statement 1 alone is sufficient!

Let's look at Statement 2. It says u = 6.
No need to plug this in the equation we derived from STEM, rather just plug the value in $$x = u^2 - v^2$$.

$$11 = 6^2 - v^2$$
$$v^2 = 6^2 - 11$$
$$v^2 = 36 - 11$$ = $$v^2 = 25$$ Therefore, $$v = +5$$ or $$v = -5$$

If $$v = +5$$
then, $$z = 6^2 + 5^2$$ (u = 6 from Statement 2)
$$z = 36 + 25$$ = $$z = 61$$

If $$v = -5$$
then, $$z = 6^2 + (-5)^2$$
$$z = 36 + 25$$ = $$z = 61$$

Statement 2 alone is sufficient!

Option 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]
I had a slightly different approach - however would love Bunuel to gauge whether or not it is broadly applicable.

S1:
y = 2uv = 60, then we know uv = 30.

Since x = u^2 - v^2 we can use the difference of squares to see that x = (u+v)(u-v) = 11. Since we know that 11 is prime - one of the factors must be 1 and the other must be 11, so we can solve that u+v = 11 and u-v =1, therefore u = 6, v=5, satisfying uv=30.

S2:
If x = u^2 - v^2 and z = u^2 + v^2, I simply added the two equations to get x+z = 2u^2 (the v^2's cancel out). Since we know that x is 11, S2 is sufficient as it gives us u = 6. Therefore z+11 = 2*6^2 and we can solve for 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]
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

Solution:

x = u^2 - v^2 = 11
y = 2uv
z = u^2 + v^2

1] y=60
2uv = 60; uv = 30

u^2 - v^2 + 2uv = 71........................(1)
u^2 + v^2 + 2uv = z + 60..................(2)

Solving the two:
z + 131 = 2z
z = 131/2

2] u=6
So v = +-5
So, z = 61

SO ans - Option (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]
KarishmaB 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

My doubt is that do we know that z+y and z-y are positive integers. how can we assume that 121 can be written in those two ways as you suggested?
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Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]
KarishmaB 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

Why are we assuming they are integers?
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Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]
bruno,

I have a question regarding your solution below :
In solving statement (1), you got to this part: (u^2 + v^2)^2 = z^2 = 121+3600. Isn't this mean there are 2 values of z: z = positive square root of (121+3600) or z = negative square root of (121+3600)? Therefore, there are 2 possible values of z based on statement (1), and therefore statement (1) is INSUFFICIENT?
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If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]
bovannu01 wrote:
bruno,

I have a question regarding your solution below :
In solving statement (1), you got to this part: (u^2 + v^2)^2 = z^2 = 121+3600. Isn't this mean there are 2 values of z: z = positive square root of (121+3600) or z = negative square root of (121+3600)? Therefore, there are 2 possible values of z based on statement (1), and therefore statement (1) is INSUFFICIENT?

First of all, who is bruno?

To address your question, z equals u^2 + v^2, so it's the sum of two non-negative values, and thus cannot be negative.
If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]
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