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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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x=u^2−v^2..........(1).......x=(u+v)(u-v).........(u+v)=x/(u-v)
y=2uv.........(2)
z=u^2+v^2.......(3)
z+y=u^2+v^2+2uv=>>>>>(u+v)^2=>>>>((x/(u-v))^2=>>>>>x^2/u^2-2uz+v^2
z+y=x^2/z-y
(z+y)(z-y)=x^2
z^2-y^2=x^2

(1) y = 60
z^2-y^2=x^2
z^2-60^2=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^2-y^2=x^2 we get the value of z.

Ans.D
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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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Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]
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nishantjayaswal wrote:
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?

Hi,

Genoa2000

It will not be the case always. It is true for sure when all are linear equations.
What is x?
x=y+3, so value of y is sufficient to answer.

But say
What is x?
x^2=y+3 and y=6...x^2=6+3=9
So x can be 3 or -3.....Not sufficient
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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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nishantjayaswal wrote:
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?

I would be wary of this approach for multiple reasons.

1. Linear equations could be equivalent in which case you get infinite solutions.

x + y = 5
2x + 2y = 10
Both are the same equation so you cannot solve for x and y

2. Also, when you have equations in higher degree, you may get multiple solutions.

x^2 = 25
gives you x = 5 or -5
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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.
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