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

Responding to a pm:



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?

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)
Adding eqn (2)+(3) results.......
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

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

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

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

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

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)

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?

Why are we assuming they are integers?

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.