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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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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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There have been a lot of posts here that focus a lot on the raw algebra, but there are not a lot of posts on quantitative reasoning. Of course, as you are studying for the GMAT, I do not want to undersell the importance of being able to use algebra. If you can't see another way to think about a Quantitative question, there is nothing wrong with some algebraic gymnastics. But the GMAT calls it "Quantitative Reasoning" for a reason. Let's look at this question from that perspective...

First of all, there is been a fair amount of discussion in this section about not needing ANY information to solve for \(u^2 + v^2\). But this assumes from the beginning that \(u\) and \(v\) must be integers because \(11 = u^2 - v^2\). However, just because \(u^2 - v^2\) is equal to an integer value, doesn't mean \(u\) and \(v\) must both be integers. For example, if \(u=100\), then \(v\) has a non-integer solution: \(v = \sqrt{89}\). This is a perfectly valid solution if there are no other additional constraints. Therefore, the math, as it stands, is not sufficient by itself. You should know that Data Sufficiency questions are NEVER sufficiently limiting without at least one of the two supplementary statements. That is not how Data Sufficiency problems work.

So, let's begin with our critical thinking takedown of this question. The problem tells us that \(11 = u^2 - v^2\) and \(y=2uv\), while asking us to solve for \(u^2 + v^2\). (Don't let those other variables get in the way. \(x\) and \(z\) are really just placeholders.) Let's analyze each statement to see if it is sufficiently limiting to get to a single answer for \(u^2 + v^2\).

Statement #1


Here is a simple algebraic approach to this statement (just to show it)...
If \(60=2uv\), then \(uv=30\) and \(v=\frac{30}{u}\). Therefore,
\(11= u^2-v^2=u^2-(\frac{30}{u})^2=u^2-(\frac{30^2}{u^2})\).

If we multiply both sides by \(u^2\) to cancel the denominator and bring everything to one side, we have:
\(u^4 - 11u^2 - 30^2 = 0\)

This is similar in shape to a quadratic that we can factor. It turns out to be:
\((u^2 - 6^2)(u^2 + 5^2) = 0\)

Since \((u^2 + 5^2) = 0\) only has imaginary solutions, we can see that \(u\) = \(6\) or \(-6\). Plugging these values into the original equation gives us two pairs of \((u,v)\) solutions: \((6,5)\) and \((-6, -5)\). And yet, because the problem is asking us for \(u^2 + v^2\), both sets of solutions result in the same solution: \(36 + 25 = 61\). Statement #1 is sufficient.

Here is the "critical thinking" approach...
However, it is possible to avoid all of that algebra. Statement #1 tells us that \(y=60\). Therefore, \(60=2uv\) and \(uv=30\). This is massively limiting. Think about it... the hypothetical solution above (\((u,v) = (100, \sqrt{89})\)) cannot be an option if \(uv=30\). Whenever you square two different numbers, the difference between these numbers will be magnified. The only exception to this is if both numbers are fractions less than 1. \((u,v)=(30,1)\) is too extreme, since \(30^2 - 1^2 = 899\). The original difference between \(u\) and \(v\) must be very small in order for \(u^2 - v^2\) to be equal to only \(11\). Plus, because the problem asks us for \(u^2 + v^2\), we don't have to worry about negative values. No matter what, the negative options for \(u\) and \(v\) would be cancelled out for purposes of this problem. We must be looking for a single solution where \(u\) and \(v\) are close together. Logically speaking, if we don't care about negatives there can be only one possibility where \(u^2-v^2 = 11\). While this does turn out to be an integer solution, (\(6^2 - 5^2=36-25=11\)), we don't even need to know that \(u\) and \(v\) must be integers for this to work. Statement #1 is definitely sufficiently limiting.

Statement #2


This statement is a lot easier to think about, and the algebra isn't that bad. If we know that \(u=6\), then:
\(11 = u^2-v^2= 6^2 - v^2\)

Solving for v gives us:
\(v^2 = 25\) and \(v=\pm 5\). We now know both \(u\) and \(v\). Therefore, we can determine \(u^2+v^2\). Statement #2 is also sufficient.

With both statements being independently sufficient, the correct answer to this question is "D".

Now, for those of you that are studying for the GMAT, let's step back a little bit to identify the strategies that got us to the answer. Thinking at the "strategy" level will allow you to solve numerous questions on the GMAT, not just this one. First, recognize that even exponents (most commonly "squares") can simultaneously hide and/or eliminate negative options. Second, many quantitative questions aren't about the algebra at all. If you find yourself slogging through a massive amount of math, there might be a quicker way to think about the problem. After all, the GMAT calls this section "Quantitative Reasoning" for a reason. Of course, if you are an algebra wiz and you can finish off a question algebraically without it costing you too much time, go for it (especially on the test, when test anxiety might cause you to miss the more holistic solutions.) But while you are studying, be on the constant lookout for quicker, holistic ways of thinking about the problems that you can add to your strategic arsenal. That is what I call "GMAT Jujitsu."
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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


Target question: What is the value of z?

Given:
u² - v² = 11
y = 2uv
z = u² + v²

Statement 1: y = 60
We can write: 60 = 2uv
Divide both sides by 2 to get: 30 = uv
Solve for u to get: u = 30/v

Take: u² - v² = 11
Replace u with 30/v to get: (30/v)² - v² = 11
Simplify to get: 900/v² - v² = 11
Multiply both sides by v² to get: 900 - v⁴ = 11v²
Rearrange to get: v⁴ + 11v² - 900 = 0
Factor to get: (v² + 36)(v² - 25) = 0
So, EITHER v² + 36 = 0 OR v² - 25 = 0
If v² + 36 = 0, then there is NO SOLUTION
If v² - 25 = 0, then v = 5 or v = -5

Statement 1 indirectly tells us that u = 30/v
If v = 5, then u = 6, in which case the answer to the target question is z = u² + v² = 6² + 5² = 61
If v = -5, then u = -6, in which case the answer to the target question is z = u² + v² = (-6)² + (-5)² = 61
NOTE, for both possible cases, we get the SAME answer to the target question
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: u = 6
We know that u² - v² = 11
Replace u with 6 to get: 6² - v² = 11
Simplify to get: 36 - v² = 11
Keep going: -v² = -25
So: v² = 25
So, EITHER v = 5 OR v = -5

Let's check each possible case:
Case a: u = 6 and v = 5. In this case, the answer to the target question is z = u² + v² = 6² + 5² = 61
Case b: u = 6 and v = -5. In this case, the answer to the target question is z = u² + v² = 6² + (-5)² = 61
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: D

Cheers,
Brent
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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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Given x = \(u^2\) – \(v^2\) , y = 2uv and z = \(u^2\) + \(v^2\) . We also know x = 11, therefore, \(u^2\) – \(v^2\) = 11.

From statement I alone, y = 60.

Therefore, 2uv = 60 or uv = 30. We can express u in terms of v as u = \(\frac{30}{v}\). Substituting this in the equation \(u^2\) – \(v^2\) = 11 and solving the resulting quadratic equation, we obtain the value of v as 5. Since uv = 30, we can now find the value of u and hence we can find the value of \(u^2\) + \(v^2\). We don’t have to go beyond this stage and find the exact value of this expression since this is a DS question.

Statement I alone is sufficient. Answer options B, C and E can be eliminated. Possible answer options are A or D.

From statement II alone, u = 6.
Therefore, \(u^2\) = 36. We also know that \(u^2\) – \(v^2\) = 11. Solving the two equations, we can find the value of \(v^2\) and hence the value of z.

Statement II alone is sufficient. Answer option A can be eliminated.

The correct answer option is D.

Hope that helps!
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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


Substituting \(x=11\) into \(x=u^2-v^2\), we get:
\(u^2-v^2=11\)
\(v^2=u^2-11\)

Statement 2:
Plugging \(u=6\) into \(v^2=u^2-11\), we get:
\(v^2=6^2-11=25\)

Since \(z=u^2 + v^2\), we get:
\(z=6^2+5^2= 61\)
SUFFICIENT.

Statement 1:
Since the condition in Statement 2 cannot contradict Statement 1, \(u=6\) must be possible in Statement 1.

Notice the chain that is yielded if \(|u|=6\):
\(v^2=u^2-11=(±6)^2-11=25\) --> \(|v|=5\) --> it is possible that \(y=2uv=2*6*5=60\)

The chain above reveals the following:
If the value of |u| increases, then the value of |v| will also increase, with the result that \(y=60\) will NOT be possible.
If the value of |u| decreases, then the value of |v| will also decrease, with the result that \(y=60\) will NOT be possible.
Implication:
\(y=60\) is possible ONLY IF \(|u|=6\).
Thus, Statement 1 -- like Statement 2 -- implies that |u|=6 and |v|=5, with the result that \(z=(±6)^2+(±5)^2 = 36+25=61\).
SUFFICIENT.

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

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