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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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BrentGMATPrepNow

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24 Feb 2019, 10:45

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vogelleblanc

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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19 Aug 2020, 19:50

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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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nishantjayaswal

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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25 Aug 2020, 05:56

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nishantjayaswal

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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Updated on: 30 Nov 2020, 14:00

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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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1

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21 Mar 2021, 00:54

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vogelleblanc

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

Show Spoiler

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

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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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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03 Apr 2021, 08:44

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vogelleblanc

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.

Show Spoiler

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26 May 2021, 07:48

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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!
Aravind B T

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31 Jul 2021, 05:31

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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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26 Jul 2022, 01:55

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vogelleblanc

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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KarishmaB

jlgdr

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)

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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06 Sep 2023, 18:34

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KarishmaB

jlgdr

Why are we assuming they are integers?

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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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27 Nov 2023, 11:01

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bovannu01

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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30 Oct 2025, 15:38

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They never say any of these values are strictly real numbers. Are we to assume no complex numbers on the GMAT?

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31 Oct 2025, 00:01

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Pranavsawant

They never say any of these values are strictly real numbers. Are we to assume no complex numbers on the GMAT?

On the GMAT, all numbers are real by default.

P.S. Pure algebraic questions are no longer a part of the DS syllabus of the GMAT.

DS questions in GMAT Focus encompass various types of word problems, such as:

Word Problems
Work Problems
Distance Problems
Mixture Problems
Percent and Interest Problems
Overlapping Sets Problems
Statistics Problems
Combination and Probability Problems

While these questions may involve or necessitate knowledge of algebra, arithmetic, inequalities, etc., they will always be presented in the form of word problems. You won’t encounter pure "algebra" questions like, "Is x > y?" or "A positive integer n has two prime factors..."

Check GMAT Syllabus for Focus Edition

You can also visit the Data Sufficiency forum and filter questions by OG 2024-2025, GMAT Prep (Focus), and Data Insights Review 2024-2025 sources to see the types of questions currently tested on the GMAT.

So, you can ignore this question.

Hope it helps.

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