Last visit was: 21 Jun 2024, 15:56 It is currently 21 Jun 2024, 15:56
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
SORT BY:
Date
Tags:
Show Tags
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 93950
Own Kudos [?]: 633721 [1]
Given Kudos: 82413
Send PM
avatar
Manager
Manager
Joined: 14 Jul 2014
Posts: 67
Own Kudos [?]: 95 [4]
Given Kudos: 49
Send PM
Tutor
Joined: 16 Oct 2010
Posts: 14982
Own Kudos [?]: 66103 [2]
Given Kudos: 435
Location: Pune, India
Send PM
Intern
Intern
Joined: 07 Jan 2017
Posts: 14
Own Kudos [?]: 14 [1]
Given Kudos: 21
Location: India
Concentration: Other, Other
Schools: UConn"19
WE:Medicine and Health (Pharmaceuticals and Biotech)
Send PM
If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]
1
Kudos
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
Intern
Intern
Joined: 30 Jul 2020
Posts: 1
Own Kudos [?]: 0 [0]
Given Kudos: 0
GMAT 1: 740 Q51 V39
Send PM
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?
RC & DI Moderator
Joined: 02 Aug 2009
Status:Math and DI Expert
Posts: 11414
Own Kudos [?]: 33652 [1]
Given Kudos: 319
Send PM
Re: If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]
1
Kudos
Expert Reply
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
Tutor
Joined: 16 Oct 2010
Posts: 14982
Own Kudos [?]: 66103 [1]
Given Kudos: 435
Location: Pune, India
Send PM
If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]
1
Kudos
Expert Reply
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
Tutor
Joined: 17 Jul 2019
Posts: 1304
Own Kudos [?]: 1746 [0]
Given Kudos: 66
Location: Canada
GMAT 1: 780 Q51 V45
GMAT 2: 780 Q50 V47
GMAT 3: 770 Q50 V45
Send PM
If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]
Expert Reply
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.
avatar
Intern
Intern
Joined: 14 May 2020
Posts: 4
Own Kudos [?]: 0 [0]
Given Kudos: 0
Send PM
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]
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
Current Student
Joined: 13 Mar 2019
Posts: 3
Own Kudos [?]: 5 [0]
Given Kudos: 8
Location: United States (UT)
Concentration: Finance, Real Estate
GMAT 1: 760 Q48 V47
GPA: 3.59
WE:Real Estate (Real Estate)
Send PM
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
Manager
Manager
Joined: 20 Aug 2018
Posts: 182
Own Kudos [?]: 35 [0]
Given Kudos: 296
Location: India
GMAT 1: 640 Q48 V28
GMAT 2: 610 Q47 V27
GPA: 3.42
Send PM
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)
Intern
Intern
Joined: 01 Sep 2021
Posts: 37
Own Kudos [?]: 15 [0]
Given Kudos: 19
Location: India
GMAT 1: 700 Q49 V35
GMAT 2: 730 Q50 V39
GPA: 08.45
Send PM
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

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?
Manager
Manager
Joined: 30 May 2013
Status:Full-time employee
Affiliations: Apple Inc
Posts: 104
Own Kudos [?]: 124 [0]
Given Kudos: 93
Location: United States
Saupayan: Mazumdar
Concentration: Economics, Leadership
GMAT 1: 760 Q51 V41
GRE 1: Q170 V160
GPA: 3.89
WE:Engineering (Computer Hardware)
Send PM
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

Answer (D)


Why are we assuming they are integers?
Intern
Intern
Joined: 11 Oct 2020
Posts: 3
Own Kudos [?]: 1 [0]
Given Kudos: 0
Send PM
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?
Math Expert
Joined: 02 Sep 2009
Posts: 93950
Own Kudos [?]: 633721 [0]
Given Kudos: 82413
Send PM
If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]
Expert Reply
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? :dontknow:

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.
GMAT Club Bot
If x = u^2 - v^2, y = 2uv and z = u^2 + v^2, and if x = 11, [#permalink]
   1   2 
Moderator:
Math Expert
93950 posts