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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

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

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28 Nov 2013, 06:46

1

This post received KUDOS

@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

Attachments

Screenshot_28.11.13_21_48.png [ 86.68 KiB | Viewed 23618 times ]

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 \(u^2 + v^2=36+25\).

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
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

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

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

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: \((z + y)(z - y) = (u + v)^2*(u - v)^2 = 11^2 = 121\)
_________________

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

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29 Jun 2014, 05:37

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.

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.

If \(x = u^2 - v^2\), \(y = 2uv\) and \(z = u^2 + v^2\), and if \(x = 11\), what is the value of z?

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.
_________________

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

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15 Sep 2014, 12:53

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..

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

Show Tags

15 Sep 2014, 14:07

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
_________________

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..

TehMoUsE wrote:

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

Both of you incorrectly assume that the variables are integers only: 11 = u^2 - v^2 has infinitely many non-integer solutions for u and v.
_________________

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