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

Question

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 :)

niyamharia1990 · Answer

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

Posted from my mobile device

vogelleblanc · Answer

@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

Bunuel · Answer

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.

vogelleblanc · Answer

... 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!

mau5 · Answer

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.

WoundedTiger · Answer

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

jlgdr · Answer

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

KarishmaB · Answer

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)

jlgdr · Answer

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

KarishmaB · Answer

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

VadimKlimenko · Answer

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.

Bunuel · Answer

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

Given:  u^2 - v^2=11  and  y = 2uv .
Question:  u^2 + v^2=?

(1) y = 60:

Thus,  2uv=60 . Square to get  4u^2v^2=3,600 .

Now, square  u^2 - v^2=11  to get  u^4-2u^2v^2+v^4=121 ;

Add  4u^2v^2  to both sides:  u^4+2u^2v^2+v^4=121+3,600 ;

Apply  a^2+2ab+b^2=(a+b)^2 :  (u^2+v^2)^2=121+3,600 .

u^2+v^2=\sqrt{121+3,600}=61 .

Sufficient.

(2) u = 6:

36 - v^2=11 ;

v^2=25 ;

u^2+v^2=36+25=61 .

Sufficient.

Answer: D.

Hope it helps.

KarishmaB · Answer

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.

tushain · Answer

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 · Answer

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

Bunuel · Answer

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.

buddyisraelgmat · Answer

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?

Given:  u^2 - v^2=11  and  y = 2uv .
Question:  u^2 + v^2=?

(1) y = 60 -->  2uv=60  -->  4u^2v^2=3,600 .

Square  u^2 - v^2=11 :  u^4-2u^2v^2+v^2=121 ;

Add  4u^2v^2  to both sides:  u^4+2u^2v^2+v^4=121+3,600 ;

Apply  a^2+2ab+b^2=(a+b)^2 :  (u^2+v^2)^2=121+3,600 .

u^2+v^2=\sqrt{121+3,600}=61 .

Sufficient.

(2) u = 6 -->  36 - v^2=11  -->  v^2=25  -->  u^2+v^2=36+25=61 . Sufficient.

Answer: D.

Hope it helps.

Hi Bunuel

Stmtn 1
Can we use  Differences of Square  approach

Property - The difference between squares grows as the squares themselves get larger

Hence given u^2 - v^2 = x = 11

Thus the only numbers that fit this equation are when u = 6 and v = 5. Hence we can find Z - Sufficient

Similalry Stmtn 2 = Suff

Hence D

Pls mention if this approach is correct

Thanks

KarishmaB · Answer

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   ^2 = 11^2 
 (u+v)^2 * (u - v)^2 = 11^2

Does this help?

Sachin07 · Answer

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

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

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